How many atoms (or ions) of each element are in 140.0 g of the following substances? a H 2 b Ca NO 3 2 c N 2 O 2 d K 2 SO 4

Question

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Answer 1

Textbook Question

Chapter 4, Problem 55QP

How many atoms (or ions) of each element are in 140.0 g of the following substances?

$a H 2 b Ca NO 3 2 c N 2 O 2 d K 2 SO 4$

(a)

Expert Solution

Interpretation Introduction

Interpretation:

The number of atoms (or ions) present in $140 g$ $H2$ is to be determined.

Explanation of Solution

For the number of formula units:

$mg→MMn→NANumber of formula units$

Here, $m$ is the given mass, $MM$ is the molar mass and $n$ is the number of moles.

For the number of atoms or ions:

$Number of formula units→formula ratioAtoms or ions$

The number of formula units present is determined as follows:

$Number of formula units=mMM×NA$ ...(1)

Here, $NA$ is Avogadroâ€™s number with a value of $6.022×1023 mol−1$ .

The number of atoms is determined as follows:

$Number of atoms=mMM×NA×Present ions$ ...(2)

By combining equations (1) and (2), the number of atoms can be determined as follows:

$Number of atoms=Number of formula units×Present ions$ ...(3)

The molar mass of one mole of $H2$ is $2.016 g$ . The weight of $H2$ is $140 g$ . In the $H2$ molecule, there are two hydrogen atoms. Substitute these values in equation (2) as follows.

$Number of H atoms=140 g2.016 g/mol×6.022×1023 mol−1×2 H atoms=8.364×1025 H atoms$

Therefore, there are $8.364×1025$ H atoms present in $140 g$ of $H2$ .

(b)

Expert Solution

Interpretation Introduction

Interpretation:

The number of atoms (or ions) of each element present in $140 g$ $CaNO32$ is to be determined.

Explanation of Solution

The molar mass of $CaNO32$ is $164.10 g/mol$ . The weight of $CaNO32$ is $140 g$ . Substitute these values in equation (1) to determine the number of formula units present in $140 g$ of $CaNO32$ as follows.

$Formula units CaNO32=140 g164.1 g/mol g/mol×6.022×1023 mol−1=5.14×1023 formula units CaNO32$

There is one $Ca2+$ in $CaNO32$ . By using equation (3), the number of $Ca2+$ ions in $140 g$ $CaNO32$ is determined as follows.

$Number of atoms=5.14×1023 × 1 Ca2+=5.14×1023 Ca2+$

There are two $NO3−$ ions in $CaNO32$ . By using equation (3), the number of $NO3−$ in $140 g$ $CaNO32$ is determined as follows.

$Number of atoms=5.14×1023 × 2 NO3−=1.028×1024 NO3−$

Therefore, there are $5.14×1023 Ca2+$ and $1.028×1024 NO3−$ present in $140 g$ of $CaNO32$ .

(c)

Expert Solution

Interpretation Introduction

Interpretation:

The number of atoms (or ions) of each element present in $140 g$ $N2O2$ is to be determined.

Explanation of Solution

The molar mass of $N2O2$ is $60.02 g/mol$ . The weight of $N2O2$ is $140 g$ . Substitute these values in equation (1) to determine the number of formula units present in $140 g$ of $N2O2$ as follows.

$Formula units N2O2=140 g60.02 g/mol g/mol×6.022×1023 mol−1=1.405×1024 formula units N2O2$

There are two N in $N2O2$ . By using equation (3), the number of N atoms in $140 g$ $N2O2$ is determined as follows.

$Number of atoms=1.405×1024 × 1 N=2.8×1024 N atoms$

There are two O ions in $N2O2$ . By using equation (3), the number of O in $140 g$ $N2O2$ is determined as follows.

$Number of atoms=1.405×1024 × 2 O=2.8×1024 O atoms$

Therefore, there are $2.8×1024 N$ and $2.8×1024 O$ atoms present in $140 g$ of $N2O2$ .

(d)

Expert Solution

Interpretation Introduction

Interpretation:

The number of atoms (or ions) of each element present in $140 g$ $K2SO4$ is to be determined.

Explanation of Solution

The molar mass of $K2SO4$ is $174.26 g/mol$ . The weight of $K2SO4$ is $140 g$ . Substitute these values in equation (1) to determine the number of formula units present in $140 g$ of $K2SO4$ as follows.

