Answer: explained below
Explanation:
Matter can change form through physical and chemical changes, but through any of these changes, matter is conserved. The same amount of matter exists before and after the change—none is created or destroyed.
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 />
the balanced chemical equation for the decomposition of H₂O₂ is as follows
2H₂O₂ ---> 2H₂O + O₂
stoichiometry of H₂O₂ to O₂ is 2:1
the number of moles of H₂O₂ decomposed is - 0.250 L x 3.00 mol/L = 0.75 mol
according to stoichiometry the number of O₂ moles is half the number of H₂O₂ moles decomposed
number of moles of O₂ - 0.75 mol / 2 = 0.375 mol
apply the ideal gas law equation to find the volume
PV = nRT
where P - standard pressure - 10⁵ Pa
V - volume
n - number of moles 0.375 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - standard temperature - 273 K
substituting the values in the equation
10⁵ Pa x V = 0.375 mol x 8.314 Jmol⁻¹K⁻¹ x 273 K
V = 8.5 L
volume of O₂ gas is 8.5 L
Answer : Option D) No charge
Explanation : An isotope undergoes radioactive decay by emitting radiation that has no mass. The radiation will not have any charge as it does not has any mass it will not emit a radiative charge.
It is known that there are some unstable radioactive isotopes which has no mass and the radiation thus has no charge in it.
According to this formula:
K= A*(e^(-Ea/RT) when we have K =1.35X10^2 & T= 25+273= 298K &R=0.0821
Ea= 85.6 KJ/mol So by subsitution we can get A:
1.35x10^2 = A*(e^(-85.6/0.0821*298))
1.35x10^2 = A * 0.03
A= 4333
by substitution with the new value of T(75+273) = 348K & A to get the new K
∴K= 4333*(e^(-85.6/0.0821*348)
= 2.16 x10^2
