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3 years ago

If 7.50 g of methane gas (CH4) is in a 3250 mL container at 25° C, what is the pressure

Chemistry

1 answer:

Andre45 [30]3 years ago

4 0

Answer: 10,452 degrees

Explanation: the temperature will rise as more methane is added

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The teacher said the volume of the liquid was 500 mL when measured a student found it was 499.7 mL what is the students percent

Natalka [10]

Answer:

Explanation:

The percentage error of a certain measurement can be found by using the formula

$P(\%) = \frac{error}{actual \: \: number} \times 100\% \\$

From the question

error = 500 - 499.7 = 0.3

actual volume = 500 mL

We have

$p(\%) = \frac{0.3}{500} \times 100 \\ = \frac{3}{50} \\$

We have the final answer as

Hope this helps you

4 0

3 years ago

How many liters of hydrogen gas will be produced at STP from the reaction of 7.179×10^23 atoms of magnesium with 54.219g of phos

Alexeev081 [22]

Answer: The volume of hydrogen gas produced will be, 12.4 L

Explanation : Given,

Mass of $H_3PO_4$ = 54.219 g

Number of atoms of $Mg$ = $7.179\times 10^{23}$

Molar mass of $H_3PO_4$ = 98 g/mol

First we have to calculate the moles of $H_3PO_4$ and $Mg$ .

$\text{Moles of }H_3PO_4=\frac{\text{Given mass }H_3PO_4}{\text{Molar mass }H_3PO_4}$

$\text{Moles of }H_3PO_4=\frac{54.219g}{98g/mol}=0.553mol$

and,

$\text{Moles of }Mg=\frac{7.179\times 10^{23}}{6.022\times 10^{23}}=1.19mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$3Mg+2H_3PO_4\rightarrow Mg(PO_4)_2+3H_2$

From the balanced reaction we conclude that

As, 3 mole of $Mg$ react with 2 mole of $H_3PO_4$

So, 0.553 moles of $Mg$ react with $\frac{2}{3}\times 0.553=0.369$ moles of $H_3PO_4$

From this we conclude that, $H_3PO_4$ is an excess reagent because the given moles are greater than the required moles and $Mg$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2$

From the reaction, we conclude that

As, 3 mole of $Mg$ react to give 3 mole of $H_2$

So, 0.553 mole of $Mg$ react to give 0.553 mole of $H_2$

Now we have to calculate the volume of $H_2$ gas at STP.

As we know that, 1 mole of substance occupies 22.4 L volume of gas.

As, 1 mole of hydrogen gas occupies 22.4 L volume of hydrogen gas

So, 0.553 mole of hydrogen gas occupies $0.553\times 22.4=12.4L$ volume of hydrogen gas

Therefore, the volume of hydrogen gas produced will be, 12.4 L

4 0

3 years ago

In what type of reaction will an acid and a base react with each other

Arada [10]

<span> When an </span>acid and a base<span> are placed together, they </span>react<span> to neutralize the </span>acid<span> and </span>base<span> properties, producing a salt. The H(+) cation of the </span>acid<span>combines with the OH(-) anion of the </span>base<span> to form water.</span>

6 0

3 years ago

Read 2 more answers

A/An ______ bond occurs when 2 atoms share electrons.

allsm [11]

Answer:

convalent???????

Explanation:

sowwy if it wrong

8 0

2 years ago

A gaseous mixture consisting of nitrogen, argon, and oxygen is in a 3.5-L vessel at 25C. Determine the number of moles of oxygen

Step2247 [10]

Answer:

Number of moles of oxygen = 0.037 mol

Explanation:

Given data:

Total pressure = 98.5 KPa

Partial pressure of nitrogen = 22.0 KPa

Partial pressure of argon = 50.0 KPa

Volume = 3.5 L

Temperature = 25°C (25+273= 298K)

Number of moles of oxygen = ?

Solution:

Total pressure = P(N₂) + P(O₂) + P(Ar)

98.5 KPa = 22.0 KPa +P(O₂) + 50.0 KPa

98.5 KPa = 72.0 KPa +P(O₂)

P(O₂) = 98.5 KPa - 72.0 KPa

P(O₂) = 26.5 KPa

KPa to atm:

26.5 KPa/ 101 = 0.262 atm

Number of moles of oxygen:

PV = nRT

n = PV/RT

n = 0.262 atm × 3.5 L / 0.0821 atm.L/mol.K × 298 K

n = 0.917atm.L /24.47atm.L/ mol

n = 0.037 mol

6 0

3 years ago