All categories

3 years ago

Thank you for all your help and support. Love you all.

Chemistry

2 answers:

iragen [17]3 years ago

8 0

Hello! Here is your answer for this question! ☺

Answer: C. 4MgCI2 --> Mg4+4CI2

Hope this helps you out! ☺

NeX [460]3 years ago

5 0

Answer:

the answer would be C. Thank you god bless you.

Explanation:

You might be interested in

A compound distributes between benzene (solvent 1) and water (solvent 2) with a distribution coefficient, K = 2.7. If 1.0g of th

mote1985 [20]

Explanation:

The given data is as follows.

Solvent 1 = benzene, Solvent 2 = water

$K_{p}$ = 2.7, $V_{S_{2}}$ = 100 mL

$V_{S_{1}}$ = 10 mL, weight of compound = 1 g

Extract = 3

Therefore, calculate the fraction remaining as follows.

$f_{n} = [1 + K_{p}(\frac{V_{S_{2}}}{V_{S_{1}}})]^{-n}$

= $[1 + 2.7(\frac{100}{10})]^{-3}$

= $(28)^{-3}$

= $4.55 \times 10^{-5}$

Hence, weight of compound to be extracted = weight of compound - fraction remaining

= 1 - $4.55 \times 10^{-5}$

= 0.00001

or, = $1 \times 10^{-5}$

Thus, we can conclude that weight of compound that could be extracted is $1 \times 10^{-5}$ .

7 0

3 years ago

Read 2 more answers

Write the net ionic equation for the reaction that occurs when a solution of hydrochloric acid (hcl) is mixed with a solution of

kap26 [50]

Ionic Equation:
H⁺(aq) + Cl⁻(aq) + Na⁺(aq) + CHO₂⁻(aq) → HCHO₂(aq) + Na⁺(aq) + Cl⁻(aq)

Net ionic equation:
H⁺(aq) + CHO₂⁻(aq) → HCHO₂(aq)

5 0

3 years ago

Read 2 more answers

Elements that are usually dull, brittle and do not conduct electricity are know as______.

Anton [14]

Answer:

nonmetal

Explanation:

.........

...

8 0

2 years ago

How do synthesis reactions and a decomposition reactions differ

Minchanka [31]

A synthesis would be A+B-->AB its combining, decomposition is AB-->A+B, its the breaking up of a substance.

8 0

3 years ago

In acidic aqueous solution, the purple complex ion Co(NH3)5Br2+ undergoes a slow reaction in which the bromide ion is replaced b

kirill115 [55]

Answer:

A) 0.065 M is its molarity after a reaction time of 19.0 hour.

B) In 52 hours $[Co(NH_3)5Br]^{2+}$ will react 69% of its initial concentration.

Explanation:

$Co(NH_3)_5(H_2O)_3+[Co(NH_3)5Br]^{2+}(Purple)(aq)+H_2O(l)\rightarrow [Co(NH_3)_5(H_2O)]^{3+}(Pinkish-orange)(aq)+Br^-(aq)$

The reaction is first order in $[Co(NH_3)5Br]^{2+}$ :

Initial concentration of $[Co(NH_3)5Br]^{2+}$ = $[A_o]=0.100 M$

a) Final concentration of $[Co(NH_3)5Br]^{2+}$ after 19.0 hours= $[A]$

t = 19.0 hour = 19.0 × 3600 seconds ( 1 hour = 3600 seconds)

Rate constant of the reaction = k = $6.3\times 10^{-6} s^{-1}$

The integrated law of first order kinetic is given as:

$[A]=[A_o]\times e^{-kt}$

$[A]=0.100 M\times e^{-6.3\times 10^{-6} s^{-1}\times 19.0\times 3600 s}$

$[A]=0.065 M$

0.065 M is its molarity after a reaction time of 19.0 h.

Initial concentration of $[Co(NH_3)5Br]^{2+}$ = $[A_o]=x$

Final concentration of $[Co(NH_3)5Br]^{2+}$ after t = $[A]=(100\%-69\%) x=31\%x=0.31x$

Rate constant of the reaction = k = $6.3\times 10^{-6} s^{-1}$

The integrated law of first order kinetic is given as:

$[A]=[A_o]\times e^{-kt}$

$0.31x=x\times e^{-6.3\times 10^{-6} s^{-1}\t}$

t = 185,902.06 s = $\frac{185,902.06 }{3600} hour = 51.64 hours$ ≈ 52 hours

In 52 hours $[Co(NH_3)5Br]^{2+}$ will react 69% of its initial concentration.

6 0

3 years ago