Answer:
She will need to add 5.6 cups of water
Explanation:
Hi there!
Marta has to dilute the citric acid to 5%.
The dilution factor will be 12%/ 5% = 2.4. Then, Marta will need to dilute the citric acid 2.4 times. If she has 4 cups of the solution, she will need to add water until she completes a volume of (4 cups ·2.4) 9.6 cups to reach the desired concentration.
Then, she will need to add 9.6 - 4 cups water = 5.6 cups of water
Another way to solve this is by using the fact that the mass of citric acid in the concentrated and diluted solution is the same. Then:
mass citric acid concentrated solution = mass citric acid in dilute solution
mass of citric acid = concentration · volume
Then:
initial concentration · volume = final concentration · volume
12% · 4 cups = 5% · volume
volume = 12% · 4 cups / 5% = 9.6 cups
The final volume of the solution at 5% will be 9.6 cups. So Marta will need 9.6 cups - 4 cups = 5.6 cups water
Since both numbers contain 2 significant digits the answer will contain 2 significant digits:
Answer: b 1.9 x 104
<span>The answer is lactic acid/lactate. This is caused by
anaerobic respiration in muscles especially during strenuous physical activity due to low oxygen supply to muscles in respect
to the demand. When there is sufficient oxygen in the muscles, pyruvate is broken down
aerobically in the Kreb’s cycle. However,
it is converted to lactate in the anaerobic
pathway. Lactic-acidosis occurs when
lactic acid accumulates in tissues and slows down metabolic pathways</span>
8.948 I’m pretty sure I hope this helped
Answer:
18.76 g of copper II nitrate
Explanation:
Now recall that we must use the formula;
n= CV
Where;
n= number of moles of copper II nitrate solid
C= concentration of copper II nitrate solution
V= volume of copper II nitrate solution
Note that;
n= m/M
Where;
m= mass of solid copper II nitrate
M= molar mass of copper II nitrate
Thus;
m/M= CV
C= 0.05 M
V= 2.00 L
M= 187.56 g/mol
m= the unknown
Substituting values;
m/ 187.56 g/mol = 0.05 M × 2.00 L
m= 0.05 M × 2.00 L × 187.56 g/mol
m= 18.76 g of copper II nitrate
Therefore, 18.76 g of copper II nitrate is required to make 0.05 M solution of copper II nitrate in 2.00 L volume.
