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2 years ago

PLEAS HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! DUE TONIGHT

Chemistry

1 answer:

Jobisdone [24]2 years ago

8 0

Answer:

It is required answer.

Explanation:

Given that :

1. using balanced chemical equation:

ammonium acetate:

The balanced equation is:

NH₃ + H₂O ===> NH₄OH

when ammonia gas dissolves in water then we get the base in form of ammonium hydroxide.

When NH₄OH reacts with CH₃COOH then we get ammonium acetate and water

NH₄OH + CH₃COOH ===> [CH₃COO]- & NH₄+ & H₂O

So, we can say that,

when we are adding an acid and a base together then we get the product of H₂O and given elements.

2. addition of barium hydroxide to sulfuric acid:

the balanced equation is

H₂SO4+ Ba(OH)₂--> BaSO₄+ 2H₂O

when acid and base reacts together than we get barium sulphate and water

when sulfuric acid and barium hydroxide.

Hence, it is required answer.

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If a pharmacist added 12 g of azelaic acid diluent should be used to prepare 8 fluidto 50 g of an ointment containing 15% ounces

natta225 [31]

Answer is: 31,45%.
mrs₁(C₉H₁₆O₄-<span>azelaic acid) = 12g.
mr</span>₂(C₉H₁₆O₄) = 50g.
ω₂(C₉H₁₆O₄) = 15% = 0,15.
mrs₂(C₉H₁₆O₄) = mr₂·ω₂ = 50g·0,15 = 7,5g.
mrs₃(C₉H₁₆O₄) = mrs₁ + mr₂ = 12g + 7,5g = 19,5g.
mr₃ = mr₂ + mr₂ = 50g + 12g = 62g.
ω₃ = mrs₃÷mr₃ = 19,5g ÷ 62g = 31,45% = 0,3145.

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2 years ago

12. The vapor pressure of water at 90°C is 0.692 atm. What is the vapor pressure (in atm) of a solution made by dissolving 3.68

luda_lava [24]

Answer : The vapor pressure (in atm) of a solution is, 0.679 atm

Explanation : Given,

Mass of $H_2O$ = 1.00 kg = 1000 g

Moles of $CsF$ = 3.68 mole

Molar mass of $H_2O$ = 18 g/mole

Vapor pressure of water = 0.692 atm

First we have to calculate the moles of $H_2O$ .

$\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=\frac{1000g}{18g/mole}=55.55mole$

Now we have to calculate the mole fraction of $H_2O$

$\text{Mole fraction of }H_2O=\frac{\text{Moles of }H_2O}{\text{Moles of }H_2O+\text{Moles of }CsF}=\frac{55.55}{55.55+3.68}=0.938$

Now we have to partial pressure of solution.

According to the Raoult's law,

$P_{Solution}=X_{H_2O}\times P^o_{H_2O}$

where,

$P_{Solution}$ = vapor pressure of solution

$P^o_{H_2O}$ = vapor pressure of water = 0.692 atm

$X_{H_2O}$ = mole fraction of water = 0.938

$P_{Solution}=X_{H_2O}\times P^o_{H_2O}$

$P_{Solution}=0.938\times 0.692atm$

$P_{Solution}=0.649atm$

Therefore, the vapor pressure (in atm) of a solution is, 0.679 atm

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