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3 years ago

What is in a banana? give me specifics!!!?

Chemistry

1 answer:

IRISSAK [1]3 years ago

5 0

Nutrients in a banana

Vitamin B6 – .5 mg
Manganese – .3 mg
Vitamin C – 9 mg 
Potassium – 450 mg Dietary Fiber – 3g
Protein – 1 g
Magnesium – 34 mg
Folate – 25.0 mcg Riboflavin – .1 mg
Niacin – .8 mg
Vitamin A – 81
IU Iron – .3 mg

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3 years ago

Read 2 more answers

A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the

baherus [9]

Answer: 1. $9.08\times 10^{-6}$ moles

2. 90 mg

Explanation:

$O_3(g)+2NaI(aq)+H_2O(l) \rightarrow O_2(g)+I_2(s)+2NaOH(aq)$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus $4.54 \times 10^{-6}$ moles of ozone is removed by = $\frac{2}{1}\times 4.54 \times 10^{-6}=9.08\times 10^{-6}$ moles of sodium iodide.

Thus $9.08\times 10^{-6}$ moles of sodium iodide are needed to remove $4.54\times 10^{-6}$ moles of $O_3$

2. $\text{Number of moles of ozone}=\frac{0.01331g}{48g/mol}=0.0003moles$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 0.0003 moles of ozone is removed by = $\frac{2}{1}\times 0.0003=0.0006$ moles of sodium iodide.

Mass of sodium iodide= $moles\times {\text {molar mass}}=0.0006\times 150g/mol=0.09g=90mg$ (1g=1000mg)

Thus 90 mg of sodium iodide are needed to remove 13.31 mg of $O_3$ .

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3 years ago

If you have 547.3 grams of Ni2O, how many molecules would be present?

jasenka [17]

Answer: 8.830418848725065

Explanation:

8.830418848725065

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3 years ago