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Andrei [34K]
4 years ago
14

If 0.680 kg of copper(I) sulfide reacts with excess oxygen, what mass of copper metal may be produced ? A) 0.680 kg B) 0.136 kg

C) 0.271 kg D) 0.543 kg E) 1.36 kg
Chemistry
1 answer:
Katen [24]4 years ago
5 0

Answer:

D. 0.543kg of copper metal is produced from 0.680kg of copper 1 sulphide.

Explanation:

First write the equation for the reaction:

Cu2S + O2 ------> 2Cu + SO2

Determine the mole ratio of the two substances:

I mole of Cu2O forms 2 moles of Copper metal

The number of moles of copper 1 sulphide used is;

n = mass of Cu2S / molar mass of Cu2S

Mass = 0.680kg = 680g

Molar mass = 159.16g/mol

n = 680g / 159.16g/mol

n = 4.272moles

Determine the number of mole of copper:

Number of moles of copper metal produced from 4.272moles of copper 1 sulphide is therefore:

n of copper = 2 * 4.272 Moles

n = 8.544moles.

Determine the mass copper:

The mass of copper metal produced is therefore = number of moles of copper * molar mass of copper

mass = 8.544 moles * 63.55g/mol

mass = 542.97grams

Mass = 0.543kg

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3 years ago
Phosphorus atomic radius is smaller than magnesium atomic radius <br> True or false
stealth61 [152]

Answer:

True

Explanation:

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<em>Hence, the atomic radius of phosphorus is smaller than the atomic radius of magnesium. Basically, the atomic radius of phosphorus is 98 pm while the atomic radius of magnesium is 145 pm.</em>

4 0
3 years ago
A solution of an unknown nonvolatile nonelectrolyte was prepared by dissolving 0.250 g of the substance in 40.0 g of ccl4. the b
Finger [1]
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Explanation : We know the formula for elevation in boiling point, which is

Δt = iK_{b}m

given that, Δt = 0.357, K_{b} = 5.02 and mass of CCl _{4} = 40,

on substituting the value we get,
0.357 = (1) X (5.02) X (x/ 0.044), on solving we get x = 2.844 X10^{-3}. 
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3 0
4 years ago
How many milliliters of 4.00 M NaOH are required to exactly neutralize 50.0 milliliters of a 2.00 M solution of HNO3 ?
kramer

Answer: The volume of NaOH required is 25.0 ml

Explanation:

According to the neutralization law,

n_1M_1V_1=n_2M_2V_2

where,

n_1 = basicity HNO_3 = 1

M_1 = molarity of HNO_3 solution = 2.00 M

V_1 = volume of  HNO_3 solution = 50.0 ml

n_2 = acidity of NaOH = 1

M_1 = molarity of NaOH solution = 4.00 M

V_1 = volume of  NaOH solution =  ?

Putting in the values we get:

1\times 2.00\times 50.0=1\times 4.00\times V_2

V_2=25.0ml

Therefore, volume of NaOH required is 25.0 ml

3 0
3 years ago
barium chloride (Ba(CIO3)2) breaks down to form barium chloride and oxygen. what is the balanced equation for this equation?
kompoz [17]

Answer:

Ba(ClO₃)₂ → BaCl₂ + 3 O₂

Explanation:

When exposed to heat, barium chlorate (Ba(ClO₃)₂ breaks down into an inorganic compound (Barium chloride - BaCl₂) and a molecule (Oxygen - O₂).

6 0
4 years ago
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