Answer:
Explanation:
a) 3.00 mol Cu x 2 mol Ag/ 1 mol Cu = 6 mol Ag
b) 3.00 mol Cu x 2 mol Ag/1 mol Cu x 107.87g Ag/ 1 mol Ag = 647.22g Ag
c) 3.00 mol Cu x 1 mol Cu(NO3)2/ 1 mol Cu = 3 mol Cu(NO3)2
d) 3.00 mol Cu x 1 mol Cu(NO3)2/ 1 mol Cu x 187.54g/ 1 mol Cu(NO3)2 =562.644g Cu(NO3)2
Answer: CrO₄⁻ and Ba²⁺
Explanation:
1) Chemical equation given:
2H⁺ + CrO₄⁻ + Ba²⁺ + 2OH⁻ → Ba²⁺ + CrO₄⁻ + 2H₂O
2) Analysis
That is an oxidation-reduction equation (some species are been oxidized and others are being reduced).
The given equation is known as total ionic equation, because it shows all the species as ions that are part of the reaction.
2) Specator ions
Spectator ions are the ions that do not change their oxidation state and are easily identified as they are the same in the reactant and product sides.
Here the ions that are the same in the reactant and product sides are:
CrO₄⁻ and Ba²⁺
3) Addtitional explanation.
Once you identify the spectator ions you can delete them from the equation to obtain the net ionic equation , which in this case turns to be:
2H⁺ + 2OH⁻ → 2H₂O
But this is not part of the question; it is some context to help you understand the use of the spectator ions concept.
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Answer : The fuel value and the fuel density of pentane is, 49.09 kJ/g and 
respectively.
Explanation :
Fuel value : It is defined as the amount of energy released from the combustion of hydrocarbon fuels. The fuel value always in positive and in kilojoule per gram (kJ/g).
As we are given that:
Fuel value = 
Molar mass of pentane = 72 g/mol
Fuel value = 
Fuel value = 49.09 kJ/g
Now we have to calculate the fuel density of pentane.
Fuel density = Fuel value × Density
Fuel density = (49.09 kJ/g) × (0.626g/mL)
Fuel density = 30.73 kJ/mL =
Thus, the fuel density of pentane is
Answer:
This is site for English speakers. Этот сайт на английском, поэтому вопрос могут удалить
Explanation:
1. 2)
2. 3)
3. 4) Sr
4. 3)
5. 4)
6. 2)
7. 1)
8. 4)
9. 3)
10. 3)
11. SO3, H2SO4, Na2SO4
12.
A) оксид меди (II) 2) CuO
Б) хлорид меди(II) 4) CuCl2
В) сульфит меди (II) 3) CuSO 3
Г) гидроксид меди (II) 1) Cu(OH)2
13.
1. Fe+HCl= б) FeCl 2 +H 2
2.Fe+O2= в) Fe 3 O 4
3. Fe(OH) 3 = г)Fe 2 O 3 +H 2O
4. FeCl 2 +NaOH= а) Fe(OH) 2 +NaCl
14. 2Ca + O2 = 2CaO
CaO + H2O = Ca(OH)2
Ca(OH)2 + 2HCl = CaCl2 + 2H2O
