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3 years ago

How many molecules of water are in 3.8 moles of water?

Chemistry

2 answers:

Salsk061 [2.6K]3 years ago

5 0

Answer:

2.29*10^24 molecules of water

Explanation:

1 molof water =6.022*10^23 molecules of water

therefore

molecules of water = 3.8*6.022*10^23

bazaltina [42]3 years ago

3 0

Answer:

one mole of water contains 6.02 x 1023 MOLECULES of water.

But each molecule of water contains 2 H and 1 O atom = 3 atoms, so there are approximately 1.8 x 1024 atoms in a mole of water.

Explanation:

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The compound consists of 40.1% carbon, 6.6% hydrogen and 53.3% oxygen. Its relative density with respect to hydrogen is -15. Wha

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Answer:

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3 years ago

Show your calculation by uploading a picture. Calculate the molar mass of ammonia, NH3

Cloud [144]

Answer:

17.04 g/mol

Explanation:

Molar Mass of NH₃

we know that

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And Hydrogen has 1.01 gram/mol

but we have 3 Hydrogens So we multiply

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3 years ago

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4. the rock cycle is the layering of eroded sediments

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3 years ago

9. What coefficient of H+ balances the atoms in this half-reaction? H+ + MnO₂ → Mn²+ + H₂O

algol13

Answer:4

Explanation:To balance the equation you need to make the number of each element equivalent in both sides.

To start add a 2 in front of the MnO2 which balances the Mn.

Then balance the oxygen by adding a 4 in front of H20.

The H then needs a 8 as it’s coefficient.

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1 year ago

A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the

baherus [9]

Answer: 1. $9.08\times 10^{-6}$ moles

2. 90 mg

Explanation:

$O_3(g)+2NaI(aq)+H_2O(l) \rightarrow O_2(g)+I_2(s)+2NaOH(aq)$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus $4.54 \times 10^{-6}$ moles of ozone is removed by = $\frac{2}{1}\times 4.54 \times 10^{-6}=9.08\times 10^{-6}$ moles of sodium iodide.

Thus $9.08\times 10^{-6}$ moles of sodium iodide are needed to remove $4.54\times 10^{-6}$ moles of $O_3$

2. $\text{Number of moles of ozone}=\frac{0.01331g}{48g/mol}=0.0003moles$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 0.0003 moles of ozone is removed by = $\frac{2}{1}\times 0.0003=0.0006$ moles of sodium iodide.

Mass of sodium iodide= $moles\times {\text {molar mass}}=0.0006\times 150g/mol=0.09g=90mg$ (1g=1000mg)

Thus 90 mg of sodium iodide are needed to remove 13.31 mg of $O_3$ .

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3 years ago