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3 years ago

Помогите срочно сделать по химии промежуточную отестацию за 8 класс

Chemistry

1 answer:

il63 [147K]3 years ago

6 0

Answer:

This is site for English speakers. Этот сайт на английском, поэтому вопрос могут удалить

Explanation:

1. 2)

2. 3)

3. 4) Sr

4. 3)

5. 4)

6. 2)

7. 1)

8. 4)

9. 3)

10. 3)

11. SO3, H2SO4, Na2SO4

12.

A) оксид меди (II) 2) CuO

Б) хлорид меди(II) 4) CuCl2

В) сульфит меди (II) 3) CuSO 3

Г) гидроксид меди (II) 1) Cu(OH)2

13.

1. Fe+HCl= б) FeCl 2 +H 2

2.Fe+O2= в) Fe 3 O 4

3. Fe(OH) 3 = г)Fe 2 O 3 +H 2O

4. FeCl 2 +NaOH= а) Fe(OH) 2 +NaCl

14. 2Ca + O2 = 2CaO

CaO + H2O = Ca(OH)2

Ca(OH)2 + 2HCl = CaCl2 + 2H2O

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The periodic table of elements is organized by the number of what?

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3 years ago

33.56 g of fructose (C6H,206) and 18.88 g of water are mixed to obtain a 40.00 ml solution a. What is this solution's density? b

Darina [25.2K]

Explanation:

Mass of fructose = 33.56 g

Mass of water = 18.88 g

Total mass of the solution = Mass of fructose + Mass of water = M

M = 33.56 g + 18.88 g =52.44 g

Volume of the solution = V = 40.00 mL

Density = $\frac{Mass}{Volume}$

a) Density of the solution:

$\frac{M}{V}=\frac{52.44 g}{40.00 mL}=1.311 g/mL$

b) Molar mass of fructose = 180.16 g/mol

Moles of fructose = $n_1=\frac{ 33.56 g}{180.16 g/mol}=0.1863 mol$

Molar mass of water = 18.02 g/mol

Moles of water= $n_2=\frac{ 18.88 g}{18.02 g/mol}=1.0477 mol$

Mole fraction of fructose in this solution: $\chi_1$

$\chi_1=\frac{n_1}{n_1+n_2}=\frac{0.1863 mol}{0.1863 mol+1.0477 mol}$

$\chi_1=0.1510$

Mole fraction of water = $\chi_2=1-\chi_1=0.8490$

c) Average molar mass of of the solution:

= $\chi_1\times 180.16 g/mol+\chi_2\times 18.02 g/mol$

$=0.1510\times 180.16 g/mol+0.8490\times 18.02 g/mol=42.50 g/mol$

d) Mass of 1 mole of solution = 42.50 g/mol

Density of the solution = 1.311 g/mL

d) Specific molar volume of the solution:

$\frac{\text{Average molar mass}}{\text{Density of the mass}}$

$=\frac{42.50 g/mol}{1.311 g/mL}=32.42 mL/mol$

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Answer:

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