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3 years ago

What do you think is the effect of acid rain on rusting ?

1 answer:

8 0

As acid rain is a mixture of pollutants and gasses. As rusting is a chemical reaction between oxygen and other gasses in the air alongside a material and a liquid. I would say they are both similar and that acid rain would speed the process of rusting/aid the process.

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What is a solution for deforestation (full sentences please)

IrinaVladis [17]

The best solution to deforestation is to curb the felling of trees by enforcing a series of rules and laws to govern it. Deforestation in the current scenario may have reduced; however, it would be too early to assume. The money-churner nature of forest resources can be tempting enough for deforestation to continue.

4 0

2 years ago

Read 2 more answers

A solution is prepared by dissolving 10.0 g of NaBr and 10.0 g of Na2SO4 in water to make a 100.0 mL solution. This solution is

Colt1911 [192]

Answer:

$M_{Na^+}=1.36M$

$M_{Br^-}=1.58M$

Explanation:

Hello,

At first, it turns out convenient to compute the total moles of sodium that will be dissolved into the solution by considering the added amounts of sodium bromide and sodium sulfate:

$n_{Na^+}=n_{Na^+,NaBr}+n_{Na^+,Na_2SO_4}\\n_{Na^+,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molNa^+}{1molNaBr}=0.0971molNa^+\\n_{Na^+,Na_2SO_4}=10.0gNa_2SO_4*\frac{1molNa_2SO_4}{142gNa_2SO_4}*\frac{2molNa^+}{1molNa_2SO_4} =0.141molNa^+\\n_{Na^+}=0.0971molNa^++0.141molNa^+\\n_{Na^+}=0.238molNa^+$

Once we've got the moles we compute the final volume via:

$V=100.0mL+75.0mL=175.0mL*\frac{1L}{1000mL}=0.1750L$

Thus, the molarity of the sodium atoms turn out into:

$M_{Na^+}=\frac{0.238mol}{0.1750L} =1.36M$

Now, we perform the same procedure but now for the bromide ions:

$n_{Br^-}=n_{Br^-,NaBr}+n_{Br^-,AlBr_3}\\n_{Br^-,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molBr^-}{1molNaBr}=0.0971molBr^-\\n_{Br^-,AlBr_3}=0.0750L*0.800\frac{molAlBr_3}{L} *\frac{3molBr^-}{1molAlBr_3}=0.180molBr^- \\n_{Br^-}=0.0971molBr^-+0.180molBr^-\\n_{Br^-}=0.277molBr^-$

Finally, its molarity results:

$M_{Br^-}=\frac{0.277molBr^-}{0.1750L}=1.58M$

Best regards.

7 0

2 years ago

Express the quantity 556.2 x 10^-12 in each units A.) ms B.) ns C.) ps D.) fs

zavuch27 [327]

The answer:
we should know the meaning of each abbreviation:
ms means millisecond, its value is 10^-3 s
ns means means nanosecond, its value is 10^-9 s
ps means picosecond, its value is 10^-12 s
fs means femtosecond, its value is 1x 10^15 s

Expressions of the quantity 556.2 x 10^-12 are
556.2 x 10^-12 =556.2 ps
556.2 x 10^-12 =556.2 x 10^-9 x 10^-3= 556.2 x 10^-9 ms
556.2 x 10^-12 = 556.2 x 10^-3 x 10^-9 = 556.2 x 10^-3 ns
556.2 x 10^-12 = 556.2 x 10^- 27 x 10^15 = 556.2 x 10^- 27 fs

6 0

3 years ago

Which of the following elements is represented by the above graphic

kolezko [41]

Answer:

Sulfur

Explanation:

Sulfur has 16 valence electrons, as shown in the diagram.

5 0

2 years ago

Estimate the ph of the resulting solution prepared by mixing 1.0 mole of solid disodium phosphate (na2hpo4) and 1.25 mole of hyd

GenaCL600 [577]

The HCl added = 1.25 moles

and the moles of Na2HPO4 = 1 mole

Now when acid is added in the given solution of Na2HPO4

One mole of H+ will react with one mole of Na2HPO4 to given one mole of NaH2PO4

Na2HPO4 + H+ ---> NaH2PO4

Now this one mole formed NaH2PO4 will further react with 0.25 moles of H+ left to form 0.25 moles of H3PO4 and 0.75 moles of NaH2PO4 will remain in the solution

So this will result into formation of a buffer of phosphoric acid and NaH2PO4

NaH2PO4 + H+ ---> H3PO4

pKa of H3PO4 = 2.1

so pH = pKa + log [salt] / [acid] = 2.1 + log [0.75 / 0.25] = 2.58

so the pH will be in between 2.1 to 7.2

7 0

3 years ago