Answer:
1. an educated guess
2. data
3. what changes in experiment
4. what stays the same in both groups
5. the group where nothing changes, normal
6. group with independent variable, what's being tested
Na₂S(aq) + Cd(NO₃)₂(aq) = CdS(s) + 2NaNO₃(aq)
v=25.00 mL
c=0.0100 mmol/mL
M(Na₂S)=78.046 mg/mmol
n(Na₂S)=n{Cd(NO₃)₂}=cv
m(Na₂S)=M(Na₂S)n(Na₂S)=M(Na₂S)cv
m(Na₂S)=78.046*0.0100*25.00≈19.5 mg
Answer:
1.022ppm is the unknown concentration of the metal
Explanation:
Based on Lambert-Beer law, the increasing in signal of a detector is directly proportional to its concentration.
The unknown concentration (X) produces a signal of 0.255
99mL * X + 1mL * 100ppm / 100mL produces a signal of 0.502
0.99X + 1ppm produce 0.502, thus, X is:
0.255 * (0.99X + 1 / 0.502) =
X = 0.503X + 0.508
0.497X = 0.508
X =
1.022ppm is the unknown concentration of the metal
They both have two electron shells. Period indicates number of shells.
Answer:
= 20.82 g of BaCl2
Explanation:
Given,
Volume = 200 mL
Molarity = 0.500 M
Therefore;
Moles = molarity × volume
= 0.2 L × 0.5 M
= 0.1 mole
But; molar mass of BaCl2 is 208.236 g/mole
Therefore; 0.1 mole of BaCl2 will be equivalent to;
= 208.236 g/mol x 0.1 mol
= 20.82 g
Therefore, the mass of BaCl2 in grams required is 20.82 g
