The final concentration of the diluted standard is 0.2 mg/dL.
<h3 /><h3>What is concentration of glucose standard after 1/5 solution?</h3>
Using the dilution formula:
where
- C1 is initial concentration
- V1 initial volume
- C2 is final concentration
- V2 is final volume.
Assuming a final volume of 100 mL, and since a 1/5 dilution is made:
C1 = 1.00 mg/dL
V1 = 20
C2 = ?
V2 = 100 mL
C2 = C1V1/V2
C2 = 20 × 1/100
C2 = 0.2 mg/dL
Therefore, the final concentration of the diluted standard is 0.2 mg/dL.
Learn more about dilution at: brainly.com/question/24881505
Answer:
Kauai
Explanation:
I look this up when I am doing math
Answer:
8.934 g
Step-by-step explanation:
We know we will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.
M_r: 192.12 44.01
H₃C₆H₅O₇ + 3NaHCO₃ ⟶ Na₃C₆H₅O₇ + 3H₂O + 3CO₂
m/g: 13.00
For ease of writing, let's write H₃C₆H₅O₇ as H₃Cit.
(a) Calculate the <em>moles of H₃Cit
</em>
n = 13.00 g × (1 mol H₃Cit /192.12 g H₃Cit)
n = 0.067 67 mol H₃Cit
(b) Calculate the <em>moles of CO₂
</em>
The molar ratio is (3 mol CO₂/1 mol H₃Cit)
n = 0.067 67 mol H₃Cit × (3 mol CO₂/1 mol H₃Cit)
n = 0.2030 mol CO₂
(c) Calculate the <em>mass of CO₂
</em>
m = 0.2030 mol CO₂ × (44.01 g CO₂/1 mol CO₂)
m = 8.934 g CO₂
KE=1/2*mass*velocity^2
So u do 1/2 * 1 * 30^2
1/2 * 1 * 900
= 450kgm/s
P.s. I'm not sure if I would have to convert kg to g.
Anyways hope this helped