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TiliK225 [7]
3 years ago
6

The compound nicl2 is an ionic compound. what are the ions of which it is composed?

Chemistry
1 answer:
densk [106]3 years ago
3 0
Nickel and chlorine
Ni+
Cl-
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Which of the
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During chemical reactions, matter is neither

created nor destroyed; it just changes form.

Explanation:

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Why is burning wood a chemical change?
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3 years ago
The density of water at 30.0 °C is 0.9956 g/mL. If the specific gravity of acetic acid is 1.040 at 30.0 °C, what is the density
mash [69]

Answer:

The density of acetic acid at 30°C = 1.0354_g/mL

Explanation:

specific gravity of acetic acid = (Density of acetic acid at 30°C) ÷ (Density of water at 30°C)

Therefore, the density of acetic acid at 30°C = (Density of water at 30°C) × (Specific gravity of acetic acid at 30°C)

= 0.9956 g/mL × 1.040

= 1.0354_g/mL

Specific gravity, which is also known as relative density, is the ratio of the density of a substance to the density of a specified standard substance.

Generally the standard substance of to which other solid and liquid substances are compared is water which has a density of 1.0 kg per litre or 62.4 pounds/cubic foot at 4 °C (39.2 °F) while gases are normally compared with dry air, with a density of 1.29 grams/litre or 1.29 ounces/cubic foot under standard conditions of a temperature of 0 °C and one standard atmospheric pressure

7 0
3 years ago
The enzyme, phosphoglucomutase, catalyzes the interconversion
Fittoniya [83]

Answer:

K_{eq = 19

ΔG° of the reaction forming glucose 6-phosphate =  -7295.06 J

ΔG° of the reaction  under cellular conditions = 10817.46 J

Explanation:

Glucose 1-phosphate     ⇄     Glucose 6-phosphate

Given that: at equilibrium, 95% glucose 6-phospate is  present, that implies that we 5% for glucose 1-phosphate

So, the equilibrium constant K_{eq can be calculated as:

K_{eq = \frac{[glucose-6-phosphate]}{[glucose-1-[phosphate]}

K_{eq= \frac{0.95}{0.05}

K_{eq = 19

The formula for calculating ΔG° is shown below as:

ΔG° = - RTinK

ΔG° = - (8.314 Jmol⁻¹ k⁻¹ × 298 k ×  1n(19))

ΔG° = 7295.05957 J

ΔG°≅ - 7295.06 J

b)

Given that; the concentration  for  glucose 1-phosphate = 1.090 x 10⁻² M

the concentration of glucose 6-phosphate is 1.395 x 10⁻⁴ M

Equilibrium constant  K_{eq can be calculated as:

K_{eq = \frac{[glucose-6-phosphate]}{[glucose-1-[phosphate]}

K_{eq}= \frac{1.395*10^{-4}}{1.090*10^{-2}}

K_{eq} = 0.01279816514  M

K_{eq} = 0.0127 M

ΔG° = - RTinK

ΔG° = -(8.314*298*In(0.0127)

ΔG° = 10817.45913 J

ΔG° = 10817.46 J

5 0
2 years ago
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