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4 years ago

How many moles of nitrogen are in 2.0×10−2mole of quinine?

Chemistry

1 answer:

galben [10]4 years ago

7 0

Answer: 4 x 10 ∧-2 moles of nitrogen.

Explanation:

The chemical formular for quinine is ; C20 H24 N2 O2

As can be seen from the chemical formular;

1 mole of quinine contains 2 moles of Nitrogen

Thus; 2.0 x 10 ∧-2 moles of quinine would contain

2.0 x 10 ∧-2 x 2 = 4 x 10 ∧-2 moles of nitrogen.

Therefore 4 x 10 ∧-2 moles of nitrogen are in 2.0×10−2mole of quinine

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Ammonium carbonate and iron(iii) nitrate are combined, solid iron(iii) carbonate and a solution of ammonium nitrate are formed.

Vanyuwa [196]

Answer:

The net ionic equation is: 3CO₃²⁻(aq) + 2Fe³⁺(aq) → Fe₂(CO₃)₃(s).

Explanation:

For the mentioned reaction:

We have the Molecular reaction:

3(NH₄)₂CO₃(aq) + 2Fe(NO₃)₃(aq) → Fe₂(CO₃)₃(s) + 6NH₄NO₃(aq)

We have the total ionic reaction:

6NH₄⁺(aq) + 3CO₃²⁻(aq) + 2Fe³⁺(aq) + 6NO₃⁻ → Fe₂(CO₃)₃(s) + 6NH₄⁺(aq) + 6NO₃⁻(aq).

NH₄⁺(aq) and NO₃⁻(aq) are spectator ions that are not changed through the reaction and still dissolved in the medium, so they can be omitted to get the net ionic equation.

So, the net ionic equation is:

3CO₃²⁻(aq) + 2Fe³⁺(aq) → Fe₂(CO₃)₃(s).

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3 years ago

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Genrish500 [490]

With an electronegativity of 0.89, Barium requires the least amount of energy to remove its valence electrons.

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3 years ago

In pressure dunt uwing frost shattering

Cerrena [4.2K]

Answer:

i dont get it???????

Explanation:

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3 years ago

Assume that your empty crucible weighs 15.98 g, and the crucible plus the sodium bicarbonate sample weighs 18.56 g. After the fi

Savatey [412]

The question is incomplete, the complete question is;

Assume that your empty crucible weighs 15.98 g, and the crucible plus the sodium bicarbonate sample weighs 18.56 g. After the first heating, your crucible and contents weighs 17.51 g. After the second heating, your crucible and contents weighs 17.50 g.

What is the theoretical yield of sodium carbonate?

What is the experimental yield of sodium carbonate?

What is the percent yield for sodium carbonate?

Which errors could cause your percent yield to be falsely high, or even over 100%?

Answer:

See Explanation

Explanation:

We have to note that water is driven away after the second heating hence we are concerned with the weight of the pure dry product.

Hence;

From the reaction;

2 NaHCO3 → Na2CO3(s) + H2O(l) + CO2(g)

Number of moles of sodium bicarbonate = 18.56 - 15.98 = 2.58 g/87 g/mol

= 0.0297 moles

2 moles of sodium bicarbonate yields 1 mole of sodium carbonate

0.0297 moles of 0.015 moles sodium bicarbonate yields 0.0297 * 1/2 = 0.015 moles

Theoretical yield of sodium carbonate = 0.015 moles * 106 g/mol = 1.59 g

Experimental yield of sodium bicarbonate = 17.50 g - 15.98 g = 1.52 g

% yield = experimental yield/Theoretical yield * 100

% yield = 1.52/1.59 * 100

% yield = 96%

The percent yield may exceed 100% if the water and CO2 are not removed from the system by heating the solid product to a constant mass.

5 0

3 years ago

If 16.29 grams of Na2SO4 is mixed with 3.697 grams of C and allowed to react according to the balanced equation: Na2SO4(aq) + 4

abruzzese [7]

Answer: The limiting reagent in the given chemical reaction is carbon metal.

Explanation:

Excess reagent is defined as the reagent which is present in large amount in a chemical reaction.

Limiting reagent is defined as the reagent which is present in small amount in a chemical reaction. Formation of product depends on the limiting reagent.

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)