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OLga [1]
3 years ago
11

Fireworks contain metals that are brightly colored when burned. They are less reactive than some other metals. Which metal is on

e example?
A. Ne
B. Li
C. Mg
D. K
Chemistry
2 answers:
Varvara68 [4.7K]3 years ago
7 0
The correct answer is C. Mg (Magnesium)
lord [1]3 years ago
3 0
<h2>Answer</h2><h3>C. Mg</h3><h2>Explanation:</h2>

When magnesium is in the metallic form it gets burn very easily in air. To start the reaction, the magnesium metal requires a source of energy. The flame provides heat so that the magnesium can overcome their activation energy. Activation energy is the minimum amount of energy needed in the system for a chemical reaction to proceed. Oxygen found in the air forms Magnesium Oxide when the magnesium metal burns in the air. A compound is in which atoms of different elements are linked or bonded to one another. Oxygen and magnesium react to form this compound and forms a white powder of the magnesium oxide that brightens it up.

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When a diprotic acid is titrated with a strong base, and the Ka1 and Ka2 are significantly different, then the pH vs. volume plo
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Complete question is;

When a diprotic acid is titrated with a strong base, and the Ka1 and Ka2 are significantly different, then the pH vs. volume plot of the titration will have

a. a pH of 7 at the equivalence point.

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c. no equivalence point.

d. one equivalence point.

e. two distinct equivalence points

Answer:

Option E - Two Distinct Equivalence points

Explanation:

I've attached a sample diprotic acid titration curve.

In diprotic acids, the titration curves assists us to calculate the Ka1 and Ka2 of the acid. Thus, the pH at the half - first equivalence point in the titration will be equal to the pKa1 of the acid while the pH at the half - second equivalence point in a titration is equal to the pKa2 of the acid.

Thus, it is clear that there are two distinct equivalence points.

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Consider the following equation: SiO2 (s) + 3C (graphite) --&gt; SiC (s) + 2CO (g) ΔH rxn = 624.6 kJ / mol rxn. Using the follow
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<u>Answer:</u> The enthalpy of the formation of SiC(s) is coming out to be -65.3 kJ/mol

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as \Delta H^o

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]

For the given chemical reaction:

SiO_2(s)+3C\text{ (graphite)}(s)\rightarrow SiC(s)+2CO(g)

The equation for the enthalpy change of the above reaction is:

\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(SiC(s))})+(2\times \Delta H^o_f_{(CO(g))})]-[(1\times \Delta H^o_f_{(SiO_2(s))})+(3\times \Delta H^o_f_{(C(s))})]

We are given:

\Delta H^o_f_{(CO(g))}=-110.5kJ/mol\\\Delta H^o_f_{(SiO_2(s))}=-910.9kJ/mol\\\Delta H^o_f_{(C(s))}=0kJ/mol\\\Delta H^o_{rxn}=624.6kJ

Putting values in above equation, we get:

624.6=[(1\times \Delta H^o_f_{(SiC(s))})+(2\times (-110.5))]-[(1\times (-910.9))+(3\times (0))]\\\\\Delta H^o_f_{(SiC(s))}=-65.3kJ/mol

Hence, the enthalpy of the formation of SiC(s) is coming out to be -65.3 kJ/mol.

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3 years ago
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