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Natali [406]
3 years ago
12

At stp, a 12.0 liter sample of CH4 has the same total number of molecules as

Chemistry
2 answers:
Elden [556K]3 years ago
5 0

Option 2: 12.0 L of CO_{2} (g) at STP.

The standard pressure and temperature values are 1 atm and 273.15 K.

Using the ideal gas equation, number of moles of gas can be calculated which is as follows:

PV=nRT...... (1)

Here, P is pressure, V is volume, n is number of moles, R is gas constant and T is temperature.

Also, in 1 mole of any gas there are 6.023\times 10^{23} molecules of the gas. This is known as Avogadro's number and denoted by symbol N_{A}

Thus,

N=n\times N_{A}

Equation (1) can be rewritten as follows:

PV=\frac{N}{N_{A}}RT

On rearranging,

N=\frac{P\times V\times N_{A}}{R\times T}

Here, all the other terms are constant except volume, thus, gas with volume equal to the volume of CH_{4} will have same number of molecules.

Volume of CH_{4} gas and CO_{2}gas is same thus, CH_{4} will have same total number of molecules as  CO_{2} gas.


Luba_88 [7]3 years ago
3 0
<span>2 the answer is 12.0 L of CO2(g) at STP</span>
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7 0
3 years ago
Please help my teacher has like given up
NikAS [45]

Answer:

24.47 L

Explanation:

Using the general gas law equation:

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = 0.0821 Latm/molK

T = temperature (K)

According to the provided information in this question,

P = 1.0 atm

V = ?

n = 1 mol

T = 25°C = 25 + 273 = 298K

Using PV = nRT

V = nRT ÷ P

V = 1 × 0.0821 × 298 ÷ 1

V = 24.465 ÷ 1

V = 24.465

V = 24.47 L

5 0
3 years ago
The chemical compound, "AB" has a molar mass of 74.548 g/mol.
sladkih [1.3K]

Answer:

47.55%

Explanation:

4 0
2 years ago
Chemistry help!<br><br>Zoom in to see better!!​
inna [77]

Answer:

11.9 g of nitrogen monoxide

Explanation:

We'll begin by calculating the number of mole in 6.75 g of NH₃. This can be obtained as follow:

Mass of NH₃ = 6.75 g

Molar mass of NH₃ = 14 + (3×1)

= 14 + 3

= 17 g/mol

Mole of NH₃ =?

Mole = mass /molar mass

Mole of NH₃ = 6.75 / 17

Mole of NH₃ = 0.397 mole

Next, we shall determine the number of mole of NO produced by the reaction of 0.397 mole of NH₃. This can be obtained as follow:

4NH₃ + 5O₂ —> 4NO + 6H₂O

From the balanced equation above,

4 moles of NH₃ reacted to produce 4 moles of NO.

Therefore, 0.397 mole of NH₃ will also react to produce 0.397 mole of NO.

Finally, we shall determine the mass of 0.397 mole of NO. This can be obtained as follow:

Mole of NO = 0.397 mole

Molar mass of NO = 14 + 16 = 30 g/mol

Mass of NO =?

Mass = mole × molar mass

Mass of NO = 0.397 × 30

Mass of NO = 11.9 g

Thus, the mass of NO produced is 11.9 g

7 0
2 years ago
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