The value of x in the molecular formula is to be calculated. Concept introduction: The ideal gas equation elaborates the physical properties of gases by relating the pressure, volume, temperature, and number of moles with each other with the help of four gas laws. This can be shown by: P V = n R T Here, n is the number of moles, R is the gas constant, T is the temperature, V is thevolume, and P is the pressure. The formula for conversion of temperature from degree Celsius to Kelvin is represented as T ( K ) = T ( ° C ) + 273.15

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 10, Problem 110AP

Interpretation Introduction

Interpretation:

The value of x in the molecular formula is to be calculated.

Concept introduction:

The ideal gas equation elaborates the physical properties of gases by relating the pressure, volume, temperature, and number of moles with each other with the help of four gas laws. This can be shown by:

$PV=nRT$

Here, $n$ is the number of moles, $R$ is the gas constant, $T$ is the temperature, $V$ is thevolume, and $P$ is the pressure.

The formula for conversion of temperature from degree Celsius to Kelvin is represented as

$T(K)=T(°C)+273.15$

Expert Solution & Answer

Answer to Problem 110AP

Solution: $MgSO4.7H2O$

Explanation of Solution

Given information:

Mass $m=54.2 g$

Pressure $P=24.8 atm$

Temperature $T=120 °C$

Volume $V=2 L$

The temperature is $120 °C$ .

The conversion of temperature from degree Celsius to Kelvin can be done by using the formula given below:

$T(K)=T(°C)+273.15=(120+273.15)=393.15 K$

The number of moles for $1 L$ of volume, that is, molar volume is

The equation for an ideal gas is as follows:

$PV=nH2ORT$

Rearrange the above equation for number of moles as follows:

$nH2O=PVRT$

Substitute $0.08206 L.atm/K.mol$ for $R$ , $393.15 K$ for $T$ , $24.8 atm$ for $P$ , and $2 L$ for $V$ in the above equation

$nH2O=(24.8 atm)(2 L)(0.08206 L.atm/K.mol)(393.15 K)=(49.632.26) mol=1.537 mol$ $nH2O=(24.8 atm)(2 L)(0.08206 L.atm/K.mol)(393.15 K)=(49.632.26) mol=1.537 mol$

Calculate the mass of water as follows:

$mH2O=nH2O×MH2O$

Substitute $1.537 mol$ for $nH2O$ and $18 g/mol$ for $MH2O$ in the above equation

$mH2O=1.537 mol×18 g/mol=27.6 g$ $mH2O=1.537 mol×18 g/mol=27.6 g$

The mass of thesample is thesum of mass of $H2O$ and $MgSO4$ , which is shown as follows:

$msample=mH2O+mMgSO4$

Rearrange the above equation for $mMgSO4$ as follows:

$mMgSO4=msample−mH2O$

Substitute $54.2 g$ for $msample$ and $27.6 g$ for $mH2O$ in the above equation

$mMgSO4=54.2 g−27.6 g=26.4 g$

Calculate the moles of $MgSO4$ as follows:

$nMgSO4=mMgSO4MMgSO4$

Substitute $26.4 g$ for $mMgSO4$ and $120.4 g/mol$ for $MMgSO4$ in the above equation

$nMgSO4=26.4 g120.4 g/mol=0.22 mol$ $nMgSO4=26.4 g120.4 g/mol=0.22 mol$

Now, the mole ratio of $H2O$ and $MgSO4$ is

$nH2OnMgSO4=1.54 mol0.22 mol=7$ $nH2OnMgSO4=1.54 mol0.22 mol=7$

As the mole ratio of $H2O$ and $MgSO4$ is $7:1$ , the value of x in the compound $MgSO4.xH2O$ will be $7$ .

So, the molecular formula of the compound is $MgSO4.7H2O$ .

