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Nina [5.8K]
3 years ago
14

A sample of gas has a volume of 125mL at 6.0 atm pressure. What will the new volume in mL be of the pressure is decreased to 0.2

atm?
Chemistry
1 answer:
Rasek [7]3 years ago
3 0
The answer is 607 your welcomeee
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Help me plz and need it bad
tatyana61 [14]

Answer:

Explanation:                         p + ci=pci

h+o=h io

4 0
3 years ago
If a gas has a volume of 15l and a temperature of 125K, and the temperature is increased to 225K, what is the new volume
aalyn [17]

Answer:

V₂ =  27 L

Explanation:

Given data:

Initial volume = 15 L

Initial temperature = 125 K

Final temperature = 225 K

Final volume = ?

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 15 L × 225K / 125 k

V₂ = 3375 L.K / 125 K

V₂ =  27 L

6 0
3 years ago
Convert 4.992×1010 g to each of the following units.<br> B. Mg<br> c. mg<br> D. metric tons
SVETLANKA909090 [29]
I think it is letter D
3 0
3 years ago
Calculate the molality of a 10.0% by mass solution of NaCl ______ m
Lubov Fominskaja [6]

Answer:

\large \boxed{\text{0.180 mol/kg }}

Explanation:

Assume 100 g of solution.

Then you have 10.0 g of NaCl and 90.0 g of water.

\text{Molal concentration} = \dfrac{\text{moles of solute}}{\text{kilograms of solvent}}

1. Moles of NaCl

\text{Moles of NaCl} = \text{10.0 g NaCl} \times \dfrac{\text{1 mol NaCl}}{\text{58.44 g NaCl}} = \text{0.1711 mol NaCl}

2. Kilograms of water

\text{Kilograms} = \text{90.0 g} \times \dfrac{\text{1 kg}}{\text{1000 g}} = \text{0.0900 kg}

3. Molal concentration

\text{Molal concentration} = \dfrac{\text{0.1711 mol}} {\text{0.0900 kg}} = \textbf{1.90 mol/kg}\\\\\text{The molal concentration of the solution is $\large \boxed{\textbf{0.180 mol/kg }}$}

5 0
3 years ago
The application of knowledge to solve practical<br> problems is known as what?
GenaCL600 [577]
Is known as “Technology”
7 0
3 years ago
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