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3 years ago

23.5 L of h2 is stored at a pressure of 58.7 Kpa what volume would the gas take up at stp

Chemistry

1 answer:

stepan [7]3 years ago

3 0

Answer:- 13.6 L

Solution:- Volume of hydrogen gas at 58.7 Kpa is given as 23.5 L. It asks to calculate the volume of hydrogen gas at STP that is standard temperature and pressure. Since the problem does not talk about the original temperature so we would assume the constant temperature. So, it is Boyle's law.

Standard pressure is 1 atm that is 101.325 Kpa.

Boyle's law equation is:

$P_1V_1=P_2V_2$

From given information:-

$P_1$ = 58.7 Kpa

$V_1$ = 23.5 L

$P_2$ = 101.325 Kpa

$V_2$ = ?

Let's plug in the values and solve it for final volume.

$58.7Kpa*23.5L=101.325Kpa*V_2$

On rearranging the equation for $V_2$

$V_2=\frac{58.7Kpa*23.5L}{101.325Kpa}$

$V_2$ = 13.6 L

So, the volume of hydrogen gas at STP for the given information is 13.6 L.

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Calculate the volume of 4.00 molar NaOH solution required to prepare 100 mL of a 0.500 molar solution of NaOH.

Kazeer [188]

Answer:

The answer to your question is V₁ = 12.5 ml

Explanation:

Data

Volume = V₁?

[NaOH] = C₁ = 4.0 M

Volume 2 = V₂ = 100 ml

[NaOH] = C₂ = 0.5 M

Formula of dilution

V₁C₁ = V₂C₂

Solve for V₁ (original solution)

V₁ = $\frac{V2C2}{C1}$

Substitution

V₁ = $\frac{(0.5)(100)}{4}$

Simplification

V₁ = $\frac{50}{4}$

Result

V₁ = 12.5 ml

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3 years ago

How many molecules of oxygen are produced when a sample of 38.9 g of water is decomposed by electricity?

statuscvo [17]

Answer:

A) $6.5\times 10^{23}\ \text{molecules}$

Explanation:

m = Mass of water = 38.9

M = Molar mass of water = 18 g/mol

$N_A$ = Avogadro's number = $6.022\times 10^{23}\ \text{mol}^{-1}$

The reaction of electrolysis would be

$2H_2O(l)\rightarrow 2H_2(g)+O_2(g)$

Number of moles of $H_2O$

$n=\dfrac{m}{M}\\\Rightarrow n=\dfrac{38.9}{18}\ \text{mol}$

From the reaction it can be seen that 2 moles of $H_2O$ gives 1 mole of $O_2$

So, number of moles of $O_2$ produced is

$\dfrac{38.9}{18}\times \dfrac{1}{2}=1.081\ \text{mol}$

Number of molecules

$1.081N_A=1.081\times 6.022\times 10^{23}\\ =6.5\times 10^{23}$

So, $6.5\times 10^{23}\ \text{molecules}$ of oxygen is produced.

8 0

3 years ago

A 151.5-g sample of a metal at 75.0°C is added to 151.5 g at 15.1°C. The temperature of the water rises to 18.7°C. Calculate the

Kryger [21]

Answer:

The specific heat capacity of the metal is 0.268 J/g°C

Explanation:

Step 1: Data given

Mass of the metal = 151.5 grams

The temperature of the metal = 75.0 °C

Temperature of water = 15.1 °C

The temperature of the water rises to 18.7°C.

The specific heat capacity of water is 4.18 J/°C*g

Step 2: Calculate the specific heat capacity of the metal

heat lost = heat gained

Q = m*c*ΔT

Qmetal = - Qwater

m(metal) * c(metal) * ΔT(metal) = m(water) * c(water) * ΔT(water)

⇒ mass of the metal = 151.5 grams

⇒ c(metal) = TO BE DETERMINED

⇒ΔT( metal) = T2 - T1 = 18.7 °C - 75.0 °C = -56.3 °C

⇒ mass of the water = 151.5 grams

⇒ c(water) = 4.184 J/g°C

⇒ ΔT(water) = 18.7° - 15.1 = 3.6 °C

151.5g * c(metal) * -56.3°C = 151.5g * 4.184 J/g°C * 3.6 °C

c(metal) = 0.268 J/g°C

The specific heat capacity of the metal is 0.268 J/g°C

5 0

4 years ago

What is the total gas pressure in a sealed flask that contains oxygen at a partial pressure of 0.35 atm and water vapor at a par

Fofino [41]

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yanalaym [24]

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