diatomic hydrogen is written as H2 (2.02 grams H2) <------- if each hydrogen atom is 1.01 grams, then two hydrogen atoms are 2.02 grams 2.0 moles H2 X 2.02 grams H2 ------------- (divide to cancel moles) = 4.04 grams/mole H2 ÷ one mole = 4.04 grams H2
D all are 1 because the left side equals the right
Answer:
We need 4.28 grams of sodium formate
Explanation:
<u>Step 1:</u> Data given
MW of sodium formate = 68.01 g/mol
Volume of 0.42 mol/L formic acid = 150 mL = 0.150 L
pH = 3.74
Ka = 0.00018
<u>Step 2:</u> Calculate [base)
3.74 = -log(0.00018) + log [base]/[acid]
0 = log [base]/[acid]
0 = log [base] / 0.42
10^0 = 1 = [base]/0.42 M
[base] = 0.42 M
<u>Step 3:</u> Calculate moles of sodium formate:
Moles sodium formate = molarity * volume
Moles of sodium formate = 0.42 M * 0.150 L = 0.063 moles
<u>Step 4:</u> Calculate mass of sodium formate:
Mass sodium formate = moles sodium formate * Molar mass sodium formate
Mass sodium formate = 0.063 mol * 68.01 g/mol
Mass sodium formate = 4.28 grams
We need 4.28 grams of sodium formate
Based on Le Chatelier's principle, if a system at equilibrium is disturbed by changes in the temperature, pressure or concentration, then the equilibrium will shift in a direction to undo the effect of the induced change.
The given reaction is endothermic i.e, heat is supplied:
CH4(g) + H2O (g) + heat ↔ 3H2(g) + CO(g)
a) When the temperature is lowered, heat is being removed from the system. The reaction will move in a direction to produce more heat i.e. to the left.
Hence, the pressure of CH4 will increase and equilibrium will shift to the left
b) When the temperature is raised, heat is being added to the system. The reaction will move in a direction to consume the added heat i.e. to the right.
Hence, the pressure of CO will increase and equilibrium will shift to the right
First, the symbol for sodium oxide is Na₂O
Each Na (sodium) has a charge of 1+, and each O has a charge of 2- :
Na₂¹⁺O²⁻
There are two Na's, however, and each one is 1+, however, so the Na₂ has a total charge of 2+. Because of this, the 2+ from the 2 Na's and the 2- from the O cancel each other out to make 0.