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3 years ago

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What is the percent composition of a copper chloride compound if a 269 g sample contains 127 g of copper and 142 grams of chlori

ne.

1 answer:

Tems11 [23]3 years ago

5 0

Answer:

Explanation:

The whole sample is 269

%copper = 127/269 * 100 = 47.2%

%chlorine = 142/269 * 100 = 52.8%

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A mixture of hydrochloric and sulfuric acids is prepared so that it contains 0.315 M HCl and 0.125 M H2SO4. What volume of 0.55

Xelga [282]

<u>Answer:</u> The volume of NaOH required is 402.9 mL

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

<u>For HCl:</u>

Molarity of HCl solution = 0.315 M

Volume of solution = 503.4 mL = 0.5034 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.315M=\frac{\text{Moles of HCl}}{0.5034L}\\\\\text{Moles of HCl}=(0.315mol/L\times 0.5034L)=0.1586mol$

<u>For sulfuric acid:</u>

Molarity of sulfuric acid solution = 0.125 M

Volume of solution = 503.4 mL = 0.5034 L

Putting values in equation 1, we get:

$0.125M=\frac{\text{Moles of }H_2SO_4}{0.5034L}\\\\\text{Moles of }H_2SO_4=(0.125mol/L\times 0.5034L)=0.0630mol$

As, all of the acid is neutralized, so moles of NaOH = [0.1586 + 0.0630] moles = 0.2216 moles

Molarity of NaOH solution = 0.55 M

Moles of NaOH = 0.2216 moles

Putting values in equation 1, we get:

$0.55M=\frac{0.2216}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{0.2216}{0.55}=0.4029L=402.9mL$

Hence, the volume of NaOH required is 402.9 mL

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3 years ago

When K- and I- combine

Charra [1.4K]

Answer:

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Explanation:

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3 years ago

If a 2,000-kilogram car accelerates at a rate of 3 meters per second squared,

Likurg_2 [28]

Answer:

Explanation:

assuming there is no friction

fnet=ma

2,000 kg*3m/s=6,000 N

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2 years ago

In a lab investigation, one group of students (group A) measures the speed of a 0.1 kg toy car to be 2.5

ANTONII [103]

Answer:

Th asner you geht whan Ypu mltiply to et

Explanation:

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2 years ago

Read 2 more answers

phosporus can be prepared from calcium phosphate by the following reaction: 2Ca3(PO4)2+6SiO2+10C → 6CaSiO3+P4+10CO Phosphorite i

morpeh [17]

Answer:

285 g of P₄

Explanation:

Let's consider the following balanced equation.

2 Ca₃(PO₄)₂ + 6 SiO₂ + 10 C → 6 CaSiO₃ + P₄ + 10 CO

We know the following relations:

100 g of phosphorite contain 75 g of Ca₃(PO₄)₂
2 moles of Ca₃(PO₄)₂ produce 1 mole of P₄
The molar mass of Ca₃(PO₄)₂ is 310 g/mol

The molar mass of P₄ is 124 g/mol

Then, for 1.9 kg of phosphorite:

$1900g(phosphorite).\frac{75gCa_{3}(PO_{4})_{2}}{100g(phosphorite)} .\frac{1molCa_{3}(PO_{4})_{2}}{310gCa_{3}(PO_{4})_{2}} .\frac{1molP_{4}}{2molCa_{3}(PO_{4})_{2}} .\frac{124gP_{4}}{1molP_{4}} =285gP_{4}$

6 0

2 years ago