Answer:
0.297 mol/L
Explanation:
<em>A chemist prepares a solution of potassium dichromate by measuring out 13.1 g of potassium dichromate into a 150 mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in mol/L of the chemist's potassium dichromate solution. Be sure your answer has the correct number of significant digits.</em>
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Step 1: Calculate the moles corresponding to 13.1 g of potassium dichromate
The molar mass of potassium dichromate is 294.19 g/mol.
13.1 g × (1 mol/294.19 g) = 0.0445 mol
Step 2: Convert the volume of solution to L
We will use the relationship 1 L = 1000 mL.
150 mL × (1 L/1000 mL) = 0.150 L
Step 3: Calculate the concentration of the solution in mol/L
C = 0.0445 mol/0.150 L = 0.297 mol/L
Answer:
63.6%
Explanation:
The given compound is:
N₂O;
The problem here is to find the percent composition of nitrogen in the compound.
First find the molar mass of the compound:
Molar mass of N₂O = 2(14) + 16 = 44g/mol
So;
Percentage composition of Nitrogen = 
x 100 = 63.6%
First, find the number of moles of UF6
Avagadro's number = 6.023 x 10^23
Number of moles = 8.0 x 10^26 / Avagadro's number = 8.0 x 10^26 / 6.023 x 10^23 = 1.328 x 10³ moles
Molecular weight of UF6 = Molecular weight of U (238.02891) + Molecular weight of F6 (6 x 18.9984032) = 238.02891 + 113.9904192 = 352.0193292 g/mol
Therefore mass of 8.0 x 10^26 UF6 molecules = 352.0193292 g/mol x 1.328 x 10³ moles = 467.481669 x 10³ grams
Answer:
Kc = [H₂S]² . [CH₄] / [ H₂O]⁴ . [CS₂]
Explanation:
The equilibrium constant indicates the % of the yield reaction and can shows where the reaction is going to be equilibrated.
It works with molar concentrations on the equilibrium and it does not consider the solids compounds
Kc also can be modified by the time of the reaction.
This reaction is:
CS₂ (g) + 4 H₂O(g) ⇌ CH₄ (g) + 2H₂S (g)
Kc = [H₂S]² . [CH₄] / [ H₂O]⁴ . [CS₂]
Answer:
B. K+cation
Explanation:
Not all that sure but hope it will help.
