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Bas_tet [7]
3 years ago
7

How many moles of H2O were produced during the combustion of CH4

Chemistry
1 answer:
Goryan [66]3 years ago
5 0
<span>Reaction of combustion ( CH4 ) :


</span><span>CH4 + 2 O2 = CO2 + 2 H2O 
</span><span>   </span>↓                                    ↓<span>
1 mole CH4 produced  2 moles of H2O

hope this helps!





</span>
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an atom of which element has the strongest attraction for electons in a chemical bond?cholorine, phosphorus, carbon, sulfur
vredina [299]

Answer:

D) Sulphur

Explanation:

Sulphur has the strongest electromagnetivity here

7 0
3 years ago
The equation represents the combustion of sucrose. C12H22O11 + 12O2 12CO2 + 11H2O If there are 10.0 g of sucrose and 8.0 g of ox
Vaselesa [24]
<span>0.0292 moles of sucrose are available. First, lookup the atomic weights of all involved elements Atomic weight Carbon = 12.0107 Atomic weight Hydrogen = 1.00794 Atomic weight Oxygen = 15.999 Now calculate the molar mass of sucrose 12 * 12.0107 + 22 * 1.00794 + 11 * 15.999 = 342.29208 g/mol Divide the mass of sucrose by its molar mass 10.0 g / 342.29208 g/mol = 0.029214816 mol Finally, round the result to 3 significant figures, giving 0.0292 moles</span>
7 0
3 years ago
A sailor on a trans-Pacific solo voyage notices one day that if he puts 735.mL of fresh water into a plastic cup weighing 25.0g,
Gennadij [26K]

Answer:

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According to Archimedes' principle, mass of freshwater and cup = mass of equal volume of seawater

mass of freshwater = density * volume

1 cm³ = 1 mL

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Therefore,  mass of equal volume of seawater = 759.265 g

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1 liter = 1000 cm³ = 1000 mL;

Density of seawater = mass / volume

Density of seawater = 759.265 g / 0.735 L = 1033.01 g/L

Density of freshwater in g/L = 0.999 g/ (1/1000) L = 999 g/L

mass of 1 Liter seawater = 1033.01 g

mass of 1 Liter freshwater = 999 g

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4 0
3 years ago
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zavuch27 [327]

Answer:

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4 0
3 years ago
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grigory [225]

Answer:

3.01 × 10^24 particles

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According to Avagadro, in one mole of a substance, there are 6.02 × 10^23 atoms or particles.

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NA = Avagadro's constant or number

This means that for 5 moles of a substance, there will be:

5 × 6.02 × 10^23

= 30.1 × 10^23

= 3.01 × 10^24 particles

8 0
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