$Formula units K2SO4=140 g174.26 g/mol g/mol×6.022×1023 mol−1=4.838×1023 formula units K2SO4$

There are 2 $K+$ in $K2SO4$ . By using equation (3), the number of $K+$ ions in $140 g$ $K2SO4$ is determined as follows.

$Number of atoms=4.838×1023 × 2 K+=9.676×1023 K+ ions$

There is one $SO24−$ in $K2SO4$ . By using equation (3), the number of $SO24−$ in $140 g$ $K2SO4$ is determined as follows.

$Number of atoms=4.838×1023 × 1 SO24−=4.838×1023 SO24−$

Therefore, there are $9.676×1023 K+$ and $4.838×1023 SO24−$ present in $140 g$ of $K2SO4$ .

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Answer 2

Textbook Question

Chapter 4, Problem 55QP

How many atoms (or ions) of each element are in 140.0 g of the following substances?

$a H 2 b Ca NO 3 2 c N 2 O 2 d K 2 SO 4$

(a)

Expert Solution

Interpretation Introduction

Interpretation:

The number of atoms (or ions) present in $140 g$ $H2$ is to be determined.

Explanation of Solution

For the number of formula units:

$mg→MMn→NANumber of formula units$

Here, $m$ is the given mass, $MM$ is the molar mass and $n$ is the number of moles.

For the number of atoms or ions:

$Number of formula units→formula ratioAtoms or ions$

The number of formula units present is determined as follows:

$Number of formula units=mMM×NA$ ...(1)

Here, $NA$ is Avogadroâ€™s number with a value of $6.022×1023 mol−1$ .

The number of atoms is determined as follows:

$Number of atoms=mMM×NA×Present ions$ ...(2)

By combining equations (1) and (2), the number of atoms can be determined as follows:

$Number of atoms=Number of formula units×Present ions$ ...(3)

The molar mass of one mole of $H2$ is $2.016 g$ . The weight of $H2$ is $140 g$ . In the $H2$ molecule, there are two hydrogen atoms. Substitute these values in equation (2) as follows.

$Number of H atoms=140 g2.016 g/mol×6.022×1023 mol−1×2 H atoms=8.364×1025 H atoms$

Therefore, there are $8.364×1025$ H atoms present in $140 g$ of $H2$ .

(b)

Expert Solution

Interpretation Introduction

Interpretation:

The number of atoms (or ions) of each element present in $140 g$ $CaNO32$ is to be determined.

Explanation of Solution

The molar mass of $CaNO32$ is $164.10 g/mol$ . The weight of $CaNO32$ is $140 g$ . Substitute these values in equation (1) to determine the number of formula units present in $140 g$ of $CaNO32$ as follows.

$Formula units CaNO32=140 g164.1 g/mol g/mol×6.022×1023 mol−1=5.14×1023 formula units CaNO32$

There is one $Ca2+$ in $CaNO32$ . By using equation (3), the number of $Ca2+$ ions in $140 g$ $CaNO32$ is determined as follows.

$Number of atoms=5.14×1023 × 1 Ca2+=5.14×1023 Ca2+$

There are two $NO3−$ ions in $CaNO32$ . By using equation (3), the number of $NO3−$ in $140 g$ $CaNO32$ is determined as follows.

$Number of atoms=5.14×1023 × 2 NO3−=1.028×1024 NO3−$

Therefore, there are $5.14×1023 Ca2+$ and $1.028×1024 NO3−$ present in $140 g$ of $CaNO32$ .

(c)

Expert Solution

Interpretation Introduction

Interpretation:

The number of atoms (or ions) of each element present in $140 g$ $N2O2$ is to be determined.

Explanation of Solution

The molar mass of $N2O2$ is $60.02 g/mol$ . The weight of $N2O2$ is $140 g$ . Substitute these values in equation (1) to determine the number of formula units present in $140 g$ of $N2O2$ as follows.

$Formula units N2O2=140 g60.02 g/mol g/mol×6.022×1023 mol−1=1.405×1024 formula units N2O2$

There are two N in $N2O2$ . By using equation (3), the number of N atoms in $140 g$ $N2O2$ is determined as follows.

$Number of atoms=1.405×1024 × 1 N=2.8×1024 N atoms$

There are two O ions in $N2O2$ . By using equation (3), the number of O in $140 g$ $N2O2$ is determined as follows.

