The molarity is the number of moles in 1 L of the solution.
The mass of NH₃ given - 2.35 g
Molar mass of NH₃ - 17 g/mol
The number of NH₃ moles in 2.35 g - 2.35 g / 17 g/mol = 0.138 mol
The number of moles in 0.05 L solution - 0.138 mol
Therefore number of moles in 1 L - 0.138 mol / 0.05 L x 1L = 2.76 mol
Therefore molarity of NH₃ - 2.76 M
Answer:
Keiko should mix 20 mL 1% solution and 80 mL 6% solution for to make 100 mL 5% solution
Explanation:
There are 2 unknown values X= mL 6% solution and Y=1% solution. So, we need 2 equations:
1. Equation acid concentration. X mL 6% + Y mL 1% = 100 mL 5%
2. Equation solvent concentration X mL 94% + Y mL 99% = 100 mL 95%
When clearing X and Y :
(X mL 6% + Y mL 1% = 100 mL 5%) (-15,7)
X mL 94% + Y mL 99% = 100 mL 95%
_______________________________
- Y 0.83 = 16.5
Y = 19.9 mL 1% solution
Replace Y in anyone equation and X = 80 mL 6% solution
I hope to see been helpful
Answer:
Reactions 1, 3 and 5
Explanation:
First thing's first, let's ensure that all the reactions given are balanced. This is given as;
CO(g) + 1/2 O2(g )→ CO2(g)
Li(s) + 1/2 F2(l) → LiF(s)
C(s) + O2(g) → CO2(g)
CaCO3(g) → CaO + CO2(g)
2Li(s) + F2(g) → 2LiF(s)
For the condition to be valid;
- There is by convention 1 mol of product made. This means we eliminate reactions with more than one mole of compound formed. This eliminates reaction 5.
- The lements haveto be in their state at room temperature. Fluorine is a gas, not a liquid, at room temperature ans pressure, so 2 is not a correct answer.
This leaves us with reactions 1, 3 and 5 as the correct reactions that satisify the condition.
Explanation:
As the given reaction is as follows.
So, according to the balanced equation, it can be seen that rate of formation of 
will be twice the rate of disappearance of 
.
And, it is known that rate of disappearance of reactant will be negative and rate of formation of products will be positive value.
This means that,
Rate of the reaction = -Rate of disappearance of
= ![k[H_{2}S][Cl_{2}]](https://tex.z-dn.net/?f=k%5BH_%7B2%7DS%5D%5BCl_%7B2%7D%5D)
= 
= 
M/s
Therefore, calculate the rate of formation of as follows.
Rate of formation of = 
= 
M/s
Thus, we can conclude that the rate of formation of is M/s.
Answer:
From the following enthalpy of reaction data and data in Appendix C, calculate ΔH∘f for CaC2(s): CaC2(s)+2H2O(l)→Ca(OH)2(s)+C2H2(g)ΔH∘=−127.2kJ
ΔHf°(C2H2) = 227.4 kJ/mol
ΔHf°(H2O) = -285.8 kJ/mol and
ΔHf°(Ca(OH)2) = -985.2 kJ/mol
(Ans)
ΔHf° of CaC2 = -59.0 kJ/mol
Explanation:
CaC2(s) + 2 H2O(l) → Ca(OH)2(s) + C2H2 (g) = −127.2kJ
ΔHrxn = −127.2kJ
ΔHrxn = ΔHf°(C2H2) + ΔHf°(Ca(OH)2) - ΔHf°(CaC2)- 2ΔHf°(H2O);
ΔHf°(CaC2) = ΔHf°(C2H2) + ΔHf°(Ca(OH)2) - 2ΔHf°(H2O) – ΔHrxn
Where
ΔHf°(C2H2) = 227.4 kJ/mol
ΔHf°(H2O) = -285.8 kJ/mol and
ΔHf°(Ca(OH)2) = -985.2 kJ/mol
ΔHf°(CaC2) =227.4 - 985.2 + 2x285.8 + 127.2 = -59.0 kJ/mol
ΔHf°(CaC2) = -59.0 kJ/mol
