Answer:
B. accepted value x 0.1
Explanation:
in the equation provided

Maximum allowed value of percentage error = 10%
put this value in the equation in stead of percentage error we get,

so maximum error = .1 x accepted value
10 % percentage error means the experimental value has 10 % error compared to accepted value.so error will be 10 % of the accepted value
or .1 times of accepted value
Energy is absorbed so, the mass is increased.
energy is released so, the mass is increased
Leonardo de Vinci, Michelangelo, and Raphael.
Answer: 1) 
Equilibrium constant is defined as the ratio of the product of concentration of products to the product of concentration of reactants each term raised to their stochiometric coefficients.
![K_{eq}=\frac{[H_2S]^2}{H_2]^2\times [S_2]}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5BH_2S%5D%5E2%7D%7BH_2%5D%5E2%5Ctimes%20%5BS_2%5D%7D)
where [] = concentration in Molarity=
Thus ![[H_2S]=\frac{68.5}{1.0}=68.5M](https://tex.z-dn.net/?f=%5BH_2S%5D%3D%5Cfrac%7B68.5%7D%7B1.0%7D%3D68.5M)
![[H_2]=\frac{0.50}{1.0}=0.50M](https://tex.z-dn.net/?f=%5BH_2%5D%3D%5Cfrac%7B0.50%7D%7B1.0%7D%3D0.50M)
![[S_2]=\frac{0.020}{1.0}=0.020M](https://tex.z-dn.net/?f=%5BS_2%5D%3D%5Cfrac%7B0.020%7D%7B1.0%7D%3D0.020M)
![K_{eq}=\frac{[68.5]^2}{0.50]^2\times [0.020]}=938450](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5B68.5%5D%5E2%7D%7B0.50%5D%5E2%5Ctimes%20%5B0.020%5D%7D%3D938450)
As the value of K is greater than 1, the reaction is product favored.
2) 
![K_{eq}=\frac{[NO_2]^2}{[N_2O_4]}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BN_2O_4%5D%7D)
![K_{eq}=\frac{[0.500]^2}{[0.0250]}=10](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5B0.500%5D%5E2%7D%7B%5B0.0250%5D%7D%3D10)
3) 
![K_{eq}=\frac{[NH_3]^2}{[N_2]\times [H_2]^3}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5BNH_3%5D%5E2%7D%7B%5BN_2%5D%5Ctimes%20%5BH_2%5D%5E3%7D)
4) Reactions which do not continue to completion are called equilibrium reactions as the rate of forward reaction is equal to the rate of backward direction.
The volume in mL of ethanol (C₂H₆O) present in the bottle is 487.5L
HOW TO CALCULATE VOLUME OF ETHANOL:
- According to this question, a particular beer is 65.0% ethanol by volume. This means that the beer contains 65/100 of ethanol.
- If a single bottle of beer contains a total of 750.0 mL of beer, the volume of ethanol in the bottle is calculated as follows:
- Therefore, volume in mL of ethanol (C₂H₆O) present in the bottle is 487.5L
Learn more at: brainly.com/question/8725232?referrer=searchResults