$Number of atoms=1.405×1024 × 2 O=2.8×1024 O atoms$

Therefore, there are $2.8×1024 N$ and $2.8×1024 O$ atoms present in $140 g$ of $N2O2$ .

(d)

Expert Solution

Interpretation Introduction

Interpretation:

The number of atoms (or ions) of each element present in $140 g$ $K2SO4$ is to be determined.

Explanation of Solution

The molar mass of $K2SO4$ is $174.26 g/mol$ . The weight of $K2SO4$ is $140 g$ . Substitute these values in equation (1) to determine the number of formula units present in $140 g$ of $K2SO4$ as follows.

$Formula units K2SO4=140 g174.26 g/mol g/mol×6.022×1023 mol−1=4.838×1023 formula units K2SO4$

There are 2 $K+$ in $K2SO4$ . By using equation (3), the number of $K+$ ions in $140 g$ $K2SO4$ is determined as follows.

$Number of atoms=4.838×1023 × 2 K+=9.676×1023 K+ ions$

There is one $SO24−$ in $K2SO4$ . By using equation (3), the number of $SO24−$ in $140 g$ $K2SO4$ is determined as follows.

$Number of atoms=4.838×1023 × 1 SO24−=4.838×1023 SO24−$

Therefore, there are $9.676×1023 K+$ and $4.838×1023 SO24−$ present in $140 g$ of $K2SO4$ .

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Answer 3

Textbook Question

Chapter 4, Problem 55QP

How many atoms (or ions) of each element are in 140.0 g of the following substances?

$a H 2 b Ca NO 3 2 c N 2 O 2 d K 2 SO 4$

(a)

Expert Solution

Interpretation Introduction

Interpretation:

The number of atoms (or ions) present in $140 g$ $H2$ is to be determined.

Explanation of Solution

For the number of formula units:

$mg→MMn→NANumber of formula units$

Here, $m$ is the given mass, $MM$ is the molar mass and $n$ is the number of moles.

For the number of atoms or ions:

$Number of formula units→formula ratioAtoms or ions$

The number of formula units present is determined as follows:

$Number of formula units=mMM×NA$ ...(1)

Here, $NA$ is Avogadroâ€™s number with a value of $6.022×1023 mol−1$ .

The number of atoms is determined as follows:

$Number of atoms=mMM×NA×Present ions$ ...(2)

By combining equations (1) and (2), the number of atoms can be determined as follows:

$Number of atoms=Number of formula units×Present ions$ ...(3)

The molar mass of one mole of $H2$ is $2.016 g$ . The weight of $H2$ is $140 g$ . In the $H2$ molecule, there are two hydrogen atoms. Substitute these values in equation (2) as follows.

$Number of H atoms=140 g2.016 g/mol×6.022×1023 mol−1×2 H atoms=8.364×1025 H atoms$

Therefore, there are $8.364×1025$ H atoms present in $140 g$ of $H2$ .

(b)

Expert Solution

Interpretation Introduction

Interpretation:

The number of atoms (or ions) of each element present in $140 g$ $CaNO32$ is to be determined.

Explanation of Solution

The molar mass of $CaNO32$ is $164.10 g/mol$ . The weight of $CaNO32$ is $140 g$ . Substitute these values in equation (1) to determine the number of formula units present in $140 g$ of $CaNO32$ as follows.

$Formula units CaNO32=140 g164.1 g/mol g/mol×6.022×1023 mol−1=5.14×1023 formula units CaNO32$

There is one $Ca2+$ in $CaNO32$ . By using equation (3), the number of $Ca2+$ ions in $140 g$ $CaNO32$ is determined as follows.

$Number of atoms=5.14×1023 × 1 Ca2+=5.14×1023 Ca2+$

There are two $NO3−$ ions in $CaNO32$ . By using equation (3), the number of $NO3−$ in $140 g$ $CaNO32$ is determined as follows.

$Number of atoms=5.14×1023 × 2 NO3−=1.028×1024 NO3−$

Therefore, there are $5.14×1023 Ca2+$ and $1.028×1024 NO3−$ present in $140 g$ of $CaNO32$ .

(c)

Expert Solution

Interpretation Introduction

Interpretation:

The number of atoms (or ions) of each element present in $140 g$ $N2O2$ is to be determined.

Explanation of Solution

The molar mass of $N2O2$ is $60.02 g/mol$ . The weight of $N2O2$ is $140 g$ . Substitute these values in equation (1) to determine the number of formula units present in $140 g$ of $N2O2$ as follows.

