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Ilia_Sergeevich [38]
2 years ago
10

How many moles of solute particles are present in 1 mL of each of the following aqueous solutions?

Chemistry
1 answer:
Tanzania [10]2 years ago
6 0

The moles of solute particles are present in 1 mL  is 3.3×10−5moles

n = 0.033 M × 1 mL

n = 3.3×10−5moles

<h3>How does an ammonium carbonate aqueous solution look?</h3>

Ammonium carbonate aqueous solution is basic.

When ammonium carbonate, (NH4)2CO3, is dissolved in water, it totally dissociates to create ammonium cations, NH+4, and carbonate anions, CO23. Thus far, so good. Hydrolysis of the ammonium ion produces ammonia, NH3, and hydronium ions, H3O+.

Ammonium carbonate has the chemical formula (NH4)2CO3. According to the formula, one mole of ammonium carbonate contains two moles of ammonium ions, NH4+.

learn more about mole refer:

brainly.com/question/21911991

#SPJ4

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12. How many moles of atoms are present in 154 g of Na2O?
tia_tia [17]

Answer:

\boxed {\boxed {\sf A. \ 2.48 \ mol \ Na_2O}}

Explanation:

We are asked to convert an amount in grams to moles. To do this, we use the molar mass. This is the number of grams in one mole of a substance. It is the same value numerically as the atomic mass on the Periodic Table, however the units are grams per mole, not atomic mass units.

Look up the molar masses for the individual elements.

  • Sodium (Na): 22.9897693 g/mol
  • Oxygen (O): 15.999 g/mol

Look back at the formula: Na₂O. Notice there is a subscript of 2 after sodium. This means there are 2 atoms of sodium in every molecule, so we have to multiply sodium's molar mass by 2 before adding oxygen's.

  • Na₂O: 2(22.9897693 g/mol)+ 15.999 g/mol = 61.9785386 g/mol

Set up a ratio using the molar mass.

\frac {61.9785386 \ g \ Na_2O}{1 \ mol \ Na_2O}

Multiply by the given number of grams.

154 \ g \ Na_2O*\frac {61.9785386 \ g \ Na_2O}{1 \ mol \ Na_2O}

Flip the ratio so the grams of sodium oxide can cancel each other out.

154 \ g \ Na_2O*\frac {1  \ mol \ Na_2O}{61.9785386 \ g \ Na_2O}

154 *\frac {1  \ mol \ Na_2O}{61.9785386 }

\frac {154}{61.9785386 }  \ mol \ Na_2O

2.48473106141 \ mol \ Na_2O

The original measurement of grams given has 3 significant figures, so our answer must have the same. For the number we calculated, that is the hundredth place.

  • 2.48<u>4</u>73106141

The 4 in the thousandth place tells us to leave the 8.

2.48 \ mol \ Na_2O

There are <u>2.48 moles of sodium oxide</u> in 154 grams, so choice A is correct.

6 0
3 years ago
How many significant figures are in 0.00102?
zaharov [31]

Answer:

3

Explanation:

102 are significant figures

7 0
2 years ago
Read 2 more answers
Balance the equation: "Sodium oxide reacts with water to form sodium hydroxide and hydrogen." I can't find the correct coefficie
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Answer:

So, sodium oxide + water = sodium hydroxide + hydrogen is written Na2O+H2O-->NaOH+H.

To balance the equation it should read NaO+H2O-->2NaOH

Explanation:

Na2O+H2O-->2NaOH+0H2.......the 0H2 is dropped because there is no value

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