Answer:
The value of Ka 
It is a weak acid
Explanation:
From the question we are told that
The concentration of ![[HClO_2]=0.24M](https://tex.z-dn.net/?f=%5BHClO_2%5D%3D0.24M)
The concentration of ![[H^+]=0.051M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.051M)
The concentration of ![[ClO_2^-]=0.051M](https://tex.z-dn.net/?f=%5BClO_2%5E-%5D%3D0.051M)
Generally the equation for the ionic dissociation of 
is
The equilibrium constant is mathematically represented as
![= \frac{[H^+][ClO_2^-]}{[HClO_2]}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B%5BH%5E%2B%5D%5BClO_2%5E-%5D%7D%7B%5BHClO_2%5D%7D)
Substituting values since all value of concentration are at equilibrium
Since the value of is less than 1 it show that in water it dose not completely
disassociated so it an acid that is weak
25% of 20
%=100
25/100 x 20
1/4 x 20
The answer is equal to 5
Answer:
about 10 pounds
Explanation:
your mass doesn't change but weight changes (due to gravity)
Answer: 1) Maximum mass of ammonia 198.57g
2) The element that would be completely consumed is the N2
3) Mass that would keep unremained, is the one of the excess Reactant, that means the H2 with 3,44g
Explanation:
- In order to calculate the Mass of ammonia , we first check the Equation is actually Balance:
N2(g) + 3H2(g) ⟶2NH3(g)
Both equal amount of atoms side to side.
- Now we verify which reagent is the limiting one by comparing the amount of product formed with each reactant, and the one with the lowest number is the limiting reactant. ( Keep in mind that we use the molecular weight of 28.01 g/mol N2; 2.02 g/mol H2; 17.03g/mol NH3)
Moles of ammonia produced with 163.3g N2(g) ⟶ 163.3g N2(g) x (1mol N2(g)/ 28.01 g N2(g) )x (2 mol NH3(g) /1 mol N2(g)) = 11.66 mol NH3
Moles of ammonia produced with 38.77 g H2⟶ 38.77 g H2 x ( 1mol H2/ 2.02 g H2 ) x (2 mol NH3 /3 mol H2 ) = 12.79 mol NH3
- As we can see the amount of NH3 formed with the N2 is the lowest one , therefore the limiting reactant is the N2 that means, N2 is the element that would be completey consumed, and the maximum mass of ammonia will be produced from it.
- We proceed calculating the maximum mass of NH3 from the 163.3g of N2.
11.66 mol NH3 x (17.03 g NH3 /1mol NH3) = 198.57 g NH3
- In order to estimate the mass of excess reagent, we start by calculating how much H2 reacts with the giving N2:
163.3g N2 x (1mol N2/28.01 g N2) x ( 3 mol H2 / 1 mol N2)x (2.02 g H2/ 1 mol H2) = 35.33 g H2
That means that only 35.33 g H2 will react with 163.3g N2 however we were giving 38.77g of H2, thus, 38.77g - 35.33 g = 3.44g H2 is left
Answer: V = 33.9 L
Explanation: We will use Charles Law to solve for the new volume.
Charles Law is expressed in the following formula. Temperatures must be converted in Kelvin.
V1 / T1 = V2 / T2 then derive for V2
V2 = V1 T2 / T1
= 35 L ( 308 K ) / 318 K
= 33.9 L
