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vazorg [7]
3 years ago
8

This is Q7 and I need help, can someone help.

Chemistry
1 answer:
andre [41]3 years ago
5 0

Answer:

Answer to A. helium, neon, argon, krypton, xenon, and radon, B. Elemental hydrogen (H, element 1), nitrogen (N, element 7), oxygen (O, element 8), fluorine (F, element 9), and chlorine (Cl, element 17) are all gases at room temperature, and are found as diatomic molecules (H2, N2, O2, F2, Cl2). C. Elements Compounds

Ar (argon) HBr (hydrogen bromide) C 3H 8 (propane)

Kr (krypton) HI (hydrogen iodide) C 4H 10 (butane)

Xe (xenon) HCN (hydrogen cyanide)* CO (carbon monoxide)

Rn (radon) H 2S (hydrogen sulfide) CO 2 (carbon dioxide)

Explanation:

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Question 22 3 p Use Bowen's Reaction Series to describe how different igneous rocks form from a single body of magma. Which spec
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The Bowen's reaction series describes how minerals form in sequential order, forming at higher temperature to a lower temperature. There are two branches of crystallization, one is the continuous branch that is on the right and the other is the discontinuous branch that is on the left.

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3 years ago
When 0.5141 g of biphenyl (C12H10) undergoes combustion in a bomb calorimeter, the temperature rises from 25.823 °C to 29.419 °C
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Answer:

\Delta_{r}U of the reaction is -6313 kJ/mol

\Delta_{r}H of the reaction is -6312 kJ/mol

Explanation:

Temperature\,\,change= \Delta U = 29.419-25.823 =3.506^{o}C

q_{cal}= C \times \Delta T

=5.861 \times 3.596 = 21.076\,kJ

q_{rxn}= -q_{cal}= -21.076\,kJ

\Delta_{r}U= -21.076 \times \frac{154}{0.5141}= -6313\, kJ/mol

Therefore, \Delta_{r}U of the reaction is -6313 kJ/mol.

The chemical reaction in bomb calorimeter  is as follows.

C_{12}H_{10}(s)+\frac{27}{2}O_{2}(g)\rightarrow 12CO_{2}(g)+5H_{2}O(g)

Number\,of\,moles\Delta n=(12+5)-\frac{27}{2}=3.5

\Delta_{r}H=\Delta E+ \Delta n. RT

=-6313+3.5\times 8.314\times 10^{-3} \times 3.596=-6312\,kJ/mol

Therefore, \Delta_{r}H of the reaction is -6312 kJ/mol.

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