Answer:
67.5%
Explanation:
Step 1: Write the balanced equation for the electrolysis of water
2 H₂O ⇒ 2 H₂ + O₂
Step 2: Calculate the theoretical yield of O₂ from 17.0 g of H₂O
According to the balanced equation, the mass ratio of H₂O to O₂ is 36.04:32.00.
17.0 g H₂O × 32.00 g O₂/36.04 g H₂O = 15.1 g O₂
Step 3: Calculate the percent yield of O₂
Given the experimental yield of O₂ is 10.2 g, we can calculate its percent yield using the following expression.
%yield = (exp yield / theoret yield) × 100%
%yield = (10.2 g / 15.1 g) × 100% = 67.5%
Answer:
Explanation:
The triple point of carbon dioxide is 5.11 atmosphere at -56.6 degree celsius . At pressure greater than 5.11 , solid carbon dioxide liquefies , as it is warmed. At pressure lesser than 5.11 atmosphere , it will go into gaseous state without liquefying . Excessive pressure helps liquification process.
So maximum pressure required is 5.11 atmosphere. Beyond this pressure , solid CO2 will liquify.
Answer:
Rate = k . [B]² . [C]
Explanation:
The dependence of the reaction rate on the concentration of the reactants is given by the reaction order of each one, as shown in the rate equation.
![Rate=k.[A]^{x} .[B]^{y} .[C]^{z}](https://tex.z-dn.net/?f=Rate%3Dk.%5BA%5D%5E%7Bx%7D%20.%5BB%5D%5E%7By%7D%20.%5BC%5D%5E%7Bz%7D)
where,
k is the rate constant
x, y, z are the reaction orders.
- <em>The rate of reaction is not affected by changing the concentration of species A.</em> This means that the reaction order for A is x = 0 since when its concentration changes, the rate stays the same.
- <em>Leaving all other factors identical, doubling the concentration of species B increases the rate by a factor of 4.</em> This means that the reaction order for B is y = 2, so when the concentration is doubled, the new rate is 2² = 4 times the initial rate.
- The rate of the reaction is linearly dependent on the concentration of C. This means that the reaction order for C is z = 1, that is, a linear dependence.
All in all, the rate equation is:
Rate = k . [B]² . [C]
Answer:
The mixing of two chemicals may result in the production of a gas which is lost to the air. This will reduce the mass of the chemical mixture, because mass is being lost in a gaseous form.