Answer:
The correct answer is Option
B. A homogeneous mixture
Explanation: Mixture is solution which is made of two or more than two substances which are combined only physically not chemically. There are two types of mixture i. e. Homogeneous mixture and heterogeneous mixture. Homogeneous mixture is a mixture in which the substances which are combined are uniform in appearance and heterogeneous mixture is a mixture in which substances are suspended in the solution and easily differentiated.
Answer:
The amount of water converted from liquid to gas with 6,768 joules is approximately 3.035 g
Explanation:
The amount of heat required to convert a given amount of liquid to gas at its boiling point is known as the latent heat of evaporation of the liquid
The latent heat of evaporation of water, Δ
≈ 2,230 J/g
The relationship between the heat supplied, 'Q', and the amount of water in grams, 'm', evaporated is given as follows
Q = m × Δ
Therefore, the amount of water, 'm', converted from liquid to gas at the boiling point temperature (100°C), when Q = 6,768 Joules, is given as follows;
6,768 J = m × 2,230 J/g
∴ m = 6,768 J /(2,230 J/g) ≈ 3.035 g
The amount of water converted from liquid to gas with 6,768 joules = m ≈ 3.035 g.
Answer:
The percent by mass of copper in the mixture was 32%
Explanation:
The ammount of HNO₃ used is:
mol HNO₃ = volume * concentration
mol HNO₃ = 0.015 l * 15.8 mol/l
mol HNO₃ = 0.237 mol
According to the reaction, 4 mol HNO₃ will react with 1 mol Cu and produce 1 mol Cu²⁺. Since we have 0.237 mol HNO₃, the amount of Cu that could react would be (0.237 mol HNO₃ * 1 mol Cu / 4 mol HNO₃) 0.06 mol. This reaction would produce 0.060 mol Cu²⁺, however, only 0.010 mol Cu²⁺ were obtained, indicating that only 0.010 mol Cu were present in the mixture. This means that the acid was in excess, so we can assume that all copper present in the mixture has reacted.
Since 0.010 mol of Cu²⁺ were produced, the amount of Cu was 0.01 mol.
1 mol of Cu has a mass of 63.5 g, then 0.01 mol has a mass of:
0.01 mol Cu * 63.5 g / 1 mol = 0.635 g.
Since this amount was present in 2.00 g mixture, the amount of copper in 100 g of the mixture will be:
100 g(mixture) * 0.635 g Cu / 2 g(mixture) = 32 g
Then, the percent by mass of Cu (which is the mass of Cu in 100 g mixture) is 32%
![\\ \tt\leadsto \dfrac{d[NH_3]}{dt}=1.50\times 10^{-6}](https://tex.z-dn.net/?f=%5C%5C%20%5Ctt%5Cleadsto%20%5Cdfrac%7Bd%5BNH_3%5D%7D%7Bdt%7D%3D1.50%5Ctimes%2010%5E%7B-6%7D)
- dt remains same for reaction
![\\ \tt\leadsto \dfrac{d[H_2]}{dt}=\dfrac{3}{2}\dfrac{d[NH_3]}{dt}](https://tex.z-dn.net/?f=%5C%5C%20%5Ctt%5Cleadsto%20%5Cdfrac%7Bd%5BH_2%5D%7D%7Bdt%7D%3D%5Cdfrac%7B3%7D%7B2%7D%5Cdfrac%7Bd%5BNH_3%5D%7D%7Bdt%7D)
![\\ \tt\leadsto \dfrac{d[H_2]}{dt}=\dfrac{3}{2}(1.5\times 10^{-6})](https://tex.z-dn.net/?f=%5C%5C%20%5Ctt%5Cleadsto%20%5Cdfrac%7Bd%5BH_2%5D%7D%7Bdt%7D%3D%5Cdfrac%7B3%7D%7B2%7D%281.5%5Ctimes%2010%5E%7B-6%7D%29)
![\\ \tt\leadsto \dfrac{d[H_2]}{dt}=2.25\times 10^{-6}Ms^{-1}](https://tex.z-dn.net/?f=%5C%5C%20%5Ctt%5Cleadsto%20%5Cdfrac%7Bd%5BH_2%5D%7D%7Bdt%7D%3D2.25%5Ctimes%2010%5E%7B-6%7DMs%5E%7B-1%7D)
M is molarity here not metre
