What has a neck but no head

Помогите срочно сделать по химии промежуточную отестацию за 8 класс

il63 [147K]

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

6 0

3 years ago

Come up with a three-step plan for your household to reduce its effect on water pollution. Your plan should include a way to edu

Marina86 [1]

Answer:

1 tell your family difrent ways to safe water

2 dont leave the water running when you brush your teeth

3 dont dump cups of water out for no reason

Explanation:

3 0

3 years ago

Read 2 more answers

What is the mass of 1.45 moles of silver sulfate?

tatuchka [14]

Answer:

449.5 g

Explanation:

Silver sulfate- Ag2SO4

M(Ag)=107 g/mol => M(Ag2)=214 g/mol

M(S)=32 g/mol

M(O)=16 g/mol => M(O4)=64 g/mol

M(Ag2SO4)=310 g/mol

n=1.45 mol

m(Ag2SO4)=M(Ag2SO4)*n=310 g/mol *1.45 mol= 449.5 g

3 0

3 years ago

Read 2 more answers

Which of the following are considered pure substance??

Shkiper50 [21]

Answer:

Explanation:

I’m 99.9% sure that it’s an element because it can’t be broken down any more than it already is.

7 0

4 years ago

What is the pOH of a 0.150 M solution of potassium nitrite? (Ka HNO2 = 4.5 x 10−4 )

yanalaym [24]

Answer:

11.9 is the pOH of a 0.150 M solution of potassium nitrite.

Explanation:

Solution : Given,

Concentration (c) = 0.150 M

Acid dissociation constant = $k_a=4.5\times 10^{-4}$

The equilibrium reaction for dissociation of $HNO_2$ (weak acid) is,

$HNO_2+H_2O\rightleftharpoons NO_2^-+H_3O^+$

initially conc. c 0 0

At eqm. $c(1-\alpha)$ $c\alpha$ $c\alpha$

First we have to calculate the concentration of value of dissociation constant $(\alpha )$ .

Formula used :

$k_a=\frac{(c\alpha)(c\alpha)}{c(1-\alpha)}$

Now put all the given values in this formula ,we get the value of dissociation constant $(\alpha )$ .

$4.5\times 10^{-4}=\frac{(0.150\alpha)(0.150\alpha)}{0.150(1-\alpha)}$

$4.5\times 10^{-4} - 4.5\times 10^{-4}\alpha =0.150\alpha ^2$

$0.150\alpha ^2+4.5\times 10^{-4}\alpha-4.5\times 10^{-4}=0$

By solving the terms, we get

$\alpha=0.0533$

No we have to calculate the concentration of hydronium ion or hydrogen ion.

$[H^+]=c\alpha=0.150\times 0.0533=0.007995 M$

Now we have to calculate the pH.

$pH=-\log [H^+]$

$pH=-\log (0.007995 M)$

$pH=2.097\approx 2.1$

pH + pOH = 14

pOH =14 -2.1 = 11.9

Therefore, the pOH of the solution is 11.9

4 0

4 years ago