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:
Itching.
Nausea and vomiting.
Weight loss.
Fatigue.
Weakness.
Jaundice.
Swelling and pain in your stomach.
Dark-colored urine and/or light-colored stool.
Explanation:
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Answer:
2 NO (g) → N2 (g) + O2 (g)
2 NOCl (g) → 2 NO (g) + Cl2 (g)
____________________________
2NOCl (g) ⟶ N2 (g) + O2 (g) + Cl2 (g)
ΔH = [90.3 kJ x 2 x -1] + [-38.6 kJ x -1 x 2] = -103.4 kJ
The ΔH for the reaction is -103.4 kJ
Answer:
1.76
Explanation:
There is some info missing. I think this is the original question.
<em>A chemist dissolves 660.mg of pure hydroiodic acid in enough water to make up 300.mL of solution. Calculate the pH of the solution. Be sure your answer has the correct number of significant digits.</em>
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Step 1: Calculate the molarity of HI(aq)
M = mass of solute / molar mass of solute × liters of solution
M = 0.660 g / 127.91 g/mol × 0.300 L
M = 0.0172 M
Step 2: Write the acid dissociation reaction
HI(aq) ⇄ H⁺(aq) + I⁻(aq)
HI is a strong acid, so [H⁺] = 0.0172 M
Step 3: Calculate the pH
pH = -log [H⁺]
pH = -log 0.0172
pH = 1.76
Answer: 5.4
Explanation:
P2 = P1V1/V2
P2 = (.60atm x 27L) / 3.0L = 5.4atm
