They are all biotic factors, meaning they were once alive or were alive. (an abiotic factor is something that has never lived.) hope this helped.
Answer:
5.67 g OF WATER WILL BE FORMED WHEN 13.7 g OF MnO2 REACTS WITH HCl GAS.
Explanation:
EQUATION FOR THE REACTION
Mn02 + 4HCl --------> MnCl2 + Cl2 + 2H2O
From the balanced reaction between manganese oxide and hydrogen chloride gas;
1 mole of MnO2 reacts to form 2 mole of water
At STP, the molecular mass of the sample is equal to the mole of the substance. So therefore:
(55 + 16 * 2) g of MnO2 reacts to form 2 * ( 1 *2 + 16) g of water
(55 + 32) g of MnO2 reacts to form 2 * 18 g of water
87 g of MnO2 reacts to form 36 g of water
If 13.7 g of MnO2 were to be used?
87 g of MnO2 = 36 g of H2O
13.7 g of MnO2 = ( 13.7 * 36 / 87) g of water
= 493.2 / 87 g of water
Mass of water = 5.669 g of water
Approximately 5.67 g of water will be formed when 13.7 g of manganese oxide reacts with excess hydrogen chloride gas.
Answer:
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Explanation:
In consideration of the conservation principle of linear momentum
Answer:
0.50 moles of Cu
Explanation:
The balanced chemical equation for given synthetic reaction is,
2 Cu + S → Cu₂S
According to balance chemical equation,
1 mole of Cu₂S is produced by = 2 moles of Cu
So,
0.25 moles of Cu₂S will be produced by = X moles of Cu
Solving for X,
X = 0.25 mol × 2 mol / 1 mol
X = 0.50 moles of Cu
Hence, as the molar ratio of Cu to Cu₂S is 2:1 hence, to produce 0.25 moles of Cu₂S we will need 0.50 moles of Cu.
The first one is D hope it helps!
