The answer is: the mass of oxygen is 16.95 grams.
The overall balanced photosynthesis reaction:
6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂.
m(C₆H₁₂O₆) = 15.90 g; mass of glucose.
n(C₆H₁₂O₆) = m(C₆H₁₂O₆) ÷ M(C₆H₁₂O₆).
n(C₆H₁₂O₆) = 15.9 g ÷ 180.18 g/mol.
n(C₆H₁₂O₆) = 0.088 mol; amount of glucose.
From chemical reaction: n(C₆H₁₂O₆) : n(O₂) = 1 : 6.
n(O₂) = 6 · 0.088 mol.
n(O₂) = 0.53 mol; amount of oxygen.
m(O₂) = 0.53 mol · 32.00 g/mol.
m(O₂) = 16.95 g; mass of oxygen.
Answer:
Nitrifying Bacteria are a group of aerobic bacteria important in the nitrogen cycle as converters of soil ammonia to nitrates, compounds usable by plants. An example is nitrosomonas or nitrobacter and species in that family.
The schematic diagram is attached below, which summarises the oxidation of ammonia or free nitrogen in the soil to nitrates for the cowpea plant's utilisation.
Answer:
Volume of water at this temperature is 27.2 mL
Explanation:
We know that 
Here density of water is 0.992 g/mL
Here mass of water is 27.0 g
So 
= 
= 27.2 mL
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
