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aliya0001 [1]
2 years ago
9

Is the supply of copper-rich ores greater or lesser than iron-rich ores?

Chemistry
1 answer:
mylen [45]2 years ago
7 0

Hi! I am not sure it is a Guess but I think it is lesser if I am wrong pls let me know.

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A 1.555-g sample of baking soda decomposes with heat to produce 0.991 g Na2CO3. Refer to Example Exercise 14.l and show the calc
Sunny_sXe [5.5K]

Answer:

a) 101%

b)59.7%

Explanation:

The equation for the thermal decomposition of baking soda is shown;

2NaHCO3 → Na2CO3 + H2O + CO2

Number of moles of baking soda= mass/molar mass= 1.555g/84.007 g/mol = 0.0185 moles

From the reaction equation;

2 moles of baking soda yields 1 mole of sodium carbonate

0.0185 moles of baking soda will yield = 0.0185 moles ×1 /2 = 9.25 ×10^-3 moles of sodium carbonate.

Therefore, mass of sodium carbonate= 9.25 ×10^-3 moles × 106gmol-1= 0.9805 g of sodium carbonate. This is the theoretical yield of sodium carbonate.

%yield = actual yield/theoretical yield ×100

% yield = 0.991/0.9805 ×100

%yield = 101%

Since ;

2NaHCO3 → Na2CO3 + H2O + CO2

And H2O + CO2 ---> H2CO3

Hence I can write, 2NaHCO3 → Na2CO3 + H2CO3

Molar mass of H2CO3= 62.03 gmol-1

Molar mass of baking soda= 84 gmol-1

Therefore, mass of baking soda=

0.325/62.03 × 2 × 84 = 0.88 g of NaHCO3

% of NaHCO3= 0.88/1.473 × 100 = 59.7%

7 0
3 years ago
Which compound lauric acid or sucrose is more soluble in water?
Greeley [361]
<span>Sucrose is more soluble in water. The reason can be attributed to hydrogen bonding between water molecules and sucrose molecules. </span>
4 0
3 years ago
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What is the specific heat of a substance if a mass of 10.0 kg increases in temperature from 10.0°C to 70.0°C when 2,520 J of hea
Ksju [112]

Answer:

A. 0.00420 J/(gi°C)

Explanation:

i got it rigt

8 0
3 years ago
How much heat is required to raise the temperature of 670g of water from 25.7"C to 66,0°C? The specific heat
inna [77]

Answer:

Explanation:

q= mc theta

where,

Q = heat gained

m = mass of the substance = 670g

c = heat capacity of water= 4.1 J/g°C    

theta =Change in temperature=( 66-25.7)

Now put all the given values in the above formula, we get the amount of heat needed.

q= mctheta

q=670*4.1*(66-25.7)

  =670*4.1*40.3

=110704.1

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3 years ago
Definition of metallurgy
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