The solubility of A g B r and F e ( O H ) 2 should be explained using the solubility product value of A g B r and F e ( O H ) 2 . Concept Introduction: Solubility is defined as the maximum quantity of solute dissolved in a given amount of solvent to make a saturated solution at a particular temperature. Molar solubility is the number of moles of a solute that can be dissolved in one liter of a solution. It is expressed as mol/L or M (molarity). Consider a general reaction: M n X m (s) ⇌ n M m+ (aq)+m X n+ (aq) The relation between solubility product and molar solubility is as follows: K sp = [ M m+ ] n [ X n- ] m Here The solubility product of salt is K sp . The molar solubility of M m+ ion is [ M m+ ] . The molar solubility of X n- ion is [ X n- ] . Molarity: The concentration for solutions is expressed in terms of molarity as follows, Molarity = No . of moles of solute Volume of solution in L Moles: One mole is equivalent to the mass of the substance consists same number of units equal to the atoms present in 12 g of 12 C . From given mass of substance moles could be calculated by using the following formula, Moles of substance = Given mass of substance Molecular mass
The solubility of A g B r and F e ( O H ) 2 should be explained using the solubility product value of A g B r and F e ( O H ) 2 . Concept Introduction: Solubility is defined as the maximum quantity of solute dissolved in a given amount of solvent to make a saturated solution at a particular temperature. Molar solubility is the number of moles of a solute that can be dissolved in one liter of a solution. It is expressed as mol/L or M (molarity). Consider a general reaction: M n X m (s) ⇌ n M m+ (aq)+m X n+ (aq) The relation between solubility product and molar solubility is as follows: K sp = [ M m+ ] n [ X n- ] m Here The solubility product of salt is K sp . The molar solubility of M m+ ion is [ M m+ ] . The molar solubility of X n- ion is [ X n- ] . Molarity: The concentration for solutions is expressed in terms of molarity as follows, Molarity = No . of moles of solute Volume of solution in L Moles: One mole is equivalent to the mass of the substance consists same number of units equal to the atoms present in 12 g of 12 C . From given mass of substance moles could be calculated by using the following formula, Moles of substance = Given mass of substance Molecular mass
Solution Summary: The author explains the solubility product value of AgBr and Fe(OH)_2.
The solubility of AgBr and Fe(OH)2 should be explained using the solubility product value of AgBr and Fe(OH)2.
Concept Introduction:
Solubility is defined as the maximum quantity of solute dissolved in a given amount of solvent to make a saturated solution at a particular temperature.
Molar solubility is the number of moles of a solute that can be dissolved in one liter of a solution. It is expressed as mol/L or M (molarity).
Consider a general reaction:
MnXm(s)⇌nMm+(aq)+mXn+(aq)
The relation between solubility product and molar solubility is as follows:
Ksp=[Mm+]n[Xn-]m
Here
The solubility product of salt is Ksp.
The molar solubility of Mm+ ion is [Mm+].
The molar solubility of Xn- ion is [Xn-].
Molarity: The concentration for solutions is expressed in terms of molarity as follows,
Molarity = No. of moles of soluteVolume of solution in L
Moles: One mole is equivalent to the mass of the substance consists same number of units equal to the atoms present in 12g of 12C.
From given mass of substance moles could be calculated by using the following formula,
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Question 59 of 70
The volume of
1
unit of plasma is 200.0 mL
If the recommended dosage
for adult patients is 10.0 mL per kg of body mass, how many units are needed for
a patient with a body mass of 80.0
kg ?
80.0
kg
10.0
DAL
1
units
X
X
4.00
units
1
1
Jeg
200.0
DAL
L
1 units
X
200.0 mL
= 4.00 units
ADD FACTOR
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ANSWER
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D
200.0
2.00
1.60 × 10³
80.0
4.00
0.0400
0.250
10.0
8.00
&
mL
mL/kg
kg
units/mL
L
unit
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19
Identify the starting material in the following reaction. Click the "draw structure" button to launch the
drawing utility.
draw structure ...
[1] 0 3
C10H18
[2] CH3SCH3
H
In an equilibrium mixture of the formation of ammonia from nitrogen and hydrogen, it is found that
PNH3 = 0.147 atm, PN2 = 1.41 atm and Pн2 = 6.00 atm. Evaluate Kp and Kc at 500 °C.
2 NH3 (g) N2 (g) + 3 H₂ (g)
K₂ = (PN2)(PH2)³ = (1.41) (6.00)³ = 1.41 x 104
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Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell