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:
-1.42, -0.375, 32.5% (.325), 3/8 (.375), √4 (2.0), 3 (3.0), 2³ (8.0)
Answer:
The range of [H⁺] is from 2.51 x 10⁻⁶ M to 6.31 x 10⁻⁶ M,
Explanation:
To answer this problem we need to keep in mind the <u>definition of pH</u>:
So now we <u>calculate [H⁺] using a pH value of 5.2 and of 5.6</u>:
-5.2 = log [H⁺]
= [H⁺]
6.31 x 10⁻⁶ M = [H⁺]
-5.6 = log [H⁺]
= [H⁺]
2.51 x 10⁻⁶ M = [H⁺]
<span>polar, Hadley, Ferrel, polar, Hadley, Ferrel. I hope this helps you.</span>