Conclusion

The value of x in the compound $MgSO4.xH2O$ is $7$ .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

Name Section Score Date EXERCISE B pH, pOH, pка, AND PKD CALCULATIONS 1. Complete the following table. Solution [H+] [OH-] PH РОН Nature of Solution A 2 x 10-8 M B 1 x 10-7 M C D 12.3 6.8 2. The following table contains the names, formulas, ka or pka for some common acids. Fill in the blanks in the table. (17 Points) Acid Name Formula Dissociation reaction Ka pka Phosphoric acid H₂PO₁ H3PO4 H++ H₂PO 7.08 x 10-3 Dihydrogen H₂PO H₂PO H+ HPO 6.31 x 10-6 phosphate Hydrogen HPO₁ 12.4 phosphate Carbonic acid H2CO3 Hydrogen HCO 6.35 10.3 carbonate or bicarbonate Acetic acid CH,COOH 4.76 Lactic acid CH₂CHOH- COOH 1.38 x 10 Ammonium NH 5.63 x 10-10 Phenol CH₂OH 1 x 10-10 Protonated form CH3NH3* 3.16 x 10-11 of methylamine

Indicate whether it is true that Co(III) complexes are very stable.

MnO2 acts as an oxidant in the chlorine synthesis reaction.

Answer 2

Question

Chapter 10, Problem 110AP

Interpretation Introduction

Interpretation:

The value of x in the molecular formula is to be calculated.

Concept introduction:

The ideal gas equation elaborates the physical properties of gases by relating the pressure, volume, temperature, and number of moles with each other with the help of four gas laws. This can be shown by:

$PV=nRT$

Here, $n$ is the number of moles, $R$ is the gas constant, $T$ is the temperature, $V$ is thevolume, and $P$ is the pressure.

The formula for conversion of temperature from degree Celsius to Kelvin is represented as

$T(K)=T(°C)+273.15$

Expert Solution & Answer

Answer to Problem 110AP

Solution: $MgSO4.7H2O$

Explanation of Solution

Given information:

Mass $m=54.2 g$

Pressure $P=24.8 atm$

Temperature $T=120 °C$

Volume $V=2 L$

The temperature is $120 °C$ .

The conversion of temperature from degree Celsius to Kelvin can be done by using the formula given below:

$T(K)=T(°C)+273.15=(120+273.15)=393.15 K$

The number of moles for $1 L$ of volume, that is, molar volume is

The equation for an ideal gas is as follows:

$PV=nH2ORT$

Rearrange the above equation for number of moles as follows:

$nH2O=PVRT$

Substitute $0.08206 L.atm/K.mol$ for $R$ , $393.15 K$ for $T$ , $24.8 atm$ for $P$ , and $2 L$ for $V$ in the above equation

$nH2O=(24.8 atm)(2 L)(0.08206 L.atm/K.mol)(393.15 K)=(49.632.26) mol=1.537 mol$ $nH2O=(24.8 atm)(2 L)(0.08206 L.atm/K.mol)(393.15 K)=(49.632.26) mol=1.537 mol$

Calculate the mass of water as follows:

$mH2O=nH2O×MH2O$

Substitute $1.537 mol$ for $nH2O$ and $18 g/mol$ for $MH2O$ in the above equation

$mH2O=1.537 mol×18 g/mol=27.6 g$ $mH2O=1.537 mol×18 g/mol=27.6 g$

The mass of thesample is thesum of mass of $H2O$ and $MgSO4$ , which is shown as follows:

$msample=mH2O+mMgSO4$

Rearrange the above equation for $mMgSO4$ as follows:

$mMgSO4=msample−mH2O$

Substitute $54.2 g$ for $msample$ and $27.6 g$ for $mH2O$ in the above equation

$mMgSO4=54.2 g−27.6 g=26.4 g$

Calculate the moles of $MgSO4$ as follows:

$nMgSO4=mMgSO4MMgSO4$

Substitute $26.4 g$ for $mMgSO4$ and $120.4 g/mol$ for $MMgSO4$ in the above equation

$nMgSO4=26.4 g120.4 g/mol=0.22 mol$ $nMgSO4=26.4 g120.4 g/mol=0.22 mol$

Now, the mole ratio of $H2O$ and $MgSO4$ is

$nH2OnMgSO4=1.54 mol0.22 mol=7$ $nH2OnMgSO4=1.54 mol0.22 mol=7$

As the mole ratio of $H2O$ and $MgSO4$ is $7:1$ , the value of x in the compound $MgSO4.xH2O$ will be $7$ .

