1.062 mol/kg.
<em>Step 1</em>. Write the balanced equation for the neutralization.
MM = 204.22 40.00
KHC8H4O4 + NaOH → KNaC8H4O4 + H2O
<em>Step 2</em>. Calculate the moles of potassium hydrogen phthalate (KHP)
Moles of KHP = 824 mg KHP × (1 mmol KHP/204.22 mg KHP)
= 4.035 mmol KHP
<em>Step 3</em>. Calculate the moles of NaOH
Moles of NaOH = 4.035 mmol KHP × (1 mmol NaOH/(1 mmol KHP)
= 4.035 mmol NaOH
<em>Step 4</em>. Calculate the mass of the NaOH
Mass of NaOH = 4.035 mmol NaOH × (40.00 mg NaOH/1 mmol NaOH)
= 161 mg NaOH
<em>Step 5</em>. Calculate the mass of the water
Mass of water = mass of solution – mass of NaOH = 38.134 g - 0.161 g
= 37.973 g
<em>Step 6</em>. Calculate the molal concentration of the NaOH
<em>b</em> = moles of NaOH/kg of water = 0.040 35 mol/0.037 973 kg = 1.062 mol/kg
Answer:
2.445 g
Explanation:
Step 1: Given and required data
- Energy in the form of heat required to boil the water (Q): 5525 J
- Latent heat of vaporization of water (∆H°vap): 2260 J/g
Step 2: Calculate the mass of water
We will use the following expression.
Q = ∆H°vap × m
m = Q / ∆H°vap
m = 5525 J / (2260 J/g)
m = 2.445 g
C. HNO₃(aq) + KOH(aq) → KNO₃(aq) + H₂O
Answer:
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Explanation:
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