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IrinaVladis [17]
4 years ago
6

Si se adiciona etilenglicol (compuesto no volátil) a agua destilada:

Chemistry
1 answer:
poizon [28]4 years ago
6 0

Answer:

Hey hi

Explanation:

Can you pls tell me which language is this pls I really want to help you... Sorry

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How many moles of sodium carbonate contain 1.773 × 1017 carbon atoms?
vesna_86 [32]

Answer:

  • 2.941 × 10⁻¹⁷ mol

Explanation:

1) Chemical formula of sodium carbonate: <em>Na₂CO₃</em>

2) Ratio of carbon atoms:

  • The number of atoms of C in the unit formula Na₂CO₃ is the subscript for the atom, which is 1 (since it is not written).

Hence, the ratio is 1 C atom / 1 Na₂CO₃ unit formula.

This is, there is 1 atom of carbon per each unit formula of sodium carbonate.

3) Calculate the number of moles in 1.773 × 10⁷ carbon atoms

  • Divide by Avogadro's number: 6.022 × 10²³ atoms / mol

  • number C moles  = 1.773 × 10⁷ atoms / (6.022 × 10²³ atoms/mol)

  • number C moles = 2.941 × 10⁻¹⁷ mol

Since, the ratio is 1: 1, the number of moles of sodium carbonate is the same number of moles of carbon atoms.

5 0
3 years ago
Which macromolecules break apart by hydrolysis?
ankoles [38]

Answer:

Dehydration synthesis reactions build molecules up and generally require energy, while hydrolysis reactions break molecules down and generally release energy. Carbohydrates, proteins, and nucleic acids are built up and broken down via these types of reactions, although the monomers involved are different in each case.

Explanation:

4 0
3 years ago
Write the name for the following molecular compounds. Remember to use the correct prefix for each compound.
Westkost [7]

Answer:

Hey there!

CS2) Carbon Disulfide.

PBr3) Phosphorus Tribromide

NO) Nitric Oxide

CF4) Carbon Tetrafluoride

P2O5) Phosphorus Pentoxide

Let me know if this helps :)

3 0
3 years ago
To determine the concentration of X in an unknown solution, 1.00 mL of 8.48 mM S was added to 3.00 mL of the unknown X solution
kogti [31]

Answer:

positif

Explanation:

3.87169.843826 = y = x = .ion \: in \: cells = y = x. >  \\  \geqslant  {8}^{2}  \times \frac{4}{3}  | \geqslant |  \times \frac{68.1 < }{3 = 8}

6 0
3 years ago
A volume of 90.0 mLmL of aqueous potassium hydroxide (KOHKOH) was titrated against a standard solution of sulfuric acid (H2SO4H2
Alja [10]

Answer:

0.823 M was the molarity of the KOH solution.

Explanation:

H_2SO_4+KOH\rightarrow K_2SO_4+2H_2O (Neutralization reaction)

To calculate the concentration of base , we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is KOH.

We are given:

n_1=2\\M_1=1.50 M\\V_1=24.7 mL\\n_2=1\\M_2=?\\V_2=90.0 mL

Putting values in above equation, we get:

2\times \1.50 M\times 24.7 mL=1\times M_2\times 90.0 mL

M_2=\frac{2\times 1.50M\times 24.7 mL}{1\times 90.0 mL}=0.823 M

0.823 M was the molarity of the KOH solution.

7 0
3 years ago
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