So, the molecular formula of the compound is $MgSO4.7H2O$ .

Conclusion

The value of x in the compound $MgSO4.xH2O$ is $7$ .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 10, Problem 110AP

Interpretation Introduction

Interpretation:

The value of x in the molecular formula is to be calculated.

Concept introduction:

The ideal gas equation elaborates the physical properties of gases by relating the pressure, volume, temperature, and number of moles with each other with the help of four gas laws. This can be shown by:

$PV=nRT$

Here, $n$ is the number of moles, $R$ is the gas constant, $T$ is the temperature, $V$ is thevolume, and $P$ is the pressure.

The formula for conversion of temperature from degree Celsius to Kelvin is represented as

$T(K)=T(°C)+273.15$

Expert Solution & Answer

Answer to Problem 110AP

Solution: $MgSO4.7H2O$

Explanation of Solution

Given information:

Mass $m=54.2 g$

Pressure $P=24.8 atm$

Temperature $T=120 °C$

Volume $V=2 L$

The temperature is $120 °C$ .

The conversion of temperature from degree Celsius to Kelvin can be done by using the formula given below:

$T(K)=T(°C)+273.15=(120+273.15)=393.15 K$

The number of moles for $1 L$ of volume, that is, molar volume is

The equation for an ideal gas is as follows:

$PV=nH2ORT$

Rearrange the above equation for number of moles as follows:

$nH2O=PVRT$

Substitute $0.08206 L.atm/K.mol$ for $R$ , $393.15 K$ for $T$ , $24.8 atm$ for $P$ , and $2 L$ for $V$ in the above equation

$nH2O=(24.8 atm)(2 L)(0.08206 L.atm/K.mol)(393.15 K)=(49.632.26) mol=1.537 mol$ $nH2O=(24.8 atm)(2 L)(0.08206 L.atm/K.mol)(393.15 K)=(49.632.26) mol=1.537 mol$

Calculate the mass of water as follows:

$mH2O=nH2O×MH2O$

Substitute $1.537 mol$ for $nH2O$ and $18 g/mol$ for $MH2O$ in the above equation

$mH2O=1.537 mol×18 g/mol=27.6 g$ $mH2O=1.537 mol×18 g/mol=27.6 g$

The mass of thesample is thesum of mass of $H2O$ and $MgSO4$ , which is shown as follows:

$msample=mH2O+mMgSO4$

Rearrange the above equation for $mMgSO4$ as follows:

$mMgSO4=msample−mH2O$

Substitute $54.2 g$ for $msample$ and $27.6 g$ for $mH2O$ in the above equation

$mMgSO4=54.2 g−27.6 g=26.4 g$

Calculate the moles of $MgSO4$ as follows:

$nMgSO4=mMgSO4MMgSO4$

Substitute $26.4 g$ for $mMgSO4$ and $120.4 g/mol$ for $MMgSO4$ in the above equation

$nMgSO4=26.4 g120.4 g/mol=0.22 mol$ $nMgSO4=26.4 g120.4 g/mol=0.22 mol$

Now, the mole ratio of $H2O$ and $MgSO4$ is

$nH2OnMgSO4=1.54 mol0.22 mol=7$ $nH2OnMgSO4=1.54 mol0.22 mol=7$

As the mole ratio of $H2O$ and $MgSO4$ is $7:1$ , the value of x in the compound $MgSO4.xH2O$ will be $7$ .

So, the molecular formula of the compound is $MgSO4.7H2O$ .

Conclusion

The value of x in the compound $MgSO4.xH2O$ is $7$ .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 10, Problem 110AP

Interpretation Introduction

Interpretation:

The value of x in the molecular formula is to be calculated.

