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natita [175]
3 years ago
13

How many moles of copper are needed to react with sulfur to produce 0.25 moles of copper (I) sulfide

Chemistry
1 answer:
lorasvet [3.4K]3 years ago
7 0
Copper  react  with  sulfur  to   produce   copper  (1)   sulfide  according  to  the  following   equation
 2Cu  +  S ---->  CU2S
from  the  equation  above  2   moles  of  copper  react   with  1  mole  of  Sulfur  to  produce  1  mole  of  copper  (1) sulfide

therefore  the  mole  ratio of  Cu  to  CU2S  is  2:1  therefore  the  moles of  copper  is   0.25  x   2=  0.5 moles
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Class material don't interact please
bagirrra123 [75]

A 0.00143 M  concentration of MnO4^- is not a reasonable solution .

<h3>Number of moles of carbonate</h3>

The ions left in solution are Na^+ and NO3^-

Number of moles of calcium nitrate  = 100/1000 L × 1 = 0.1 moles

Since;

1 mole of sodium carbonate reacts with 1 mole of calcium nitrate  then 0.1 moles of sodium carbonate were used.

<h3>Conductivity of filtrate</h3>

The claim of the student that the concentration of sodium carbonate is too low is wrong because the value was calculated from concentration  and volume of calcium nitrate  and not using the precipitate. If the filtrate is tested for conductivity, it will be found to conduct electricity because it contains sodium and NO3 ions.

2) In the reaction as shown, the MnO4^- ion was reduced.

The initial volume is 3.4 mL while the final volume is 29.6 mL.

Number of moles of MnO4^- ion = (29.6 mL - 3.4 mL)/1000 × 0.0235 M = 0.0006157 moles

<h3>The calculations are performed as follows</h3>

  • If 2 moles of MnO4^- reacted with 5 moles of acid

0.0006157 moles of MnO4^- reacted with  0.0006157 moles ×  5 moles/ 2 moles

= 0.0015 moles

  • In this case, number of moles of acid = 0.139 g/90 g/mol = 0.0015 moles

Number of moles of MnO4^-  = 0.00143 M × (29.6 mL - 3.4 mL)/1000

= 0.000037 moles

  • If 2 moles of MnO4^- reacts with 5 moles of acid

0.000037 moles of MnO4^- reacts with 0.000037 moles × 5 moles/ 2 moles

= 0.000093 moles

  • Hence, this is not a reasonable amount of solution.

Learn more about MnO4^- : brainly.com/question/10887629

8 0
2 years ago
At 850K, 65L of gas has a pressure of 450kPa. What is the volume (in liters) if the gas is cooled to 430K and the pressure decre
Novosadov [1.4K]

Answer:

V₂ =  45.53 L

Explanation:

Given data:

Initial temperature = 850 K

Initial volume = 65 L

Initial pressure = 450 KPa

Final temperature = 430 K

Final pressure = 325 KPa

Final volume = ?

Solution:

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

V₂ = P₁V₁ T₂/ T₁ P₂  

V₂ = 450 KPa× 65 L × 430 K / 850 K × 325KPa  

V₂ = 12577500 KPa .L. K / 276250 K. KPa

V₂ =  45.53 L

8 0
3 years ago
Rank the following in order of increasing (lowest to highest) reactivity:
choli [55]

Answer:

a)  Li<Na<Rb

b) P<S

c) Rn<Xe

d) Hg<W<Cs

4 0
2 years ago
The standard molar heat of fusion of ice is 6020 j/mol. calculate q, w, and ∆e for melting 1.00 mol of ice at 0◦c and 1.00 atm p
zysi [14]

Answer :    q = 6020 J, w = -6020 J, Δe = 0

Solution : Given,

Molar heat of fusion of ice = 6020 J/mole

Number of moles = 1 mole

Pressure = 1 atm

Molar heat of fusion : It is defined as the amount of energy required to melt 1 mole of a substance at its melting point. There is no temperature change.

The relation between heat and molar heat of fusion is,

q=\Delta H_{fusion}(\frac{Mass}{\text{ Molar mass}})  (in terms of mass)

or, q=\Delta H_{fusion}\times Moles     (in terms of moles)

Now we have to calculate the value of q.

q=6020J/mole\times 1Mole=6020J

When temperature is constant then the system behaves isothermally and Δe is a temperature dependent variable.

So, the value of \Delta e=0

Now we have to calculate the value of w.

Formula used :    \Delta e=q+w

where, q is heat required, w is work done and \Delta e is internal energy.

Now put all the given values in above formula, we get

0=6020J+w

w = -6020 J

Therefore, q = 6020 J, w = -6020 J, Δe = 0

3 0
3 years ago
The "particles" of gases are actually atoms, or molecules. truefalse
kirill [66]
They are atoms. a molecule is made up of atoms. 
8 0
3 years ago
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