The concentration of acetic acid in the vinegar is found to be 0.5857 M.
<u>Explanation:</u>
Here one mole of acetic acid is used to neutralize one mole of sodium hydroxide. According to the law of Volumetric analysis, we can write the equation as,
V1M1 = V2M2
V1 and M1 being the volume and molarity of NaOH
V2 and M2 being the volume and molarity of CH₃COOH
M2 = 
= 
= 0.5857 M
So the concentration of acetic acid in the vinegar is 0.5857 M.
22. Mass=Force/acceleration or M=F/a. So the force of this equation would be 88N and the acceleration would be 4m/s^2 so the mass in the problem would be 22.
<u>Answer:</u> The equilibrium constant for the given reaction is 1.33
<u>Explanation:</u>
We are given:
Equilibrium concentration of ammonia = 2 M
Equilibrium concentration of nitrogen gas = 3 M
Equilibrium concentration of hydrogen gas = 1 M
For the given chemical equation:
The expression of 
for above equation follows:
![K_{eq}=\frac{[NH_3]^2}{[N_2][H_3]^3}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5BNH_3%5D%5E2%7D%7B%5BN_2%5D%5BH_3%5D%5E3%7D)
Putting values in above equation, we get:
Hence, the equilibrium constant for the given reaction is 1.33
Answer:
82.24% percent composition of N and 17.76% percent composition of H .
Explanation:
Answer:
46 g of sodium
Explanation:
Sodium reacts vigorously with fluoride gas to form NaF as shown
2Na (s) + F2 (g) ------> 2NaF (s) (Na = 23, F = 19)
2 moles of Na reacts with 1 mole of F2 to produce NaF
By calculating the molar mass of the elements involved in the reaction then multiplied by the mole, the mass can be obtained.
23 * 2 g/mol of Na reacts with 1 * 19 g/mol of F2
46 g/mol of Na reacts with 19 g/mol of F2 to produce NaF
Since the mole ratio is 2 to 1 and 19 g of F2 is used for the reaction, 46 g of sodium will be consumed for the reaction to be achieved.