Concept introduction:

The ideal gas equation elaborates the physical properties of gases by relating the pressure, volume, temperature, and number of moles with each other with the help of four gas laws. This can be shown by:

$PV=nRT$

Here, $n$ is the number of moles, $R$ is the gas constant, $T$ is the temperature, $V$ is thevolume, and $P$ is the pressure.

The formula for conversion of temperature from degree Celsius to Kelvin is represented as

$T(K)=T(°C)+273.15$

Expert Solution & Answer

Answer to Problem 110AP

Solution: $MgSO4.7H2O$

Explanation of Solution

Given information:

Mass $m=54.2 g$

Pressure $P=24.8 atm$

Temperature $T=120 °C$

Volume $V=2 L$

The temperature is $120 °C$ .

The conversion of temperature from degree Celsius to Kelvin can be done by using the formula given below:

$T(K)=T(°C)+273.15=(120+273.15)=393.15 K$

The number of moles for $1 L$ of volume, that is, molar volume is

The equation for an ideal gas is as follows:

$PV=nH2ORT$

Rearrange the above equation for number of moles as follows:

$nH2O=PVRT$

Substitute $0.08206 L.atm/K.mol$ for $R$ , $393.15 K$ for $T$ , $24.8 atm$ for $P$ , and $2 L$ for $V$ in the above equation

$nH2O=(24.8 atm)(2 L)(0.08206 L.atm/K.mol)(393.15 K)=(49.632.26) mol=1.537 mol$ $nH2O=(24.8 atm)(2 L)(0.08206 L.atm/K.mol)(393.15 K)=(49.632.26) mol=1.537 mol$

Calculate the mass of water as follows:

$mH2O=nH2O×MH2O$

Substitute $1.537 mol$ for $nH2O$ and $18 g/mol$ for $MH2O$ in the above equation

$mH2O=1.537 mol×18 g/mol=27.6 g$ $mH2O=1.537 mol×18 g/mol=27.6 g$

The mass of thesample is thesum of mass of $H2O$ and $MgSO4$ , which is shown as follows:

$msample=mH2O+mMgSO4$

Rearrange the above equation for $mMgSO4$ as follows:

$mMgSO4=msample−mH2O$

Substitute $54.2 g$ for $msample$ and $27.6 g$ for $mH2O$ in the above equation

$mMgSO4=54.2 g−27.6 g=26.4 g$

Calculate the moles of $MgSO4$ as follows:

$nMgSO4=mMgSO4MMgSO4$

Substitute $26.4 g$ for $mMgSO4$ and $120.4 g/mol$ for $MMgSO4$ in the above equation

$nMgSO4=26.4 g120.4 g/mol=0.22 mol$ $nMgSO4=26.4 g120.4 g/mol=0.22 mol$

Now, the mole ratio of $H2O$ and $MgSO4$ is

$nH2OnMgSO4=1.54 mol0.22 mol=7$ $nH2OnMgSO4=1.54 mol0.22 mol=7$

As the mole ratio of $H2O$ and $MgSO4$ is $7:1$ , the value of x in the compound $MgSO4.xH2O$ will be $7$ .

So, the molecular formula of the compound is $MgSO4.7H2O$ .

Conclusion

The value of x in the compound $MgSO4.xH2O$ is $7$ .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 10, Problem 110AP

Interpretation Introduction

Interpretation:

The value of x in the molecular formula is to be calculated.

Concept introduction:

The ideal gas equation elaborates the physical properties of gases by relating the pressure, volume, temperature, and number of moles with each other with the help of four gas laws. This can be shown by:

$PV=nRT$

Here, $n$ is the number of moles, $R$ is the gas constant, $T$ is the temperature, $V$ is thevolume, and $P$ is the pressure.

