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2 years ago

How many molecules are there in 122 grams of NO2

Chemistry

1 answer:

AnnyKZ [126]2 years ago

3 0

Thus, there are 1.47 × 10^(23) molecules present in 122 grams of NO2.

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creativ13 [48]

I am sure the answer is D

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3 years ago

Round each number to four significant figures

ira [324]

Answer:

84,791 » 8479

256.75 » 256.8

431,801 » 4318

0.00078100 » 0.0007810

Explanation:

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2 years ago

Determine which law is appropriate for solving the following problem. Carbon dioxide is in a steel tank at 20°C, 10 liters and 1

patriot [66]

<u>Answer:</u>

Law used: Combined Gas Law

<u>Explanation:</u>

We are given the following problem:

Carbon dioxide is in a steel tank at 20°C, 10 liters and 1 atm. What is the pressure on the gas when the tank is heated to 100°C?

To solve this, the most appropriate law that can be used it Combined Gas Law, which is the result of combining the Boyle's law, Charles' law, and Gay-Lussac's law together.

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3 years ago

Read 2 more answers

A sample of a compound is analyzed and found to contain 0.420 g nitrogen, 0.480g oxygen, 0.540 g carbon and 0.135 g hydrogen. Wh

RSB [31]

Answer:

C2H5NO

Explanation:

constituent elements N O C H

Mass composition 0.420 0.480 0.540 0.135

mole ratio 0.42/14 0.48/16 0.54/12 0.135/1

= 0.03 0.03 0.045 0.135

dividing by the smallest 0.03/0.03 0.03/0.03 0.045/0.03 0.135/0.03

ratio = 1 1 1.5 4.5

= 1 1 2 5

EMPERICAL FORMULA = C2H5NO

7 0

2 years ago

A 0.98 gram sample of a volatile liquid was heated to 348 k. the gas occupied 265 ml of space at a pressure of 0.95 atm. what is

arlik [135]

Answer:

The molecular weight is $Z = 111.2 \ g/mol$

Explanation:

From the question we are told that

The mass of the sample is $m = 0.98 \ g$

The temperature is $T = 348 K$

The volume which the gas occupied is $V = 265 \ ml = 265 *10^{-3} L$

The pressure is $P = 0.95 \ atm$

Generally from the ideal gas equation we have that

$PV = n RT$

Here n is the number of moles of the gas while the R is the gas constant with value $R = 0.0821 \ atm \cdot L \cdot mol^{-1} \cdot K^{-1}$

$n = \frac{PV}{ RT}$

=> $n = \frac{ 0.95 * 265 *10^{-3} }{ 0.0821 * 348}$

=> $n = 0.00881 \ mol$

Generally the molecular weight is mathematically represented as

$Z = \frac{m}{n}$

=> $Z = \frac{0.98 }{0.00881}$

=> $Z = 111.2 \ g/mol$

8 0

2 years ago