$Formula units N2O2=140 g60.02 g/mol g/mol×6.022×1023 mol−1=1.405×1024 formula units N2O2$

There are two N in $N2O2$ . By using equation (3), the number of N atoms in $140 g$ $N2O2$ is determined as follows.

$Number of atoms=1.405×1024 × 1 N=2.8×1024 N atoms$

There are two O ions in $N2O2$ . By using equation (3), the number of O in $140 g$ $N2O2$ is determined as follows.

$Number of atoms=1.405×1024 × 2 O=2.8×1024 O atoms$

Therefore, there are $2.8×1024 N$ and $2.8×1024 O$ atoms present in $140 g$ of $N2O2$ .

(d)

Expert Solution

Interpretation Introduction

Interpretation:

The number of atoms (or ions) of each element present in $140 g$ $K2SO4$ is to be determined.

Explanation of Solution

The molar mass of $K2SO4$ is $174.26 g/mol$ . The weight of $K2SO4$ is $140 g$ . Substitute these values in equation (1) to determine the number of formula units present in $140 g$ of $K2SO4$ as follows.

$Formula units K2SO4=140 g174.26 g/mol g/mol×6.022×1023 mol−1=4.838×1023 formula units K2SO4$

There are 2 $K+$ in $K2SO4$ . By using equation (3), the number of $K+$ ions in $140 g$ $K2SO4$ is determined as follows.

$Number of atoms=4.838×1023 × 2 K+=9.676×1023 K+ ions$

There is one $SO24−$ in $K2SO4$ . By using equation (3), the number of $SO24−$ in $140 g$ $K2SO4$ is determined as follows.

$Number of atoms=4.838×1023 × 1 SO24−=4.838×1023 SO24−$

Therefore, there are $9.676×1023 K+$ and $4.838×1023 SO24−$ present in $140 g$ of $K2SO4$ .

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Answer 4

Textbook Question

Chapter 4, Problem 55QP

How many atoms (or ions) of each element are in 140.0 g of the following substances?

$a H 2 b Ca NO 3 2 c N 2 O 2 d K 2 SO 4$

(a)

Expert Solution

Interpretation Introduction

Interpretation:

The number of atoms (or ions) present in $140 g$ $H2$ is to be determined.

Explanation of Solution

For the number of formula units:

$mg→MMn→NANumber of formula units$

Here, $m$ is the given mass, $MM$ is the molar mass and $n$ is the number of moles.

For the number of atoms or ions:

$Number of formula units→formula ratioAtoms or ions$

The number of formula units present is determined as follows:

$Number of formula units=mMM×NA$ ...(1)

Here, $NA$ is Avogadroâ€™s number with a value of $6.022×1023 mol−1$ .

The number of atoms is determined as follows:

$Number of atoms=mMM×NA×Present ions$ ...(2)

By combining equations (1) and (2), the number of atoms can be determined as follows:

$Number of atoms=Number of formula units×Present ions$ ...(3)

The molar mass of one mole of $H2$ is $2.016 g$ . The weight of $H2$ is $140 g$ . In the $H2$ molecule, there are two hydrogen atoms. Substitute these values in equation (2) as follows.

$Number of H atoms=140 g2.016 g/mol×6.022×1023 mol−1×2 H atoms=8.364×1025 H atoms$

Therefore, there are $8.364×1025$ H atoms present in $140 g$ of $H2$ .

(b)

Expert Solution

Interpretation Introduction

Interpretation:

The number of atoms (or ions) of each element present in $140 g$ $CaNO32$ is to be determined.

Explanation of Solution

The molar mass of $CaNO32$ is $164.10 g/mol$ . The weight of $CaNO32$ is $140 g$ . Substitute these values in equation (1) to determine the number of formula units present in $140 g$ of $CaNO32$ as follows.

$Formula units CaNO32=140 g164.1 g/mol g/mol×6.022×1023 mol−1=5.14×1023 formula units CaNO32$

There is one $Ca2+$ in $CaNO32$ . By using equation (3), the number of $Ca2+$ ions in $140 g$ $CaNO32$ is determined as follows.

$Number of atoms=5.14×1023 × 1 Ca2+=5.14×1023 Ca2+$

There are two $NO3−$ ions in $CaNO32$ . By using equation (3), the number of $NO3−$ in $140 g$ $CaNO32$ is determined as follows.