The formula for conversion of temperature from degree Celsius to Kelvin is represented as

$T(K)=T(°C)+273.15$

Expert Solution & Answer

Answer to Problem 110AP

Solution: $MgSO4.7H2O$

Explanation of Solution

Given information:

Mass $m=54.2 g$

Pressure $P=24.8 atm$

Temperature $T=120 °C$

Volume $V=2 L$

The temperature is $120 °C$ .

The conversion of temperature from degree Celsius to Kelvin can be done by using the formula given below:

$T(K)=T(°C)+273.15=(120+273.15)=393.15 K$

The number of moles for $1 L$ of volume, that is, molar volume is

The equation for an ideal gas is as follows:

$PV=nH2ORT$

Rearrange the above equation for number of moles as follows:

$nH2O=PVRT$

Substitute $0.08206 L.atm/K.mol$ for $R$ , $393.15 K$ for $T$ , $24.8 atm$ for $P$ , and $2 L$ for $V$ in the above equation

$nH2O=(24.8 atm)(2 L)(0.08206 L.atm/K.mol)(393.15 K)=(49.632.26) mol=1.537 mol$ $nH2O=(24.8 atm)(2 L)(0.08206 L.atm/K.mol)(393.15 K)=(49.632.26) mol=1.537 mol$

Calculate the mass of water as follows:

$mH2O=nH2O×MH2O$

Substitute $1.537 mol$ for $nH2O$ and $18 g/mol$ for $MH2O$ in the above equation

$mH2O=1.537 mol×18 g/mol=27.6 g$ $mH2O=1.537 mol×18 g/mol=27.6 g$

The mass of thesample is thesum of mass of $H2O$ and $MgSO4$ , which is shown as follows:

$msample=mH2O+mMgSO4$

Rearrange the above equation for $mMgSO4$ as follows:

$mMgSO4=msample−mH2O$

Substitute $54.2 g$ for $msample$ and $27.6 g$ for $mH2O$ in the above equation

$mMgSO4=54.2 g−27.6 g=26.4 g$

Calculate the moles of $MgSO4$ as follows:

$nMgSO4=mMgSO4MMgSO4$

Substitute $26.4 g$ for $mMgSO4$ and $120.4 g/mol$ for $MMgSO4$ in the above equation

$nMgSO4=26.4 g120.4 g/mol=0.22 mol$ $nMgSO4=26.4 g120.4 g/mol=0.22 mol$

Now, the mole ratio of $H2O$ and $MgSO4$ is

$nH2OnMgSO4=1.54 mol0.22 mol=7$ $nH2OnMgSO4=1.54 mol0.22 mol=7$

As the mole ratio of $H2O$ and $MgSO4$ is $7:1$ , the value of x in the compound $MgSO4.xH2O$ will be $7$ .

So, the molecular formula of the compound is $MgSO4.7H2O$ .

Conclusion

The value of x in the compound $MgSO4.xH2O$ is $7$ .

Answer 6

Chapter 10, Problem 110AP

Interpretation Introduction

Interpretation:

The value of x in the molecular formula is to be calculated.

Concept introduction:

The ideal gas equation elaborates the physical properties of gases by relating the pressure, volume, temperature, and number of moles with each other with the help of four gas laws. This can be shown by:

$PV=nRT$

Here, $n$ is the number of moles, $R$ is the gas constant, $T$ is the temperature, $V$ is thevolume, and $P$ is the pressure.

The formula for conversion of temperature from degree Celsius to Kelvin is represented as

$T(K)=T(°C)+273.15$

Expert Solution & Answer

Answer to Problem 110AP

Solution: $MgSO4.7H2O$

Explanation of Solution

Given information:

Mass $m=54.2 g$

Pressure $P=24.8 atm$

Temperature $T=120 °C$

Volume $V=2 L$

The temperature is $120 °C$ .

