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Kisachek [45]
3 years ago
5

lead can react with oxygen gas. If lead (IV) oxide is the product of the reaction, how would the reaction be classified

Chemistry
1 answer:
slamgirl [31]3 years ago
7 0

Answer:

this reaction is an oxidation reaction

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Solution X is a strong base with a pH of 12. Solution X is mixed with solution Y,and the pH of the resulting mixture is 8.Based
myrzilka [38]

The strong Base with a pH of 12 is reduced by 4 units upon being added with solution Y. If you added a strong acid to the strong base, all ions are present in the solution, yes? So every OH- is neutralised by every H+ for example, meaning the resultant pH should be 7. The resultant pH is only 8 however, so solution Y must be a <em>weak acid </em>only!

5 0
3 years ago
Which of the Atoms shown has an atomic number four
dimulka [17.4K]

Answer:

B

Explanation:

Atomic # = Protons

it says 4 p in the inside of the orbital

4 0
3 years ago
If 20 grams of Zinc phosphate reacts with excess hydrochloric acid and produces 18 grams of Zinc chloride what is the percent yi
fenix001 [56]

Answer:

Y=85\%

Explanation:

Hello!

In this case, since we know the balanced chemical reaction, we are first able to realize there is a 1:3 mole ratio between zinc phosphate and zinc chloride; it means that we can first compute the moles of the desired product via stoichiometry:

n_{ZnCl_2}=20gZn_3(PO_4)_2*\frac{1molZn_3(PO_4)_2}{386.11gZn_3(PO_4)_2}*\frac{3molZnCl_2}{1molZn_3(PO_4)_2}=0.16gZnCl_2

Next, since those moles are associated with the theoretical yield of zinc chloride, we obtain the corresponding mass:

m_{ZnCl_2}^{theoretical}=0.16molZnCl_2*\frac{136.29gZnCl_2}{1molZnCl_2} =21gZnCl_2

Finally, we compute the percent yield by diving the actual yield (18 g) by the theoretical yield:

Y=\frac{18g}{21g}*100\%\\\\Y=85\%

Best regards!

4 0
2 years ago
What is the only nonmetal in group 14?
Fantom [35]

Answer:

The only nonmetal in group 14 is carbon.

Explanation:

7 0
3 years ago
Read 2 more answers
Predict the missing product of this equation<br><br><br>1 MgF2 + 1 Li2CO3 -&gt; 1 ______ +2LiF
ch4aika [34]

Answer:

MgCO₃

Explanation:

From the question given above, we obtained:

MgF₂ + Li₂CO₃ —> __ + 2LiF

The missing part of the equation can be obtained by writing the ionic equation for the reaction between MgF₂ and Li₂CO₃. This is illustrated below:

MgF₂ (aq) —> Mg²⁺ + 2F¯

Li₂CO₃ (aq) —> 2Li⁺ + CO₃²¯

MgF₂ + Li₂CO₃ —>

Mg²⁺ + 2F¯ + 2Li⁺ + CO₃²¯ —> Mg²⁺CO₃²¯ + 2Li⁺F¯

MgF₂ + Li₂CO₃ —> MgCO₃ + 2LiF

Now, we share compare the above equation with the one given in the question above to obtain the missing part. This is illustrated below:

MgF₂ + Li₂CO₃ —> __ + 2LiF

MgF₂ + Li₂CO₃ —> MgCO₃ + 2LiF

Therefore, the missing part of the equation is MgCO₃

8 0
2 years ago
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