Answer:
Mass = 121 g
Explanation:
Given data:
Mass in gram of CO₂ = ?
Volume = 61.8 L
Pressure = standard = 1 atm
Temperature = 273.15 K
Solution:
Formula:
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
1 atm × 61.8 L = n ×0.0821 atm.L/ mol.K × 273.15 k
61.8 L.atm = 22.42 atm.L/ mol × n
n = 61.8 L.atm /22.42 atm.L/ mol
n = 2.76 mol
Mass in gram:
Mass = number of moles × molar mass
Mass = 2.76 mol × 44 g/mol
Mass = 121 g
Answer:Since there is virtually nothing in space to scatter or re-radiate the light to our eye, we see no part of the light and the sky appears to be black.
Explanation:
Answer:
8. the answer is B.
9. the answer is A.
Explanation:
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8. In this case, by bearing to mind that the limiting reactant is always completely consumed and the excess one remain as a leftover at the end of the reaction, we can also infer that as all the limiting reactant is consumed, it must determine the maximum amount of product as the excess reactant will hypothetically produce a greater mass than expected; thus, the answer to this question is B.
9. In this case, since the mole ratio of oxygen to water is 1:2, the following proportional factor is used to calculate the produced mass of water:
Thus, the answer is this case is A.
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a. AgBr(s)⇒ Ag⁺(aq) + Br⁻(aq)
b. Ksp AgBr = s²
c. 5 x 10⁻¹³ mol/L
<h3>Further explanation</h3>
Given
solubility AgBr = 7.07 x 10⁻⁷ mol/L
Required
The dissolution reaction
Ksp
The solubility product constant
Solution
a. dissolution reaction of AgBr
AgBr(s)⇒ Ag⁺(aq) + Br⁻(aq)
b. Ksp
Ksp AgBr = [Ag⁺] [Br⁻]
Ksp AgBr = (s) (s)
Ksp AgBr = s²
c. Ksp AgBr = (7.07 x 10⁻⁷)² = 5 x 10⁻¹³ mol/L
