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kow [346]
3 years ago
6

Under constant temperature, the volume occupied by a gas varies inversely to the pressure applied. If the gas occupies a volume

of 24 cubic inches under a pressure of 5 pounds per square inch, find the volume when the gas is subjected to a pressure of 10 pounds per square inch.
Chemistry
1 answer:
solniwko [45]3 years ago
7 0

Answer:

not sure, but you know what?

Explanation:

you are amazing! keep doing what youre doing!

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Be sure to answer all parts.The combustion of a 40.0−g gaseous mixture of H2 and CH4 releases 3766 kJ of heat. Calculate the amo
Svetach [21]

Answer:H2=11.4g

CH4=28.6g

Explanation:The complete combustion of the two gases can be represented by a balanced reaction below

1. CH4 +2O2___CO2+2H2O

2.2H2+O2___2H2O

Combining the two we have CH4 +2H2+3O2___

CO2+4H2O

Since the mixture contains 40gof CH4 and 2, therefore 20g of CH4 and 8g of H2 combines.

Calculated from their molecular Mass i.e CH4 12+4×2)=20 and 2H2= 2×2×2=8g

Mass of CH4=20/28×40=28.6g

2H2=8/28×40=11.4g

7 0
3 years ago
0:00 一<br> what is 18,000 divided by 450,000 in scientific notation
ankoles [38]

Answer:

4 × 10-2

(scientific notation)

= 4e-2

(scientific e notation)

= 40 × 10-3

(engineering notation)

(thousandth; prefix milli- (m))

Explanation:

All of the anwsers

3 0
2 years ago
What’s the pattern of a Gas and Plasma?!!
Nataly [62]

Answer:

Gas turns into plasma when heat or energy is added to it. The atoms that make up the gas start to lose their electrons and become positively charged ions. The lost electrons are then able to float freely. This process is called ionization..

Hope it help!

3 0
2 years ago
Controlled nuclear chain reactions
Lunna [17]
The nuclear reactions which are under experimenter's control are said to be controlled nuclear reactions. In this, you can maintain the speed of the incident particle. α and β-decay process are examples of non-controlled nuclear reactions. 
8 0
3 years ago
in a simulation mercury removal from industrial wastewater, 0.020 L of 0.10 M sodium sulfide reacts with 0.050 L of 0.010 M merc
Margarita [4]

Answer:  0.1161 grams of mercury(II) sulfide) form.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}     .....(1)

a) Molarity of Na_2S solution = 0.10 M

Volume of solution = 0.020 L

Putting values in equation 1, we get:

0.10M=\frac{\text{Moles of }Na_2S}{0.020L}\\\\\text{Moles of Na_2S}={0.10mol/L\times 0.020}=0.002mol

\text {Moles of}Na_2S=0.10M\times 0.020L=0.002mol

b) Molarity of Hg(NO_3)_2 solution = 0.010 M

Volume of solution = 0.050 L

Putting values in equation 1, we get:

0.010M=\frac{\text{Moles of }Hg(NO_3)_2}{0.050L}\\\\\text{Moles of }Hg(NO_3)_2={0.010mol/L\times 0.050}=0.0005mol

Na_2S+Hg(NO_3)_2\rightarrow HgS+2NaNO_3

According to stoichiometry :

1 mole of Hg(NO_3)_2 reacts with 1 mole of Na_2S

Thus 0.0005 moles of HgNO_3 reacts with=\frac{1}{1}\times 0.0005=0.0005 moles of Hg(NO_3)_2

Thus Hg(NO_3)_2 is the limiting reagent and Na_2S is the excess reagent.

According to stoichiometry :

1 mole of Hg(NO_3)_2 forms=  1 mole of Hg_2S

Thus 0.0005 moles of Hg(NO_3)_2 forms=\frac{1}{1}\times 0.0005=0.0005 moles of Hg_2S

mass of H_2S=moles\times {\text {Molar mass}}=0.0005mol\times 232.2g/mol=0.1161g

Thus 0.1161 grams of mercury(II) sulfide) form.

5 0
3 years ago
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