$Number of atoms=5.14×1023 × 2 NO3−=1.028×1024 NO3−$

Therefore, there are $5.14×1023 Ca2+$ and $1.028×1024 NO3−$ present in $140 g$ of $CaNO32$ .

(c)

Expert Solution

Interpretation Introduction

Interpretation:

The number of atoms (or ions) of each element present in $140 g$ $N2O2$ is to be determined.

Explanation of Solution

The molar mass of $N2O2$ is $60.02 g/mol$ . The weight of $N2O2$ is $140 g$ . Substitute these values in equation (1) to determine the number of formula units present in $140 g$ of $N2O2$ as follows.

$Formula units N2O2=140 g60.02 g/mol g/mol×6.022×1023 mol−1=1.405×1024 formula units N2O2$

There are two N in $N2O2$ . By using equation (3), the number of N atoms in $140 g$ $N2O2$ is determined as follows.

$Number of atoms=1.405×1024 × 1 N=2.8×1024 N atoms$

There are two O ions in $N2O2$ . By using equation (3), the number of O in $140 g$ $N2O2$ is determined as follows.

$Number of atoms=1.405×1024 × 2 O=2.8×1024 O atoms$

Therefore, there are $2.8×1024 N$ and $2.8×1024 O$ atoms present in $140 g$ of $N2O2$ .

(d)

Expert Solution

Interpretation Introduction

Interpretation:

The number of atoms (or ions) of each element present in $140 g$ $K2SO4$ is to be determined.

Explanation of Solution

The molar mass of $K2SO4$ is $174.26 g/mol$ . The weight of $K2SO4$ is $140 g$ . Substitute these values in equation (1) to determine the number of formula units present in $140 g$ of $K2SO4$ as follows.

$Formula units K2SO4=140 g174.26 g/mol g/mol×6.022×1023 mol−1=4.838×1023 formula units K2SO4$

There are 2 $K+$ in $K2SO4$ . By using equation (3), the number of $K+$ ions in $140 g$ $K2SO4$ is determined as follows.

$Number of atoms=4.838×1023 × 2 K+=9.676×1023 K+ ions$

There is one $SO24−$ in $K2SO4$ . By using equation (3), the number of $SO24−$ in $140 g$ $K2SO4$ is determined as follows.

$Number of atoms=4.838×1023 × 1 SO24−=4.838×1023 SO24−$

Therefore, there are $9.676×1023 K+$ and $4.838×1023 SO24−$ present in $140 g$ of $K2SO4$ .

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

Interpretation Introduction

Interpretation:

The number of atoms (or ions) present in $140 g$ $H2$ is to be determined.

Explanation of Solution

For the number of formula units:

$mg→MMn→NANumber of formula units$

Here, $m$ is the given mass, $MM$ is the molar mass and $n$ is the number of moles.

For the number of atoms or ions:

$Number of formula units→formula ratioAtoms or ions$

The number of formula units present is determined as follows:

$Number of formula units=mMM×NA$ ...(1)

Here, $NA$ is Avogadroâ€™s number with a value of $6.022×1023 mol−1$ .

The number of atoms is determined as follows:

$Number of atoms=mMM×NA×Present ions$ ...(2)

By combining equations (1) and (2), the number of atoms can be determined as follows:

$Number of atoms=Number of formula units×Present ions$ ...(3)

The molar mass of one mole of $H2$ is $2.016 g$ . The weight of $H2$ is $140 g$ . In the $H2$ molecule, there are two hydrogen atoms. Substitute these values in equation (2) as follows.

$Number of H atoms=140 g2.016 g/mol×6.022×1023 mol−1×2 H atoms=8.364×1025 H atoms$

Therefore, there are $8.364×1025$ H atoms present in $140 g$ of $H2$ .

(b)

Expert Solution

Interpretation Introduction

Interpretation:

The number of atoms (or ions) of each element present in $140 g$ $CaNO32$ is to be determined.

Explanation of Solution

The molar mass of $CaNO32$ is $164.10 g/mol$ . The weight of $CaNO32$ is $140 g$ . Substitute these values in equation (1) to determine the number of formula units present in $140 g$ of $CaNO32$ as follows.

$Formula units CaNO32=140 g164.1 g/mol g/mol×6.022×1023 mol−1=5.14×1023 formula units CaNO32$

There is one $Ca2+$ in $CaNO32$ . By using equation (3), the number of $Ca2+$ ions in $140 g$ $CaNO32$ is determined as follows.

