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kvv77 [185]
1 year ago
10

There are 2 gasses, A, B. They weigh 2.46g and 0.5g respectively, and the Volume of A is 3 times the volume of B. A has a molecu

lar mass of 28. What could A be?
(A) NO2
(B) N2O
(C) N2O4
(D) N2O5
Chemistry
2 answers:
dangina [55]1 year ago
8 0

Answer:

B

Explanation:

molecular mass of B is 28

Mandarinka [93]1 year ago
7 0

Answer:

(B) N2O is the answer.....................

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1.00 L of gas at a standard temperature and pressure is compressed to 473mL. What is the new pressure of the gas?
Nadusha1986 [10]

Answer:

The pressure of the gas is 2.11 atm.

Explanation:

From the given,

V_{1}=1.00\,L

P_{1}=1\,atm

P_{2}=?

V_{2}=473\,ml=0.473\,L

P_{1}V_{1}=P_{2}V_{2}

(1)(1)=(0.473)(P_{2})

P_{2}=\frac{(1)(1)}{0.473}=2.11

Therefore, The pressure of the gas is 2.11 atm.

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In the mathematical equation showing that pressure (P) multiplied by volume (V) remains equal to a constant (k) pressure and vol
hram777 [196]

Answer:

Inversely

Explanation:

pV = k

When p increases, V must decrease for k to remain constant.

When V increases, p must decrease for k to remain constant.

When the product of two variables is a constant, they are inversely proportional  to each other.

3 0
3 years ago
A balloon containing helium gas expands from 230
Anit [1.1K]

The answer for the following problem is mentioned below.

  • <u><em>Therefore the final  moles of the gas is 14.2 × </em></u>10^{-4}<u><em> moles.</em></u>

Explanation:

Given:

Initial volume (V_{1}) = 230 ml

Final volume (V_{2}) = 860 ml

Initial moles (n_{1}) = 3.8 ×10^{-4} moles

To find:

Final moles (n_{2})

We know;

According to the ideal gas equation;

    P × V = n × R × T

where;

P represents the pressure of the gas

V represents the volume of the gas

n represents the no of the moles of the gas

R represents the universal gas constant

T represents the temperature of the gas

So;

    V ∝ n

\frac{V_{1} }{V_{2} } = \frac{n_{1} }{n_{2} }

where,

(V_{1}) represents the initial volume of the gas

(V_{2}) represents the final volume of the gas

(n_{1}) represents the initial  moles of the gas

(n_{2}) represents the final moles of the gas

Substituting the above values;

   \frac{230}{860} = \frac{3.8 * 10^-4}{n_{2} }

  n_{2} = 14.2 × 10^{-4} moles

<u><em>Therefore the final  moles of the gas is 14.2 × </em></u>10^{-4}<u><em> moles.</em></u>

7 0
3 years ago
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