The conversion of temperature from degree Celsius to Kelvin can be done by using the formula given below:

$T(K)=T(°C)+273.15=(120+273.15)=393.15 K$

The number of moles for $1 L$ of volume, that is, molar volume is

The equation for an ideal gas is as follows:

$PV=nH2ORT$

Rearrange the above equation for number of moles as follows:

$nH2O=PVRT$

Substitute $0.08206 L.atm/K.mol$ for $R$ , $393.15 K$ for $T$ , $24.8 atm$ for $P$ , and $2 L$ for $V$ in the above equation

$nH2O=(24.8 atm)(2 L)(0.08206 L.atm/K.mol)(393.15 K)=(49.632.26) mol=1.537 mol$ $nH2O=(24.8 atm)(2 L)(0.08206 L.atm/K.mol)(393.15 K)=(49.632.26) mol=1.537 mol$

Calculate the mass of water as follows:

$mH2O=nH2O×MH2O$

Substitute $1.537 mol$ for $nH2O$ and $18 g/mol$ for $MH2O$ in the above equation

$mH2O=1.537 mol×18 g/mol=27.6 g$ $mH2O=1.537 mol×18 g/mol=27.6 g$

The mass of thesample is thesum of mass of $H2O$ and $MgSO4$ , which is shown as follows:

$msample=mH2O+mMgSO4$

Rearrange the above equation for $mMgSO4$ as follows:

$mMgSO4=msample−mH2O$

Substitute $54.2 g$ for $msample$ and $27.6 g$ for $mH2O$ in the above equation

$mMgSO4=54.2 g−27.6 g=26.4 g$

Calculate the moles of $MgSO4$ as follows:

$nMgSO4=mMgSO4MMgSO4$

Substitute $26.4 g$ for $mMgSO4$ and $120.4 g/mol$ for $MMgSO4$ in the above equation

$nMgSO4=26.4 g120.4 g/mol=0.22 mol$ $nMgSO4=26.4 g120.4 g/mol=0.22 mol$

Now, the mole ratio of $H2O$ and $MgSO4$ is

$nH2OnMgSO4=1.54 mol0.22 mol=7$ $nH2OnMgSO4=1.54 mol0.22 mol=7$

As the mole ratio of $H2O$ and $MgSO4$ is $7:1$ , the value of x in the compound $MgSO4.xH2O$ will be $7$ .

So, the molecular formula of the compound is $MgSO4.7H2O$ .

Conclusion

The value of x in the compound $MgSO4.xH2O$ is $7$ .

Answer 7

Chapter 10, Problem 110AP

Interpretation Introduction

Interpretation:

The value of x in the molecular formula is to be calculated.

Concept introduction:

The ideal gas equation elaborates the physical properties of gases by relating the pressure, volume, temperature, and number of moles with each other with the help of four gas laws. This can be shown by:

$PV=nRT$

Here, $n$ is the number of moles, $R$ is the gas constant, $T$ is the temperature, $V$ is thevolume, and $P$ is the pressure.

The formula for conversion of temperature from degree Celsius to Kelvin is represented as

$T(K)=T(°C)+273.15$

Answer 8

Expert Solution & Answer

Answer to Problem 110AP

Solution: $MgSO4.7H2O$

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given information:

Mass $m=54.2 g$

Pressure $P=24.8 atm$

Temperature $T=120 °C$

Volume $V=2 L$

The temperature is $120 °C$ .

The conversion of temperature from degree Celsius to Kelvin can be done by using the formula given below:

$T(K)=T(°C)+273.15=(120+273.15)=393.15 K$

The number of moles for $1 L$ of volume, that is, molar volume is

The equation for an ideal gas is as follows:

$PV=nH2ORT$

Rearrange the above equation for number of moles as follows:

$nH2O=PVRT$

Substitute $0.08206 L.atm/K.mol$ for $R$ , $393.15 K$ for $T$ , $24.8 atm$ for $P$ , and $2 L$ for $V$ in the above equation

$nH2O=(24.8 atm)(2 L)(0.08206 L.atm/K.mol)(393.15 K)=(49.632.26) mol=1.537 mol$ $nH2O=(24.8 atm)(2 L)(0.08206 L.atm/K.mol)(393.15 K)=(49.632.26) mol=1.537 mol$