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3241004551 [841]
3 years ago
6

What kind of plant growth hormone to regular plant growth. We dissolve 5.0 mg of the solution into 2.0 kg of water. What is the

dosage of the solution in ppm? _ppm
Chemistry
1 answer:
In-s [12.5K]3 years ago
3 0

Answer:

The dosage of the solution is 2.5 parts per million.

Explanation:

The solvent proportion in parts per million is equal to the ratio of solution in miligrams to ratio of water in miligrams multiplied by a million. (A kilogram is equivalent to a million miligrams) That is:

p.p.m. = \frac{5\,mg}{2000000\,mg}\times 1000000

p.p.m. = 2.5\, ppm

The dosage of the solution is 2.5 parts per million.

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How many grams of a stock solution that is 92.5 percent H2SO4 by mass would be needed to make 250 grams of a 35.0 percent by mas
IrinaVladis [17]

94.6 g.  You must use 94.6 g of 92.5 % H_2SO_4 to make 250 g of 35.0 % H_2SO_4.

We can use a version of the <em>dilution formula</em>

<em>m</em>_1<em>C</em>_1 = <em>m</em>_2<em>C</em>_2

where

<em>m</em> represents the mass and

<em>C</em> represents the percent concentrations

We can rearrange the formula to get

<em>m</em>_2= <em>m</em>_1 × (<em>C</em>_1/<em>C</em>_2)

<em>m</em>_1 = 250 g; <em>C</em>_1 = 35.0 %

<em>m</em>_2 = ?; _____<em>C</em>_2 = 92.5 %

∴ <em>m</em>_2 = 250 g × (35.0 %/92.5 %) = 94.6 g

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List four energy resources​
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Read 2 more answers
A 13 g sample of P4010 contains how many
Alex777 [14]

Answer:

\large \boxed{5.5 \times 10^{22}\text{ molecules of P$_{2}$O}_{5}}

Explanation:

You must calculate the moles of P₄O₁₀, convert to moles of P₂O₅,  then convert to molecules of P₂O₅.

1. Moles of P₄O₁₀

\text{Moles of P$_{4}$O}_{10} = \text{13 g P$_{4}$O}_{10} \times \dfrac{\text{1 mol P$_{4}$O}_{10}}{\text{283.89 g P$_{4}$O}_{10}} = \text{0.0458 mol P$_{4}$O}_{10}

2. Moles of P₂O₅

P₄O₁₀ ⟶ 2P₂O₅

The molar ratio is 2 mol P₂O₅:1 mol P₄O₁₀

\text{Moles of P$_{2}$O}_{5} = \text{0.0458 mol P$_{4}$O}_{10} \times \dfrac{\text{2 mol P$_{2}$O}_{5}}{\text{1 mol P$_{4}$O}_{10}} = \text{0.0916 mol P$_{2}$O}_{5}

3. Molecules of P₂O₅

\text{No. of molecules} = \text{0.0916 mol P$_{2}$O}_{5} \times \dfrac{6.022 \times 10^{23}\text{ molecules P$_{2}$O}_{5}}{\text{1 mol P$_{2}$O}_{5}}\\\\= \mathbf{5.5 \times 10^{22}}\textbf{ molecules P$_{2}$O}_{5}\\\text{There are $\large \boxed{\mathbf{5.5 \times 10^{22}}\textbf{ molecules of P$_{2}$O}_{5}}$}

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