Answer is: <span> two samples have in common same amount of substance and same number of particles.
1) There are same amount of substance in both beakers:
n(Zn) = 1 mol.
n(ZnCl</span>₂) = 1 mol.
2) There are same number of particles (atoms, molecules, ions) in both beakers:
N(Zn) = n(Zn) · Na.
N(Zn) = 1 mol · 6.023·10²³ 1/mol = 6.023·10²³ atoms of zinc.
N(ZnCl₂) = n(ZnCl₂) · Na.
N(ZnCl₂) = 1 mol · 6.023·10²³ 1/mol = 6.023·10²³ molecules of zinc(II) chloride.
Na - Avogadro number.
<span>Calculating average atomic mass is exactly like calculating a weighted average. Perform the following calculation:
(mass1)(percentage1) + (mass2)(percentage2) = average atomic mass
(35.0)(0.75) + (37.0)(0.25) = average atomic mass
Make sure your percentages are in decimal form for this calculation.
One of the other answers given is correct, though the explanation is lacking a bit. Two of the answers can be eliminated immediately: 35.0 amu and 37.0 amu cannot be the average. If the mixture of isotopes was 50% and 50%, then 36.0 amu would be correct; however, the mixture is 75/25. This leaves only one possible answer choice.</span>
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:
Explanation:
Here we have the mass of CO₂ added = 340 g
From
We have, where the molar mass of CO₂ is 44.01 g/mol
Therefore,
71. Included drawing attached
72. Here we have the pressure of the gas given by Charles law which can be resented as follows;
Where:
P₁ = Initial pressure = 6.1 atmospheres
P₂ = Final pressure
T₁ = Initial Temperature = 293 K
T₂ = Initial Temperature = 313 K
Therefore,
