Answer:
0.190 M
Explanation:
Let's consider the neutralization reaction between HCl and NaOH.
HCl + NaOH = NaCl + H2O
11.9 mL of 0.160 M NaOH were used. The reacting moles of NaOH were:
0.0119 L × 0.160 mol/L = 1.90 × 10⁻³ mol
The molar ratio of HCl to NaOH is 1:1. The reacting moles of HCl are 1.90 × 10⁻³ moles.
1.90 × 10⁻³ moles of HCl are in 10.0 mL of solution. The molarity of HCl is:
M = 1.90 × 10⁻³ mol / 10.0 × 10⁻³ L = 0.190 M
In this question, you are given the average cofactor mass per cell (41.5pg) and the total cells count(105 cells). You are asked how much cofactor that will be found from those cells(microgram= 10^6 picogram). Then the calculation would be:
Cofactor mass= cofactor per cell * cell count= 41.5pg/cell * 105 cells= 4357.5pg= 4.36 x 10^3pg
Then convert the picogram(pg) into microgram: 4.36 x 10^3pg/ (10^6pg/microgram)= 4.36x10^-3 microgram or 0.00436 microgram
if 105 cells mean 10^5 cells, the answer should be 4.15 microgram
Answer:
6.72M of HNO3
Explanation:
In the problem you are diluting the original HNO3 solution by the addition of some water. The final volume is:
290.7mL + 350.0mL = 640.7mL
And you are diluting the solution:
640.7mL / 350.0mL = 1.8306 times
As the original concentration was 12.3M, the final concentration will be:
12.3M / 1.8306 =
<h3>6.72M of HNO3</h3>
Answer:
2.4 litters of water are required.
Explanation:
Given data:
Mass of LiF = 19.6 g
Molarity of solution = 0.320 M
Volume of water used = ?
Solution:
Number of moles = mass/molar mass
Number of moles = 19.6 g/ 26 g/mol
Number of moles = 0.75 mol
Volume required:
Molarity = number of moles/ volume in L
Now we will put the values in above given formula.
0.320 M = 0.75 mol / volume in L
Volume in L = 0.75 mol /0.320 M
M = mol/L
Volume in L = 2.4 L
Answer:
THE CURRENT REQUIRED TO PRODUCE 193000 C OF ELECTRICITY IS 35.74 A.
Explanation:
Equation:
Al3+ + 3e- -------> Al
3 F of electricity is required to produce 1 mole of Al
3 F of electricity = 27 g of Al
If 18 g of aluminium was used, the quantity of electricity to be used up will be:
27 g of AL = 3 * 96500 C
18 G of Al = x C
x C = ( 3 * 96500 * 18 / 27)
x C = 193 000 C
For 18 g of Al to be produced, 193000 C of electricity is required.
To calculate the current required to produce 193 000 C quantity of electricity, we use:
Q = I t
Quantity of electricity = Current * time
193 00 = I * 1.50 * 60 * 60 seconds
I = 193 000 / 1.50 * 60 *60
I = 193 000 / 5400
I = 35.74 A
The cuurent required to produce 193,000 C of electricity by 18 g of aluminium is 35.74 A
