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wolverine [178]
4 years ago
13

50 points - PLEASE ANSWER ENTIRELY - WILL REWARD BRAINLIEST

Chemistry
1 answer:
miss Akunina [59]4 years ago
5 0

If a compound is made from a metal and a non-metal, its bonding will be ionic.

If a compound is made from two non-metals, its bonding will be covalent.

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Since the configuration contains the number 2s and 2p, you can assume that it is in period 2 of the periodic table.  By noble gas shorthand, you start with helium, which is atomic number 2.  2s²2p^4 has 2+4 = 6 valence

Add 2+6 = 8, which is Oxygen.
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What kind of data would you need to collect to carry out this experiment?
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Did entropy increase or decrease?
seropon [69]

Answer:

1) increase

2) increase

Explanation:

Entropy is the degree of disorderliness or randomness of a system. It is the measure of the unavailable energy in a system.

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8 0
3 years ago
Kc for the reaction of hydrogen and iodine to produce hydrogen iodide, H2(g) I2(g) ⇌ 2HI(g) is 54.3 at 430°C. Determine the init
Jlenok [28]

Answer : The initial concentration of HI and concentration of HI at equilibrium is, 0.27 M and 0.386 M  respectively.

Solution :  Given,

Initial concentration of H_2 and I_2 = 0.11 M

Concentration of H_2 and I_2 at equilibrium = 0.052 M

Let the initial concentration of HI be, C

The given equilibrium reaction is,

                          H_2(g)+I_2(g)\rightleftharpoons 2HI(g)

Initially               0.11   0.11            C

At equilibrium  (0.11-x) (0.11-x)   (C+2x)

As we are given that:

Concentration of H_2 and I_2 at equilibrium = 0.052 M  = (0.11-x)

0.11 - x = 0.052

x = 0.11 - 0.052

x = 0.058 M

The expression of K_c will be,

K_c=\frac{[HI]^2}{[H_2][I_2]}

54.3=\frac{(C+2(0.058))^2}{(0.052)\times (0.052)}

By solving the terms, we get:

C = 0.27 M

Thus, initial concentration of HI = C = 0.27 M

Thus, the concentration of HI at equilibrium = (C+2x) = 0.27 + 2(0.058) = 0.386 M

3 0
3 years ago
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