The amount of matter stays the same between the substances,
Answer:
a. The original temperature of the gas is 2743K.
b. 20atm.
Explanation:
a. As a result of the gas laws, you can know that the temperature is inversely proportional to moles of a gas when pressure and volume remains constant. The equation could be:
T₁n₁ = T₂n₂
<em>Where T is absolute temperature and n amount of gas at 1, initial state and 2, final states.</em>
<em />
<em>Replacing with values of the problem:</em>
T₁n₁ = T₂n₂
X*7.1g = (X+300)*6.4g
7.1X = 6.4X + 1920
0.7X = 1920
X = 2743K
<h3>The original temperature of the gas is 2743K</h3><h3 />
b. Using general gas law:
PV = nRT
<em>Where P is pressure (Our unknown)</em>
<em>V is volume = 2.24L</em>
<em>n are moles of gas (7.1g / 35.45g/mol = 0.20 moles)</em>
R is gas constant = 0.082atmL/molK
And T is absolute temperature (2743K)
P*2.24L = 0.20mol*0.082atmL/molK*2743K
<h3>P = 20atm</h3>
<em />
Answer:
Two
Explanation:
The law of conservation of mass states that in an isolated system the mass present is neither destroyed nor created by chemical changes or physical changes.
This tells us that the mass of reactants must be equal to the product mass.
If 1 atom of Zn react with one atoms of sulfur, the product will be 1 molecule of zinc sulfide according to the equation below
Zn(s) + S(s) ⇒ ZnS(s)
therefore two atoms each f zinc and sulfur will product two molecules of Zn sulfide
Answer : The vapor pressure (in atm) of a solution is, 0.679 atm
Explanation : Given,
Mass of
= 1.00 kg = 1000 g
Moles of
= 3.68 mole
Molar mass of
= 18 g/mole
Vapor pressure of water = 0.692 atm
First we have to calculate the moles of
.

Now we have to calculate the mole fraction of 

Now we have to partial pressure of solution.
According to the Raoult's law,

where,
= vapor pressure of solution
= vapor pressure of water = 0.692 atm
= mole fraction of water = 0.938



Therefore, the vapor pressure (in atm) of a solution is, 0.679 atm
Answer:
= 12 mL H202
Explanation:
Given that, the concentration of H2O2 is given antiseptic = 3.0 % v/v
It implies that, 3ml H2O2 is present in 100 ml of solution.
Therefore, to calculate the amount of H202 in 400.0 mL bottle of solution;
we have;
(3.0 mL/ 100 mL) × 400 mL
= 12 mL H202