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2 years ago

13

!!!PLEASE HELP!!!!

1 answer:

Lynna [10]2 years ago

4 0

Answer: B

Explanation: molarity = concentration c= n/V = 0.5 mol/ 0.05 l = 10 mol/l

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Thank you for all your help and support.<br><br><br><br> Love you all.

NeX [460]

Answer:

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

Explanation:

5 0

3 years ago

Read 2 more answers

Some commercially available algaecides for swimming pools claim to contain 7% copper. Could the method used in this experiment t

AleksandrR [38]

Answer:

Explanation:

7% copper implies 7 w/v%(weight/volume %) of copper. This implies a 100 mL algaecide arrangement contains 7 grams of copper.
Henceforth we have a thought of the measure of copper that ought to be available in a given example of algaecide.
Indeed, even a 1 mL algaecide test is sufficient to discover the percentage of copper in it.
so a 25mL sample of algaecide must have around 1.75g of copper.

4 0

3 years ago

Complete the following equations

hoa [83]

The answer is NaCl

-Hope this helps

5 0

2 years ago

CsBr formula name???

olganol [36]

Answer

PubChem CID/molecular formula

Explanation:

Cesium bromide

PubChem CID 24592

Molecular Formula CsBr or BrCs

Synonyms CESIUM BROMIDE 7787-69-1 Caesium bromide Cesiumbromide Cesium bromide (CsBr) More...

Molecular Weight 212.81 g/mol

Component Compounds CID 260 (Hydrogen bromide) CID 5354618 (Cesium)

have a good day /night

may i please have a branllist

4 0

2 years ago

⦁ Find the concentration of H+, OH-, PH and POH of 0.03 M of magnesium hydroxide which ionizes to the extent of only 1 /3 in aqu

lions [1.4K]

Answer:

$pH=12.3\\\\pOH=1.7\\$

$[H^+]=5x10^{-13}M$

$[OH^-]=0.02M$

Explanation:

Hello there!

In this case, according to the given ionization of magnesium hydroxide, it is possible for us to set up the following reaction:

$Mg(OH)_2(s)\rightleftharpoons Mg^{2+}(aq)+2OH^-(aq)$

Thus, since the ionization occurs at an extent of 1/3, we can set up the following relationship:

$\frac{1}{3} =\frac{x}{[Mg(OH)_2]}$

Thus, x for this problem is:

$x=\frac{[Mg(OH)_2]}{3}=\frac{0.03M}{3}\\\\x= 0.01M$

Now, according to an ICE table, we have that:

$[OH^-]=2x=2*0.01M=0.02M$

Therefore, we can calculate the H^+, pH and pOH now:

$[H^+]=\frac{1x10^{-14}}{0.02}=5x10^{-13}M$

$pH=-log(5x10^{-13})=12.3\\\\pOH=14-pH=14-12.3=1.7$

Best regards!

4 0

3 years ago