Answer:
2NaCN + CaCO3 --> Na2CO3 + Ca(CN)2
Explanation:
Knowing the names gets us: NaCN + CaCO3 --> Na2CO3 + Ca(CN)2
Balance: there are two sodiums and cyanides on the product side so add a 2 to the reactant side.
Usually, elements with more similar properties share the same column on the periodic table.
Therefore, if you look at the periodic table, you can see that Oxygen, Sulfur, and Selenium all share the same column. Therefore, they all have similar properties
Answer: #3 ( O, S, Se )
Hope this helped!!! :D
Answer:
A precipitate will begin to form at [Cu+] = 3.0 *10^-10 M
The precipitate formed is CuI
Explanation:
Step 1: Data given
The solution contains 0.021 M Cl- and 0.017 M I-.
Ksp(CuCl) = 1.0 × 10-6
Ksp(CuI) = 5.1 × 10-12.
Step 2: Calculate [Cu+]
Ksp(CuCl) = [Cu+] [Cl-]
1.0 * 10^-6 = [Cu+] [Cl-]
1.0 * 10^-6 = [Cu+] [0.021]
[Cu+] = 1.0 * 10^-6 / 0.021
[Cu+] = 4.76 *10^-5 M
Ksp(CuI) = [Cu] [I]
5.1 * 10^-12 = [Cu+] [I-]
5.1 * 10^-12 =[Cu+] [0.017]
[Cu+] = 5.1 * 10^-12 / 0.017
[Cu+] = 3.0 *10^-10 M
[Cu+]from CuI hast the lowest concentration
A precipitate will begin to form at [Cu+] = 3.0 *10^-10 M
The precipitate formed is CuI
The acid dissociation constant is defined as Ka = [H+][A-]/[HA] where [H+], [A-] and [HA] are the concentrations of protons, conjugate base, and acid in solution respectively. Assuming this is a weak acid as the pH is quite high for a 1.35 M solution, we can assume that the change in [HA] is negligible and therefore [HA] = 1.35 M.
To calculate [H+] we can use the relationship pH = -log[H+], rearranging to give: [H+] = 10^(-pH) = 10^(-2.93) = 1.17 x 10^(-3).
Since the acid is relatively concentrated we can assume therefore that [H+] = [A-] as for each proton dissociated, a conjugate base is formed.
Therefore, we can calculate Ka as:
Ka = [H+]^2/[HA] = (1.17 x 10^-3 M)^2/1.35 = 1.01 x 10^-6 M
Answer:
1.44mole of CO
Explanation:
The reaction equation is given as:
5C + 2SO₂ → CS₂ + 4CO
We check to see if the expression is balanced and it is so;
Now;
Given;
1.8mole of C reacted; how many moles of CO are produced;
From the balanced reaction equation:
5 mole of C is expected to produce 4 mole of CO
1.8 mole of C will then produce 
= 1.44mole of CO
