The mass of carbon dioxide that would be needed to weigh is 462 grams.
Given:
10.5 moles of carbon dioxide.
The periodic table
To find:
The mass of carbon dioxide would be needed to weigh out to make 10.5 moles of carbon dioxide.
Solution:
Mass of carbon dioxide needed = m
Moles of carbon dioxide needed = 10.5 mol
The molar mass of carbon dioxide = M
The atomic mass of carbon atom = 12.011 g/mol
The atomic mass of oxygen atom = 15.999 g/mol
The moles of carbon dioxide will be given as:
The mass of carbon dioxide that would be needed to weigh is 462 grams.
Learn more about the moles of substance here:
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Answer:
1.35 moles of O²⁻
21.6 grams of O²⁻
Explanation:
We know that the charge on Aluminium ion is +3 (i.e. Al³⁺) while, the charge on Oxide ion is -2 (i.e. O²⁻). Therefore, the overall neutral Al₂O₃ compound has 2 Al³⁺ ions and 3 O²⁻ ions. Since, we can say that,
1 mole of Al₂O3 contains = 3 moles of O²⁻ ions
So,
0.450 moles of Al₂O₃ will have = X g of O²⁻
Solving for X,
X = 0.450 mol × 3 mol ÷ 1 mol
X = 1.35 moles of O²⁻
As the mass of an atom is mainly due to the presence of protons and neutrons hence, the addition of two electrons (-ve 2 shows two gained electron) to Oxygen will make a negligible change to the atomic masss of Oxygen because electron is said to be almost 1800 times lighter than proton. Hence, the ionic mass of O²⁻ will be 16 g/mol and the mass of given moles is calculated as,
Mass = Moles × Ionic Mass
Mass = 1.35 mol × 16 g/mol
Mass = 21.6 g
Answer:
C₂H₂ + 3H₂ ⟶ 2CH₄
Explanation:
The initial concentrations are:
[CH₄] = 6.30 ÷ 6.00 = 1.05 mol·L⁻¹
[C₂H₂] = 4.20 ÷ 6.00 = 0.700 mol·L⁻¹
[H₂] = 11.15 ÷ 6.00 = 1.858 mol·L⁻¹
2CH₄ ⇌ C₂H₂ + 3H₂
I/mol·L⁻¹: 1.05 0.700 1.858
![Q = \dfrac{\text{[C$_{2}$H$_{2}$][H$_{2}$]}^{3}}{\text{[CH$_{4}$]}^{2}} = \dfrac{ 0.700\times 1.858^{3}}{1.05^{2}}= 4.07](https://tex.z-dn.net/?f=Q%20%3D%20%5Cdfrac%7B%5Ctext%7B%5BC%24_%7B2%7D%24H%24_%7B2%7D%24%5D%5BH%24_%7B2%7D%24%5D%7D%5E%7B3%7D%7D%7B%5Ctext%7B%5BCH%24_%7B4%7D%24%5D%7D%5E%7B2%7D%7D%20%3D%20%5Cdfrac%7B%200.700%5Ctimes%201.858%5E%7B3%7D%7D%7B1.05%5E%7B2%7D%7D%3D%204.07)
Q > K
That means we have too many products.
The reaction will go to the left to get rid of the excess products.
C₂H₂ + 3H₂ ⟶ 2CH₄
