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Ad libitum [116K]
3 years ago
7

A sample of methane gas has a volume of 10 l at a pressure of 1.5 atm and a temperature of 20 °c. what volume does the gas occup

y if the pressure is lowered to 0.9 atm at a temperature of 40 °c?
Chemistry
1 answer:
a_sh-v [17]3 years ago
7 0
We can use combined gas laws to solve for the volume of the gas
 \frac{PV}{T}
where P - pressure, V - volume , T - temperature and k - constant
\frac{P1V1}{T1} = \frac{P2V2}{T2}
parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation 
T1 - temperature in Kelvin - 20 °C + 273 = 293 K
T2 - 40 °C + 273 = 313 K
substituting the values 
\frac{1.5 atm*10L}{293 K} =  \frac{0.9 atm*V}{313K}
V = 17.8 L
volume of the gas is 17.8 L 
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What type of reaction is shown below? Check all that apply. Ca + 2H2O → Ca(OH)2 + H2 synthesis decomposition combustion single r
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Ca + 2H2O → Ca(OH)2 + H2

Here this reaction can be compared with

A + BC  ---> AB + C

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So the correct answer is

Single replacement also known as single displacement


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1. Find the masses of the following amounts.
In-s [12.5K]

The mass of 2.15 mol of hydrogen sulphide (H₂S) will be 73.272 gm and the mass of  3.95 × 10⁻³ mol of lead(II) iodide, (PbI₂) will be 1.82 gm

<h3>What is Mole ?</h3>

A mole is a very important unit of measurement that chemists use.

A mole of something means you have 6.023 x 10 ²³ of that thing.

  • For 2.15 mol of hydrogen sulphide (H₂S) :

1 mole hydrogen sulphide (H₂S) = 34.08088 grams

Therefore,

2.15 mol of hydrogen sulphide (H₂S) = 34.08088 grams x 2.15 mol

                                                              = 73.272 gm

  • For 3.95 × 10⁻³ mol of lead(II) iodide, (PbI₂) ;

1 mol of lead(II) iodide, (PbI₂) = 461.00894 grams

Therefore,

3.95 × 10⁻³ mol of lead(II) iodide, (PbI₂) = 461.00894 grams x 3.95 × 10⁻³ mol

                                                                  = 1.82 gm

Hence,The mass of 2.15 mol of hydrogen sulphide (H₂S) will be 73.272 gm and the mass of  3.95 × 10⁻³ mol of lead(II) iodide, (PbI₂) will be 1.82 gm

Learn more about mole here ;

brainly.com/question/21323029

#SPJ1

7 0
2 years ago
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