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ANTONII [103]
3 years ago
10

A solution of cough syrup contains 5.00% active ingredient by volume. If the total volume of the bottle is 31.0 mL, how many mil

liliters of active ingredients are in the bottle?
Chemistry
2 answers:
Phantasy [73]3 years ago
5 0

Answer:

The answer to your question is 1.55 ml

Explanation:

Data

Percent by volume = 5%

Total volume = 31 ml

Volume of solute = ?

Formula

Percent by volume = \frac{Volume of solute}{volume of solution} x 100

Solve for volume of solute

Volume of solute = \frac{percent by volume x volume of solution}{100}

Substitution

Volume of solute = \frac{5 x 31}{100}

Simplification

Volume of solute = 1.55 ml

AlekseyPX3 years ago
4 0

Answer:

Total volume of the syrup is 31.0ml

% of active ingredient = 5% = 0.05

The volume of the activ ingredient in the syrup is 31*0.05 = 1.55ml

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Provide a balanced molecular equation, total ionic, and net ionic equation for sodium phosphate and zinc acetate.
vivado [14]

Answer: Balanced molecular equation :

2Na_3PO_4(aq)+3(CH_3COO)_2Zn(aq)\rightarrow 6CH_3COONa(aq)+Zn_3(PO_4)_2(s)

Total ionic equation:

6Na^+(aq)+3PO_4^{2-}(aq)+6CH_3COO^-(aq)+3Zn^{2+}(aq)\rightarrow 6CH_3COO^-(aq)+6Na^+(aq)+Zn_3(PO_4)_2(s)  

The net ionic equation:

2PO_4^{3-}(aq)+3Zn^{2+}(aq)\rightarrow Zn_3(PO_4)_2(s)

Explanation:

Complete ionic equation : In complete ionic equation, all the substances that are strong electrolyte are present in an aqueous state as ions.

Net ionic equation : In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

When sodium phosphate and zinc acetate then it gives zinc phosphate and sodium acetate as  product.

The balanced molecular equation will be,

2Na_3PO_4(aq)+3(CH_3COO)_2Zn(aq)\rightarrow 6CH_3COONa(aq)+Zn_3(PO_4)_2(s)

The total ionic equation in separated aqueous solution will be,

6Na^+(aq)+2PO_4^{3-}(aq)+6CH_3COO^-(aq)+3Zn^{2+}(aq)\rightarrow 6CH_3COO^-(aq)+6Na^+(aq)+Zn_3(PO_4)_2(s)

In this equation, and  are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

2PO_4^{3-}(aq)+3Zn^{2+}(aq)\rightarrow Zn_3(PO_4)_2(s)

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3 years ago
True or False: A heterozygous genotype produces the dominant phenotype because the dominant allele masks the recessive allele.
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A 3.452 g sample containing an unknown amount of a Ce(IV) salt is dissolved in 250.0-mL of 1 M H2SO4. A 25.00 mL aliquot is anal
SOVA2 [1]

Answer:

1,812 wt%

Explanation:

The reactions for this titration are:

2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻

I₃⁻ + 2S₂O₃⁻ → 3I⁻ + S₄O₆²⁻

The moles in the end point of S₂O₃⁻ are:

0,01302L×0,03428M Na₂S₂O₃ = 4,463x10⁻⁴ moles of S₂O₃⁻. As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:

4,463x10⁻⁴ moles of S₂O₃⁻×\frac{1molI_{3}^-}{2molS_{2}O_{3}^-} = 2,2315x10⁻⁴ moles of I₃⁻

As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:

2,2315x10⁻⁴ moles of I₃⁻×\frac{2molCe^{4+}}{1molI_{3}^-} = 4,463x10⁻⁴ moles of Ce(IV). These moles are:

4,463x10⁻⁴ moles of Ce(IV)×\frac{140,116g}{1mol} = <em>0,0625 g of Ce(IV)</em>

As the sample has a 3,452g, the weight percent is:

0,0625g of Ce(IV) / 3,452g × 100 = <em>1,812 wt%</em>

I hope it helps!

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