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olya-2409 [2.1K]
3 years ago
6

What is the pressure of 0.540 mol of an ideal gas at 35.5 L and 223 K?

Chemistry
1 answer:
Vladimir79 [104]3 years ago
4 0
P: ?
V: 35,5L = 35,5 dm³
n: 0,54 mol
R: 83,14 hPa·dm³/mol·K
T: 223K
..................
pV = nRT
p = nRT/V
p = (0,54×83,14×223)/35,5
p = 282,02 hPa
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Klio2033 [76]

<em>ANSWER</em>

The number of moles of methane is 905.32 moles

STEP-BY-STEP EXPLANATION:

Given information

The number of particles of methane = 5.45 x 10^26 particles

Let x represents the number of moles of methane

To calculate the number of moles, we will be using the below formula

\text{Number of particles = number of moles x Avogadro's constant}

Recall that, the Avogadro's constant is given as

6.02\cdot10^{23}\begin{gathered} 5.45\cdot10^{26}\text{ = x }\cdot\text{ 6.02 }\cdot10^{23} \\ \text{Divide both sides by 6.02 }\cdot10^{23} \\ x\text{ = }\frac{5.45\cdot10^{26}}{6.02\cdot10^{23}} \\ x\text{ = }\frac{5.45}{6.02}\cdot10^{26\text{ - 23}} \\ x\text{ = 0.9053 }\cdot10^3 \\ x\text{ = 905.32 moles} \end{gathered}

Therefore, the number of moles of methane is 905.32 moles

6 0
1 year ago
Find the no. of atoms present in 40g of nitrogen gas at STP?
nika2105 [10]

Answer:

8.60 * 10^{23} atoms N2

Explanation:

We want to convert grams to moles and then moles to atoms.

First, we convert grams of nitrogen gas (which is N2) to moles. To do so, we need the molar mass of N2, which is just 14.01 * 2 = 28.02 g.

40 g N2 * \frac{1 mol N2}{28.02 g N2} = 1.43 mol N2

Now, we need to convert moles to atoms by using Avogadro's number, which is 6.022*10^{23}:

1.43 mol N2 * \frac{6.022*10^{23} atoms N2 }{1 mol N2} = 8.60 * 10^{23} atoms N2

Thus, the answer is 8.60 * 10^{23} atoms N2.

6 0
3 years ago
What is a substance that reduces the amount of activation energy in a chemical reaction called?
FinnZ [79.3K]

The answer is catalyst

3 0
3 years ago
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What type of bonds are these?
gulaghasi [49]
Jeez this is middle school really. covalently bond
3 0
3 years ago
when 824 J of heat is added to 9 grams of oil at 30°c , the temperature increases to 87°C. what is the specific heat of the oil
sattari [20]

Answer:

c = 1.61 j/g.°C

Explanation:

Given data:

Mass of oil = 9 g

Heat added = 824 j

Initial temperature =  30°C

Final temperature = 87°C

Specific heat of oil = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature - final temperature

ΔT = 87°C - 30°C

ΔT = 57 °C

Q = m.c. ΔT

824 j = 9 g × c ×  57 °C

824 j = 513 g. °C × c

c = 1.61 j/g.°C

5 0
3 years ago
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