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Aleksandr-060686 [28]
2 years ago
12

How do you find the amount of moles is .032 grams of water and whats the answer

Chemistry
1 answer:
masha68 [24]2 years ago
6 0

Answer:

\boxed {\boxed {\sf 0.0018 \ mol \ H_2 O }}

Explanation:

First, we need to find the molecular mass of water (H₂O).

H₂O has:

  • 2 Hydrogen atoms (subscript of 2)
  • 1 Oxygen atom (implied subscript of 1)

Use the Periodic Table to find the mass of hydrogen and oxygen. Then, multiply by the number of atoms of the element.

  • Hydrogen: 1.0079 g/mol
  • Oxygen: 15.9994 g/mol

There are 2 hydrogen atoms, so multiply the mass by 2.

  • 2 Hydrogen: (1.0079 g/mol)(2)= 2.0158 g/mol

Now, find the mass of H₂O. Add the mass of 2 hydrogen atoms and 1 oxygen atom.

  • 2.0158 g/mol + 15.9994 g/mol = 18.0152 g/mol

Next, find the amount of moles using the molecular mass we just calculated. Set up a ratio.

0.032 \ g  \ H_2 O* \frac{ 1 \ mol \ H_2 O}{18.0152 \ g \ H_2 O}

Multiply. The grams of H₂O will cancel out.

0.032 * \frac{1 \ mol \ H_2 O}{18.0152 }

\frac{0.032 *1 \ mol \ H_2 O}{18.0152 }

0.00177627781 \ mol \ H_2 O

The original measurement given had two significant figures (3,2). We must round to have 2 significant figures. All the zeroes before the 1 are not significant. So, round to the ten thousandth.

The 7 in the hundred thousandth place tells us to round up.

0.0018 \ mol \ H_2 O

There are about <u>0.0018 moles in 0.032 grams.</u>

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2 years ago
The molar mass of glucose is 180.2 g/mol. How many grams of glucose will be produced when 132.0 g of CO2 reacts with an excess o
andriy [413]

Answer:

The mass in grams of glucose produced when 132.0 g of CO2 reacts with an excess of water is 90.1 grams

Explanation:

The chemical equation for the reaction is

6H₂O + 6CO₂  → C₆H₁₂O₆ + 6O₂

From the reaction, it is seen that 6 moles of H₂O reacts ith 6 moles of CO₂ to produce 1 mole of glucose  C₆H₁₂O₆ and 6 moles oxygen gas

The molar mass of CO₂ = 44.01 g/mol

There fpre 132.0 g contains 132.0/44.01 moles or ≅ 3 moles

However since 6 moles of CO₂ produces 1 mole of O₂, then 3 moles of CO₂ will prduce 1/6×3 or 0.5 moles of C₆H₁₂O₆

and since the molar mass (or the mass of one mole) of C₆H₁₂O₆ is 180.2 grams/mole then 0.5 mole of C₆H₁₂O₆ will have a mass of

mass of 1 mole C₆H₁₂O₆ = 180.2 g

mass of 0.5 mole C₆H₁₂O₆ = 180.2 g × 0.5 = 90.1 grams

Mass of glucose produced = 90.1 grams

7 0
3 years ago
What is the empirical formula for a compound that is 94.1% oxygen and 5.90 % hydrogen?
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Answer:

The empirical formula would be N₂Os * Page 2 Calculate the empirical formulaof a compound that is 94.1% oxygen, 5.9% hydrogen.

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A box of paper clips weighs 1.1 kg. If each paper clip weighs 0.88 grams, how many paper clips are in the box? (Hint: 1 kg = 100
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1.1kg = 1100g
If each paper clip is 0.88g : 1100 / 0.88 = 1250
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2 years ago
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Answer:

For this experiment we are going to take plate 1 as the control plate, so, in it there will be just E. coli in LB/agar; in plate 2, we are going to put E. coli in LB/agar and some ampicillin.  Then, we have to wait for the E. coli colonies to form. After a while, the E. coli growth can be compared on both plates and determine if ampicillin affects or not the E. coli colonies.

Explanation:

If the ampicillin affects negatively E. coli colonies, we are going to observe that in plate 1 (control plate) there are E. coli colonies growing, but in plate 2, there is no E. coli colonies or, at least, there is a fewer number of colonies on it. If ampicillin doesn't affect E.coli, plate 1 (control) and plate 2 (ampicillin experiment) are going to be similar in number of colonies.

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