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inna [77]
3 years ago
6

In the reaction between copper sulphate solution and sodium sulphide solution, 15.9 g copper sulphate completely reacts with 7.8

g of sodium sulphide. It is observed that 9.5 g of copper sulphide is formed. What is the mass of sodium sulphate solution formed?
Chemistry
1 answer:
nikitadnepr [17]3 years ago
8 0

Answer:

The mass of formed sodium sulphate solution is 14.2 g.

Explanation:

From the given,

Mass of copper sulphate = 15.9 g

Mass of sodium sulphide = 7.8 g

Total mass of reactant  =  Mass of copper sulphate + Mass of sodium sulphide

                                     = 15.9 + 7.8g = 23.7g

Mass of copper sulphide formed = 9.5 g

Mass of sodium sulphate = Total mass - Mass of copper sulphide

                                          = 23.7 g - 9.5 g = 14.2 g

Therefore, The mass of formed sodium sulphate solution is 14.2 g.

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Fe2O3 + 2Al = 2Fe + Al2O3 is this a redox reaction
katovenus [111]

Answer:

<h3>2Al+ Fe2O3 gives 2Fe + Al2O3. The given reaction is a redox reaction. As oxidation and reduction are taking place simultaneously.</h3>

Explanation:

like this...Identify oxidation and reduction with their agents:

<h3>•2Al+ Fe2O3 →2Fe + Al2O3</h3>

<h3>•Fe2O3 is reduced to Fe whereas Al is oxidized to Al2O3</h3>

<h3>In the above reaction:</h3>

<h3>Oxidizing agent:Fe2O3</h3>

<h3>Reducing agent:Al</h3>

I hope it's help you (◠‿・)—☆

5 0
3 years ago
A student heats 10.52 g of sodium hydrogen carbonate in a crucible until the compound completely decomposes to sodium carbonate
saveliy_v [14]

Answer:

m_{Na_2CO_3}^{theoretical}=6.636gNa_2CO_3

Explanation:

Hello!

In this case, since the decomposition of sodium hydrogen carbonate is:

2NaHCO_3(s)\rightarrow Na_2CO_3(s)+H_2O(g)+CO_2(g)

Thus, since there is a 2:1 mole ratio between the sodium hydrogen carbonate and sodium carbonate, and the molar masses are 84.01 and 105.99 g/mol respectively, we obtain the following theoretical yield:

m_{Na_2CO_3}^{theoretical}=10.52gNaHCO_3*\frac{1molNaHCO_3}{84.01gNaHCO_3}*\frac{1molNa_2CO_3}{2molNaHCO_3}  *\frac{105.99gNa_2CO_3}{1molNa_2CO_3}\\\\ m_{Na_2CO_3}^{theoretical}=6.636gNa_2CO_3

Best regards!

4 0
3 years ago
Pressure is this sample?
faltersainse [42]

Answer:

P = 164 Atm

Explanation:

PV = nRT => P = nRT/V

n = 10.0 moles

R = 0.08206 L·Atm/mol·K

T = 27.0°C = 300 K

V = 1.50 Liters

P = (10.0 mol)(0.08206 L·Atm/mol·K )(300 K)/(1.50 Liters) = 164.12 Atm ≅ 164 Atm (3 sig. figs.)

5 0
3 years ago
How many significant figures are in the following number? 0.000485
aliina [53]

Answer:

4 or 3

Explanation:

8 0
3 years ago
Read 2 more answers
I begin the reaction with 0.45 g of beryllium. If my actual yield of beryllium chloride (mm = 79.91 g/mol) was 3.5 grams, what w
trasher [3.6K]

The percentage of yield was 777.78%

<u>Explanation:</u>

We have the equation,

Be [s]  +  2 HCl [aq]  →  BeCl 2(aq]  + H 2(g]  ↑ Be (s]  + 2 HCl [aq]  →  BeCl 2(aq]  + H 2(g] ↑

To find the percent yield we have the formula

Percentage of Yield= what you actually get/ what you should theoretically get  x 100

                                   =3.5 g/0.45 g 100

                                    = 777.78 %

The percentage of yield was 777.78%

5 0
4 years ago
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