Answer:
We could do two 1:50 dilutions and one 1:4 dilutions.
Explanation:
Hi there!
A solution that is 1000 ug/ ml (or 1000 mg / l) is 1000 ppm.
Knowing that 1 ppm = 1000 ppb, 100 ppb is 0.1 ppm.
Then, we have to dilute the stock solution (1000 ppm / 0.1 ppm) 10000 times.
We could do two 1:50 dilutions and one 1:4 dilutions (50 · 50 · 4 = 10000). Since the first dilution is 1:50, you will use the smallest quantity of the stock solution (if we use the 10.00 ml flask):
First step (1:50 dilution):
Take 0.2 ml of the stock solution using the third dispenser (20 - 200 ul), and pour it in the 10.00 ml flask. Fill with water to the mark (concentration : 1000 ppm / 50 = 20 ppm).
Step 2 (1:50 dilution):
Take 0.2 ml of the solution made in step 1 and pour it in another 10.00 ml flask. Fill with water to the mark. Concentration 20 ppm/ 50 = 0.4 ppm)
Step 3 (1:4 dilution):
Take 2.5 ml of the solution made in step 3 (using the first dispenser 1 - 5 ml) and pour it in a 10.00 ml flask. Fill with water to the mark. Concentration 0.4 ppm / 4 = 0.1 ppm = 100 ppb.
In this item, we are simply to find the ions that may bond and are able to form a formula unit. We are also instructed to give out their name. There are numerous possible combinations of ions to form a compound. Some answers are given in the list below.
1. Na⁺ , Cl⁻ , NaCl ---> sodium chloride (this is most commonly known as table salt)
2. C⁴⁺ , O²⁻ , CO₂ ---> carbon dioxide
3. Al³+ , Cl⁻ , AlCl₃ ----> aluminum chloride
4. Ca²⁺ , Cl⁻ , CaCl₂ ---> calcium chloride
5. Li⁺ , Br⁻ , LiBr ---> lithium bromide
6. Mg³⁺ , O²⁻ , Mg₂O₃ ----> magnesium oxide
7. K⁺ , I⁻ , KI ---> potassium iodide
8. H⁺ , Cl⁻ , HCl --> hydrogen chloride
9. H⁺ , Br⁻ , HBr ----> hydrogen bromide
10. Na⁺ , Br⁻ , NaBr ---> sodium bromide
The answer to this is Drinking Waer
Answer:
20619.4793 years
Explanation:
The half life of carbon-14 = 5730 years
The formula for the half life for a first order kinetic reaction is:
Where,
is the half life
k is the rate constant.
Thus rate constant is:
5730 years=ln(2)/k
k = 1.21×10⁻⁴ years ⁻¹
Using integrated rate law as:
Where,
is the concentration at time t
is the initial concentration
Given that the final concentration contains 8.25 % of the original quantity which means that:
So,
ln(.0825)= -1.21×10⁻⁴×t
<u>
t = 20619.4793 years</u>
<u></u>
Answer:
1) Se2O5
2) I2O6
3)Zn3n2
4) Cr(HCO3)3
Explanation:
selenium pentaoxide (= also called diselenium pentoxide)
= Se2O5
⇒ Se = 78.97 g/mol
⇒ O = 16 g/mol
⇒ 2*78.97 + 5*16 = 237.94 g/mol
iodine trichloride
= I2O6
⇒ I = 126.9 g/mol
⇒ Cl = 35.45 g/mol
⇒ 2* 126.9 + 6 * 35.45 = 466.5 g/mol
zinc (1) nitride does not exist (it's Zinc(ii)nitride
The oxidation number for zinc is always 2
Zn3n2
⇒ Zn = 65.38 g/mol
⇒ N = 14 g/mol
⇒3*65.38 + 2* 14 = 224.14 g/mol
chromium (III) bicarbonate
Cr(HCO3)3
⇒ Cr = 52 g/mol
⇒ H = 1.01 g/mol
⇒ C = 12 g/mol
⇒ O = 16 g/mol
52 + 3*1.01 + 3*12 + 6*16 = 235.03 g/mol
