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leva [86]
3 years ago
15

An iron ore was found to contain 92.0% pyrite (fool's gold). Pyrite has the formula FeS2and is 46.5% iron. How many grams of iro

n are there in 50.8 grams of iron ore?
Chemistry
1 answer:
Liula [17]3 years ago
4 0
Firstly calculate the grams in the last 8 percent before moving onto the pyrite section.
50.8x0.08=4.064g
We know that iron ore in this case has 92 percent pyrite which contains 46.5 percent iron so we do 50.8x0.92=46.736g from this we need to find 46.5 percent of the iron content in the 92 percent pyrite section then add this answer to the 8 percent of iron ore we found at the start.46.736x0.465=21.73224g
21.73224g+4.064=25.79624g of iron ore 25.8g(3sf)
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Conversation 13kg into gram<br>4 poi ts<br>​
Olenka [21]

Answer:

13000

Explanation:

The kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side.

A gram is a unit of weight equal to 1/1000th of a kilogram. A gram is the approximate weight of a cubic centimeter of water.

5 0
3 years ago
URGENT CHEMISTRY EXPERT!
vovangra [49]

Answer:

Part 1: 7.42 mL; Part 2: 3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ 2Cu₃(PO₄)₂(s)

Explanation:

Part 1. Volume of reactant

(a) Balanced chemical equation.

\rm 2Na_{3}PO_{4} + 3CuCl_{2} \longrightarrow Cu_{3}(PO_{4})_{2} + 6NaCl

(b) Moles of CuCl₂

\text{Moles of CuCl}_{2} =\text{ 16.7 mL CuCl}_{2} \times \dfrac{\text{0.200 mmol CCl}_{2}}{\text{1 mL CuCl}_{2}} =  \text{3.340 mmol CuCl}_{2}

(c) Moles of Na₃PO₄

The molar ratio is 2 mmol Na₃PO₄:3 mmol CuCl₂

\text{Moles of Na$_{3}$PO}_{4} =  \text{3.340 mmol CuCl}_{2} \times \dfrac{\text{2 mmol Na$_{3}$PO}_{4}}{\text{3 mmol CuCl}_{2}} =\text{2.227 mmol Na$_{3}$PO}_{4}

(d) Volume of Na₃PO₄

V = \text{2.227 mmol Na$_{3}$PO}_{4}\times \dfrac{\text{1 mL Na$_{3}$PO}_{4}}{\text{0.300 mmol Na$_{3}$PO}_{4}} = \text{7.42 mL Na$_{3}$PO}_{4} \\\\\text{The reaction requires $\large \boxed{\textbf{7.42 mL Na$_{3}$PO}_{4}}$}

Part 2. Net ionic equation

(a) Molecular equation

\rm 2Na_{3}PO_{4}(\text{aq}) + 3CuCl_{2}(\text{aq}) \longrightarrow Cu_{3}(PO_{4})_{2}(\text{s}) + 6NaCl(\text{aq})

(b) Ionic equation

You write molecular formulas for the solids, and you write the soluble ionic substances as ions.

According to the solubility rules, metal phosphates are insoluble.

6Na⁺(aq) + 2PO₄³⁻(aq) + 3Cu²⁺(aq) + 6Cl⁻(aq) ⟶ Cu₃(PO₄)₂(s) + 6Na⁺(aq) + 6Cl⁻(aq)  

(c) Net ionic equation

To get the net ionic equation, you cancel the ions that appear on each side of the ionic equation.

<u>6Na⁺(aq)</u> + 2PO₄³⁻(aq) + 3Cu²⁺(aq) + <u>6Cl⁻(aq)</u> ⟶ Cu₃(PO₄)₂(s) + <u>6Na⁺(aq)</u> + <u>6Cl⁻(aq)</u>  

The net ionic equation is

3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ Cu₃(PO₄)₂(s)

7 0
3 years ago
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2 years ago
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Answer:I think it could be A or B but I would choose A.

Explanation:

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3 years ago
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pV=nRT \Rightarrow p=\frac{nRT}{V}
p - pressure, n - number of moles, R - the gas constant, T - temperature, V - volume

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The mass of gases is 50 g.

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The greatest number of moles is in the container with Ar, so there is the highest pressure.
4 0
3 years ago
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