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vlabodo [156]
3 years ago
15

A gas occupies 100 mL at 150. kPa. Find its volume at 200. kPa. You must show all your work to receive credit. Be sure to identi

fy which of the Gas Laws you will be using.
please help this question is making me want to drop out of school
Chemistry
1 answer:
PtichkaEL [24]3 years ago
6 0

The Boyle-Mariotte's law or Boyle's law is one of the laws of gases that <u>relates the volume (V) and pressure (P) of a certain amount of gas maintained at constant temperature</u>, as follows:

PV = k

where k is a constant.

We can relate the state of a gas at a specific pressure and volume to another state in which the same gas is at different P and V since the product of both variables is equal to a constant, according to the Boyle's law, which will be the same regardless of the state of the gas. In this way,

P₁V₁ = P₂V₂

Where P₁ and V₁ is the pressure and volume of the gas to a state 1 and P₂ and V₂ is the pressure and volume of the same gas in a state 2.

In this case, in the state 1 the gas occupies a volume V₁ = 100 mL at a pressure of P₁ = 150 kPa. Then, in the state 2 the gas occupies a volume V₂ (that we must calculate through the boyle's law) at a pressure of P₂ = 200 kPa.  Substituting these values in the previous equation and clearing V₂, we have,

P₁V₁ = P₂V₂ → V₂ = \frac{P1V1}{P2}  

→ V₂ = \frac{100 mL x 150 kPa}{200 kPa}

→ V₂ = 75 mL

Then, the volume occupied by the gas at 200 kPa is V₂ = 75 mL

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Alexeev081 [22]
Molar mass 

C₂H₄O₂  = 60.0 g/mol

n = mass / molar mass

3.00 = mass / 60.0

m = 3.00 * 60.0

m = 180 g of <span>C₂H₄O₂ 

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5 0
3 years ago
For the simple decomposition reactionAB(g)→ A(g) + B(g)Rate =k[AB]2 and k=0.2 L/mol*s . How long will it takefor [AB] to reach 1
mixer [17]

Answer:

6.66 s will it take for [AB] to reach 1/3 of its initial concentration 1.50 mol/L.

Explanation:

Rate = k[AB]^2

The order of the reaction is 2.

Integrated rate law for second order kinetic is:

\frac{1}{[A_t]} = \frac{1}{[A]_0}+kt

Where, [A_0] is the initial concentration  = 1.50 mol/L

[A_t] is the final concentration  = 1/3 of initial concentration = \frac{1}{3}\times 1.50\ mol/L = 0.5 mol/L

Rate constant, k = 0.2 L/mol*s

Applying in the above equation as:-

\frac{1}{0.5} = \frac{1}{1.50}+0.2t

\frac{1}{1.5}+0.2x=\frac{1}{0.5}

t = 6.66\ s

<u>6.66 s will it take for [AB] to reach 1/3 of its initial concentration 1.50 mol/L.</u>

5 0
3 years ago
An analytical chemist is titrating 118.3 mL of a 0.3500 M solution of butanoic acid (HC3H7CO2) with a 0.400 M solution of KOH. T
TEA [102]

Answer:

pH = 12.33

Explanation:

Lets call HA = butanoic acid and A⁻ butanoic acid and its conjugate base butanoate respectively.

The titration reaction is

HA + KOH ---------------------------- A⁻ + H₂O + K⁺

number of moles of HA :   118.3 ml/1000ml/L x 0.3500 mol/L = 0.041 mol HA

number of  moles of OH  : 115.4 mL/1000ml/L x 0.400 mol/L  = 0.046 mol A⁻

therefore the weak acid will be completely consumed and what we have is  the unreacted strong base KOH which will drive the pH of the solution since the contribution of the conjugate base is negligible.

n unreacted KOH = 0.046 - 0.041 = 0.005 mol KOH

pOH = - log (KOH)

M KOH = 0.005 mol / (0.118.3 +0.1154)L = 0.0021 M

pOH = - log (0.0021) = 1.66

pH = 14 - 1.96 = 12.33

Note: It is a mistake to ask for the pH of the <u>acid solutio</u>n since as the above calculation shows we have a basic solution the moment all the acid has been consumed.

4 0
2 years ago
Titration of a 20.0-mL sample of acid rain required 1.7 mL of 0.0811 M NaOH to reach the end point. If we assume that the acidit
Rasek [7]

Answer:

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

Explanation:

Volume of NaOH = 1.7 ml = 0.0017 L

Molarity of NaOH = 0.0811 M

Moles of NaOH = n

0.0811 M=\frac{n}{0.0017 L}

n = 0.0001378 mol

H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O

According to reaction, 2 mol of NaOH neutralize 1 mol of sulfuric acid.

Then 0.0001378 mol of NaOH will neutralize:

\frac{1}{2}\times 0.0001378 mol=6.8935\times 10^{-5} mol of sulfuric acid.

Concentration of sulfuric acid in the acid rain sample: x

x=\frac{6.8935\times 10^{-5}}{0.02 L}=0.0034467 mol/L

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

7 0
3 years ago
How many moles in 3.69 x 10^30 molecules of carbon dioxide?
elixir [45]

Answer:

four million two hundred and thirty

Explanation:

7 0
3 years ago
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