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Katarina [22]
3 years ago
5

What are some essential things someone must do to make sure he or she is in good health?

Chemistry
2 answers:
Rainbow [258]3 years ago
8 0
It is very important to eat healthy, and exercise. A healthy diet has many health benefits. Exercising can help prevent heart diseases, blood pressure, diabetes, osteoporosis, and depression.

Hope this helps! You can do more research online!
Sedaia [141]3 years ago
3 0
Keep fit, lots of sleep and healthy diet.
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Calculate the percent composition of Ca3P2
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Calculate the molar mass of Ca3P2 in grams per mole or search for a chemical formula or substance.
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3 years ago
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Can you take picture of the whole question? I can’t see.
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Write down three use of chlorine​
melomori [17]

Answer:

Chlorine is commonly used as an antiseptic and is used to make drinking water safe and to treat swimming pools. Large amounts of chlorine are used in many industrial processes, such as in the production of paper products, plastics, dyes, textiles, medicines, antiseptics, insecticides, solvents and paints.

Explanation:

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4 0
3 years ago
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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
3 years ago
if 0.3 moles of carbon reacts completely with 0.3 moles of molecular oxygen, what is the yield in Percent of 6.6 g of CO2 are fo
Anettt [7]
Answer is:<span>the yield is 50%.
</span>
Chemical reaction: C + O₂ → CO₂.
n(C) = 0.3 mol; amount of substance.
n(O₂) = 0.3 mol.
From chemical reaction: n(C) : n(CO₂) = 1 : 1.
n(CO₂) = 0.3 mol.
M(CO₂) = 44 g/mol; molar mass of caron(IV) oxide.
m(CO₂) = n(CO₂) · M(CO₂).
m(CO₂) =0.3 mol · 44 g/mol.
m(CO₂) = 13.2 g; mass of carbon(IV) oxide.
the yield = 6.6 g ÷ 13.2 g · 100%.
the yield = 50%.
8 0
3 years ago
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