Answer:
moles of glucose
<u>2.3166 moles of glucose</u>
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Explanation:
The balance reaction for the formation of glucose is :
here , CO2 = carbon dioxide
H2O = water
C6H12O6 = glucose
O2 = Oxygen
According to this equation :
6 mole of CO2 = 6 mole of H2O = 1 mole of C6H12O6 = 6 mole of O2
We are asked to calculate the mole of Glucose from carbon dioxide.
So,
6 mole of CO2 produce = 1 mole of C6H12O6
1 mole of CO2 will produce =
moles of glucose
13.9 moles of CO2 will produce :
=2.3166 moles of glucose
Note : first , Always calculate for one mole (By dividing)
. After this , multiply the answer with the moles given.
Always write the substance whose amount is asked(glucose) to the right hand side
Answer:
1.35 g
Explanation:
Data Given:
mass of Potassium Permagnate (KMnO₄) = 3.34 g
Mass of Oxygen: ?
Solution:
First find the percentage composition of Oxygen in Potassium Permagnate (KMnO₄)
So,
Molar Mass of KMnO₄ = 39 + 55 + 4(16)
Molar Mass of KMnO₄ = 158 g/mol
Calculate the mole percent composition of Oxygen in Potassium Permagnate (KMnO₄).
Mass contributed by Oxygen (O) = 4 (16) = 64 g
Since the percentage of compound is 100
So,
Percent of Oxygen (O) = 64 / 158 x 100
Percent of Oxygen (O) = 40.5 %
It means that for ever gram of Potassium Permagnate (KMnO₄) there is 0.405 g of Oxygen (O) is present.
So,
for the 3.34 grams of Potassium Permagnate (KMnO₄) the mass of Oxygen will be
mass of Oxygen (O) = 0.405 x 3.34 g
mass of Oxygen (O) = 1.35 g
Answer:
38.3958 °C
Explanation:
As,
1 gram of carbohydrates on burning gives 4 kilocalories of energy
1 gram of protein on burning gives 4 kilocalories of energy
1 gram of fat on burning gives 9 kilocalories of energy
Thus,
27 g of fat on burning gives 9*27 = 243 kilocalories of energy
20 g of protein on burning gives 4*20 = 80 kilocalories of energy
48 gram of carbohydrates on burning gives 4*48 = 192 kilocalories of energy
Total energy = 515 kilocalories
Using,
Given: Volume of water = 23 L = 23×10⁻³ m³
Density of water= 1000 kg/m³
So, mass of the water:
Mass of water = 23 kg
Initial temperature = 16°C
Specific heat of water = 0.9998 kcal/kg°C
Solving for final temperature as:
<u>Final temperature = 38.3958 °C </u>
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