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saveliy_v [14]
3 years ago
10

Fireworks, KNO3 Give only the names of the elements alphabetically, separating them with commas.

Chemistry
2 answers:
Alborosie3 years ago
4 0
Answer : The  question is to write the elements alphabetically and seperating them with commas.

So in this case for your question of Fireworks , KNO_{3},

The answer will be K, N, O.

You have to just differentiate the elements and write them according to the alphabetical order.
Ksenya-84 [330]3 years ago
4 0

<u>Answer:</u> The elements present are potassium, nitrogen and oxygen.

<u>Explanation:</u>

Elements are defined as the substances in which same type of atoms are present. <u>For Example:</u> Sodium, Chlorine etc.

The chemical formula of the compound used in fireworks is KNO_3

The chemical name for the compound is potassium nitrate.

This compound is formed by the combination of three elements. In 1 mole of potassium nitrate, 1 mole of potassium atom, 1 mole of nitrogen atom and 3 moles of oxygen atoms are present.

Hence, the elements present are potassium, nitrogen and oxygen.

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What this means is that the two have the same number of protons (9), but Fluoride has 10 electrons compared to Fluorine's 9.

So the answers are:
Protons - 9
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An asteroid hit the Yucatán Peninsula about 65 million years ago,
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Answer:

C. It blocked out sunlight

Explanation:

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7 0
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8 0
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The estimated heat of vaporization of diethyl ether using the Chen's rule is A. 29.7 KJ/mol B. 33.5 KJ/mol C. 26.4 KJ/mol D. 36.
Brums [2.3K]

Answer:

C. 26.4 kJ/mol

Explanation:

The Chen's rule for the calculation of heat of vaporization is shown below:

\Delta H_v=RT_b\left [ \frac{3.974\left ( \frac{T_b}{T_c} \right )-3.958+1.555lnP_c}{1.07-\left ( \frac{T_b}{T_c} \right )} \right ]

Where,

\Delta H_v is the Heat of vaoprization (J/mol)

T_b is the normal boiling point of the gas (K)

T_c is the Critical temperature of the gas (K)

P_c is the Critical pressure of the gas (bar)

R is the gas constant (8.314 J/Kmol)

For diethyl ether:

T_b=307.4\ K

T_c=466.7\ K

P_c=36.4\ bar

Applying the above equation to find heat of vaporization as:

\Delta H_v=8.314\times307.4 \left [ \frac{3.974\left ( \frac{307.4}{466.7} \right )-3.958+1.555ln36.4}{1.07-\left ( \frac{307.4}{466.7} \right )} \right ]

\Delta H_v=26400 J/mol

The conversion of J into kJ is shown below:

1 J = 10⁻³ kJ

Thus,

\Delta H_v=26.4 kJ/mol

<u>Option C is correct</u>

6 0
4 years ago
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