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:
D
Explanation:
The electromagnetic radiation is emitted due to a particle moves from a higher to a lower energy state
Can you show the question that goes with those answer pls
To give 33.6 dm³ hydrogen gas at STP, 18.06 x 10²³ atoms of Na must react completely.
<h3>What is Mole concept ?</h3>
A mole is a unit of measurement used to measure the amount of any fundamental entity (atoms, molecules, ions) present in the substance.
As according to the given equation, 2 moles (ie 12.04 x 10²³ atoms) of Na-atoms produces 1 mole (22.4 ltr) of H₂-gas.
Hence, to produce 33.6 ltr (equivalent to 33.6 dm³) of H₂-gas produced by ;
= 12.04 x 10²³ atoms of Na / 22.4 ltr of H₂-gas x 33.6 ltr
= 18.06 x 10²³ atoms of Na
Hence, To give 33.6 dm³ hydrogen gas at STP, 18.06 x 10²³ atoms of Na must react completely.
Learn more about Mole concept here ;
brainly.com/question/20483253
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Firstly we need to determine the partial pressure of O2:
We will now use the Henry's Law equation to determine the solubility of the gas:
Answer: Solubility is 2.7x10^-3 M