$Number of atoms=5.14×1023 × 1 Ca2+=5.14×1023 Ca2+$

There are two $NO3−$ ions in $CaNO32$ . By using equation (3), the number of $NO3−$ in $140 g$ $CaNO32$ is determined as follows.

$Number of atoms=5.14×1023 × 2 NO3−=1.028×1024 NO3−$

Therefore, there are $5.14×1023 Ca2+$ and $1.028×1024 NO3−$ present in $140 g$ of $CaNO32$ .

(c)

Expert Solution

Interpretation Introduction

Interpretation:

The number of atoms (or ions) of each element present in $140 g$ $N2O2$ is to be determined.

Explanation of Solution

The molar mass of $N2O2$ is $60.02 g/mol$ . The weight of $N2O2$ is $140 g$ . Substitute these values in equation (1) to determine the number of formula units present in $140 g$ of $N2O2$ as follows.

$Formula units N2O2=140 g60.02 g/mol g/mol×6.022×1023 mol−1=1.405×1024 formula units N2O2$

There are two N in $N2O2$ . By using equation (3), the number of N atoms in $140 g$ $N2O2$ is determined as follows.

$Number of atoms=1.405×1024 × 1 N=2.8×1024 N atoms$

There are two O ions in $N2O2$ . By using equation (3), the number of O in $140 g$ $N2O2$ is determined as follows.

$Number of atoms=1.405×1024 × 2 O=2.8×1024 O atoms$

Therefore, there are $2.8×1024 N$ and $2.8×1024 O$ atoms present in $140 g$ of $N2O2$ .

(d)

Expert Solution

Interpretation Introduction

Interpretation:

The number of atoms (or ions) of each element present in $140 g$ $K2SO4$ is to be determined.

Explanation of Solution

The molar mass of $K2SO4$ is $174.26 g/mol$ . The weight of $K2SO4$ is $140 g$ . Substitute these values in equation (1) to determine the number of formula units present in $140 g$ of $K2SO4$ as follows.

$Formula units K2SO4=140 g174.26 g/mol g/mol×6.022×1023 mol−1=4.838×1023 formula units K2SO4$

There are 2 $K+$ in $K2SO4$ . By using equation (3), the number of $K+$ ions in $140 g$ $K2SO4$ is determined as follows.

$Number of atoms=4.838×1023 × 2 K+=9.676×1023 K+ ions$

There is one $SO24−$ in $K2SO4$ . By using equation (3), the number of $SO24−$ in $140 g$ $K2SO4$ is determined as follows.

$Number of atoms=4.838×1023 × 1 SO24−=4.838×1023 SO24−$

Therefore, there are $9.676×1023 K+$ and $4.838×1023 SO24−$ present in $140 g$ of $K2SO4$ .

Answer 8

(a)

Expert Solution

Interpretation Introduction

Interpretation:

The number of atoms (or ions) present in $140 g$ $H2$ is to be determined.

Explanation of Solution

For the number of formula units:

$mg→MMn→NANumber of formula units$

Here, $m$ is the given mass, $MM$ is the molar mass and $n$ is the number of moles.

For the number of atoms or ions:

$Number of formula units→formula ratioAtoms or ions$

The number of formula units present is determined as follows:

$Number of formula units=mMM×NA$ ...(1)

Here, $NA$ is Avogadroâ€™s number with a value of $6.022×1023 mol−1$ .

The number of atoms is determined as follows:

$Number of atoms=mMM×NA×Present ions$ ...(2)

By combining equations (1) and (2), the number of atoms can be determined as follows:

$Number of atoms=Number of formula units×Present ions$ ...(3)

The molar mass of one mole of $H2$ is $2.016 g$ . The weight of $H2$ is $140 g$ . In the $H2$ molecule, there are two hydrogen atoms. Substitute these values in equation (2) as follows.

$Number of H atoms=140 g2.016 g/mol×6.022×1023 mol−1×2 H atoms=8.364×1025 H atoms$

Therefore, there are $8.364×1025$ H atoms present in $140 g$ of $H2$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Interpretation Introduction

Interpretation:

The number of atoms (or ions) present in $140 g$ $H2$ is to be determined.

Answer 13

Explanation of Solution

For the number of formula units:

$mg→MMn→NANumber of formula units$

Here, $m$ is the given mass, $MM$ is the molar mass and $n$ is the number of moles.

