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vova2212 [387]
4 years ago
15

How can water be decomposed into its elements, hydrogen and oxygen

Chemistry
1 answer:
algol [13]4 years ago
4 0
Water can be decomposed with electrolysis. Which gives you oxygen and hydrogen with their separate types of atoms.
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That is a alkazeltser shot
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1. The hydrogenation of ethene gas at 298 K shows a decrease in disorder (AS-120.7 J/mol-K) during an exothermic reaction ( Delt
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K shows a decrease in disorder (AS = -120.7 J/(mol-K)) during an exothermic reaction (AH' = -136.9 kJ/mol). Determine whether the reaction is spontaneous or ...
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A ___________________________ is a mixture in which the properties of thesubstances are spread evenly.
wel

Answer:

homogenous mixture

Explanation:

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3 years ago
How many moles each element are in a mole of methane?
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\large \mathcal{ANSWER}

This means that the mass of one methane molecule is 12.011 u + (4 × 1.008u), or 16.043 u. This means that one mole of methane has a mass of 16.043 grams. A mole can be thought of as two bags of different sized balls. One bag contains 3 tennis balls and the other 3 footballs.

Samples of helium (He), nitrogen (N 2 ), and methane (CH 4 ) are at STP. Each contains 1 mole or 6.02 × 10 23 particles. However, the mass of each gas is different and corresponds to the molar mass of that gas: 4.00 g/mol for He, 28.0 g/mol for N 2 , and 16.0 g/mol for CH 4 .

3 0
3 years ago
In the laboratory you are asked to make a 0.175 m barium iodide solution using 13.9 grams of barium iodide. How much water shoul
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<u>Answer:</u> The mass of water that should be added in 203.07 grams

<u>Explanation:</u>

To calculate the molality of solution, we use the equation:

\text{Molality}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

Where,

m = molality of barium iodide solution = 0.175 m

m_{solute} = Given mass of solute (barium iodide) = 13.9 g

M_{solute} = Molar mass of solute (barium iodide) = 391.14 g/mol

W_{solvent} = Mass of solvent (water) = ? g

Putting values in above equation, we get:

0.175=\frac{13.9\times 1000}{391.14\times W_{solvent}}\\\\W_{solvent}=\frac{13.9\times 1000}{391.14\times 0.175}=203.07g

Hence, the mass of water that should be added in 203.07 grams

4 0
3 years ago
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