Answer:11.19
Explanation:
And one mole of a hydrogen atom is of 1.008 grams. So, 2 hydrogen moles weighs 2.016 grams. Hence, one mole of water has 2.016 grams of hydrogen mole. Therefore, the percentage composition of hydrogen would be 2.016/18.0152 = 11.19%.
Answer:
hmmmm
Explanation: Balance the reaction of KOH + H3PO4 = K3PO4 + H2O using this chemical equation balancer!
Answer:
10.5g
Explanation:
First, let us calculate the number of mole of NaHCO3 present in the solution. This is illustrated below:
Volume = 250mL = 250/1000 = 0.25L
Molarity = 0.5M
Mole =?
Molarity = mole /Volume
Mole = Molarity x Volume
Mole = 0.5 x 0.25
Mole = 0.125 mole
Now, we shall be converting 0.125 mole of NaHCO3 to grams to obtain the desired result. This can be achieved by doing the following:
Molar Mass of NaHCO3 = 23 + 1 + 12 +(16x3) = 23 + 1 +12 +48 = 84g/mol
Number of mole of NaHCO3 = 0.125 mole
Mass of NaHCO3 =?
Mass = number of mole x molar Mass
Mass of NaHCO3 = 0.125 x 84
Mass of NaHCO3 = 10.5g
Therefore, 10.5g of NaHCO3 is needed.
Answer:
13.20
Explanation:
Step 1: Calculate the moles of Ba(OH)₂
The molar mass of Ba(OH)₂ is 171.34 g/mol.
0.797 g × 1 mol/171.34 g = 4.65 × 10⁻³ mol
Step 2: Calculate the molar concentration of Ba(OH)₂
Molarity is equal to the moles of solute divided by the liters of solution.
[Ba(OH)₂] = 4.65 × 10⁻³ mol/60 × 10⁻³ L = 0.078 M
Step 3: Calculate [OH⁻]
Ba(OH)₂ is a strong base according to the following equation.
Ba(OH)₂ ⇒ Ba²⁺ + 2 OH⁻
The concentration of OH⁻ is 2/1 × 0.078 M = 0.16 M
Step 4: Calculate the pOH
pOH = -log OH⁻ = -log 0.16 = 0.80
Step 5: Calculate the pH
We will use the following expression.
pH + pOH = 14
pH = 14 - 0.80 = 13.20