For the number of atoms or ions:

$Number of formula units→formula ratioAtoms or ions$

The number of formula units present is determined as follows:

$Number of formula units=mMM×NA$ ...(1)

Here, $NA$ is Avogadroâ€™s number with a value of $6.022×1023 mol−1$ .

The number of atoms is determined as follows:

$Number of atoms=mMM×NA×Present ions$ ...(2)

By combining equations (1) and (2), the number of atoms can be determined as follows:

$Number of atoms=Number of formula units×Present ions$ ...(3)

The molar mass of one mole of $H2$ is $2.016 g$ . The weight of $H2$ is $140 g$ . In the $H2$ molecule, there are two hydrogen atoms. Substitute these values in equation (2) as follows.

$Number of H atoms=140 g2.016 g/mol×6.022×1023 mol−1×2 H atoms=8.364×1025 H atoms$

Therefore, there are $8.364×1025$ H atoms present in $140 g$ of $H2$ .

Answer 14

(b)

Expert Solution

Interpretation Introduction

Interpretation:

The number of atoms (or ions) of each element present in $140 g$ $CaNO32$ is to be determined.

Explanation of Solution

The molar mass of $CaNO32$ is $164.10 g/mol$ . The weight of $CaNO32$ is $140 g$ . Substitute these values in equation (1) to determine the number of formula units present in $140 g$ of $CaNO32$ as follows.

$Formula units CaNO32=140 g164.1 g/mol g/mol×6.022×1023 mol−1=5.14×1023 formula units CaNO32$

There is one $Ca2+$ in $CaNO32$ . By using equation (3), the number of $Ca2+$ ions in $140 g$ $CaNO32$ is determined as follows.

$Number of atoms=5.14×1023 × 1 Ca2+=5.14×1023 Ca2+$

There are two $NO3−$ ions in $CaNO32$ . By using equation (3), the number of $NO3−$ in $140 g$ $CaNO32$ is determined as follows.

$Number of atoms=5.14×1023 × 2 NO3−=1.028×1024 NO3−$

Therefore, there are $5.14×1023 Ca2+$ and $1.028×1024 NO3−$ present in $140 g$ of $CaNO32$ .

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

Interpretation Introduction

Interpretation:

The number of atoms (or ions) of each element present in $140 g$ $CaNO32$ is to be determined.

Answer 19

Explanation of Solution

The molar mass of $CaNO32$ is $164.10 g/mol$ . The weight of $CaNO32$ is $140 g$ . Substitute these values in equation (1) to determine the number of formula units present in $140 g$ of $CaNO32$ as follows.

$Formula units CaNO32=140 g164.1 g/mol g/mol×6.022×1023 mol−1=5.14×1023 formula units CaNO32$

There is one $Ca2+$ in $CaNO32$ . By using equation (3), the number of $Ca2+$ ions in $140 g$ $CaNO32$ is determined as follows.

$Number of atoms=5.14×1023 × 1 Ca2+=5.14×1023 Ca2+$

There are two $NO3−$ ions in $CaNO32$ . By using equation (3), the number of $NO3−$ in $140 g$ $CaNO32$ is determined as follows.

$Number of atoms=5.14×1023 × 2 NO3−=1.028×1024 NO3−$

Therefore, there are $5.14×1023 Ca2+$ and $1.028×1024 NO3−$ present in $140 g$ of $CaNO32$ .

Answer 20

(c)

Expert Solution

Interpretation Introduction

Interpretation:

The number of atoms (or ions) of each element present in $140 g$ $N2O2$ is to be determined.

Explanation of Solution

The molar mass of $N2O2$ is $60.02 g/mol$ . The weight of $N2O2$ is $140 g$ . Substitute these values in equation (1) to determine the number of formula units present in $140 g$ of $N2O2$ as follows.

$Formula units N2O2=140 g60.02 g/mol g/mol×6.022×1023 mol−1=1.405×1024 formula units N2O2$

There are two N in $N2O2$ . By using equation (3), the number of N atoms in $140 g$ $N2O2$ is determined as follows.

$Number of atoms=1.405×1024 × 1 N=2.8×1024 N atoms$

There are two O ions in $N2O2$ . By using equation (3), the number of O in $140 g$ $N2O2$ is determined as follows.

$Number of atoms=1.405×1024 × 2 O=2.8×1024 O atoms$

Therefore, there are $2.8×1024 N$ and $2.8×1024 O$ atoms present in $140 g$ of $N2O2$ .

Answer 21

(c)

Expert Solution

Answer 22

(c)

Expert Solution

Answer 23

Expert Solution

Answer 24

Interpretation Introduction

Interpretation:

The number of atoms (or ions) of each element present in $140 g$ $N2O2$ is to be determined.

Answer 25

Explanation of Solution

The molar mass of $N2O2$ is $60.02 g/mol$ . The weight of $N2O2$ is $140 g$ . Substitute these values in equation (1) to determine the number of formula units present in $140 g$ of $N2O2$ as follows.

$Formula units N2O2=140 g60.02 g/mol g/mol×6.022×1023 mol−1=1.405×1024 formula units N2O2$

There are two N in $N2O2$ . By using equation (3), the number of N atoms in $140 g$ $N2O2$ is determined as follows.

$Number of atoms=1.405×1024 × 1 N=2.8×1024 N atoms$

There are two O ions in $N2O2$ . By using equation (3), the number of O in $140 g$ $N2O2$ is determined as follows.

$Number of atoms=1.405×1024 × 2 O=2.8×1024 O atoms$

Therefore, there are $2.8×1024 N$ and $2.8×1024 O$ atoms present in $140 g$ of $N2O2$ .

Answer 26

(d)

Expert Solution

Interpretation Introduction

Interpretation:

The number of atoms (or ions) of each element present in $140 g$ $K2SO4$ is to be determined.

Explanation of Solution

The molar mass of $K2SO4$ is $174.26 g/mol$ . The weight of $K2SO4$ is $140 g$ . Substitute these values in equation (1) to determine the number of formula units present in $140 g$ of $K2SO4$ as follows.

$Formula units K2SO4=140 g174.26 g/mol g/mol×6.022×1023 mol−1=4.838×1023 formula units K2SO4$

There are 2 $K+$ in $K2SO4$ . By using equation (3), the number of $K+$ ions in $140 g$ $K2SO4$ is determined as follows.

$Number of atoms=4.838×1023 × 2 K+=9.676×1023 K+ ions$

There is one $SO24−$ in $K2SO4$ . By using equation (3), the number of $SO24−$ in $140 g$ $K2SO4$ is determined as follows.

$Number of atoms=4.838×1023 × 1 SO24−=4.838×1023 SO24−$

Therefore, there are $9.676×1023 K+$ and $4.838×1023 SO24−$ present in $140 g$ of $K2SO4$ .

Answer 27

(d)

Expert Solution

Answer 28

(d)

Expert Solution

Answer 29

Expert Solution

Answer 30

Interpretation Introduction

Interpretation:

The number of atoms (or ions) of each element present in $140 g$ $K2SO4$ is to be determined.

Answer 31

Explanation of Solution

The molar mass of $K2SO4$ is $174.26 g/mol$ . The weight of $K2SO4$ is $140 g$ . Substitute these values in equation (1) to determine the number of formula units present in $140 g$ of $K2SO4$ as follows.

$Formula units K2SO4=140 g174.26 g/mol g/mol×6.022×1023 mol−1=4.838×1023 formula units K2SO4$

There are 2 $K+$ in $K2SO4$ . By using equation (3), the number of $K+$ ions in $140 g$ $K2SO4$ is determined as follows.

$Number of atoms=4.838×1023 × 2 K+=9.676×1023 K+ ions$

There is one $SO24−$ in $K2SO4$ . By using equation (3), the number of $SO24−$ in $140 g$ $K2SO4$ is determined as follows.

$Number of atoms=4.838×1023 × 1 SO24−=4.838×1023 SO24−$

Therefore, there are $9.676×1023 K+$ and $4.838×1023 SO24−$ present in $140 g$ of $K2SO4$ .

Answer 32

Ch. 4 - Prob. 1QC Ch. 4 - Prob. 2QC Ch. 4 - Prob. 3QC Ch. 4 - Prob. 4QC Ch. 4 - Prob. 1PP Ch. 4 - Prob. 2PP Ch. 4 - Prob. 3PP Ch. 4 - Prob. 4PP Ch. 4 - Prob. 5PP Ch. 4 - Prob. 6PP

How many atoms (or ions) of each element are in 140.0 g of the following substances? a H 2 b Ca NO 3 2 c N 2 O 2 d K 2 SO 4

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