In this item, we calculate the moles of NaCl with the basis of 1L of solution.
First, calculate the mass of the solution using the density.
m = (1L) x (1000g/1000mL)(1000mL/1L) = 1000g
Then, we calculate mass of NaCl by multiplying the calculate value with the percentage.
m of NaCl = (1000 g)(0.0090) = 9 g
Then, calculate the moles of NaCl by the molar mass.
molarity = (9g)(1 mole/ 58.44 g) = 0.1539 mole / L solution or 0.1529M. Then, the answer to this item is letter C.
Answers are:
Catabolism:
- g<span>enerally exergonic (spontaneous): In this reactions energy is released.
- </span><span>convert NAD+ to NADH. Electrons and protons released in reactions are attached to NAD+.
- </span><span>generation of ATP. ATP is synthesis from ADP.
- </span><span>convert large compounds to smaller compounds. Foe example starch to monosaccaharides.
Anabolism:
</span><span>- convert NADPH to NADP+. Protons and electrons are used to make chemical bonds.
</span>- <span>convert small compounds to larger compounds.</span>
Answer is: <span>the molarity of HCl is </span>0.097 M.
Chemical reaction: LiOH + HCl → LiCl + H₂O.
V(HCl) = 13.60 mL - 1.25 mL = 12.35 mL.
V(LiOH) = 11.20 mL - 2.65 mL = 8.55 mL.
c(LiOH) = 0.140 M.
From chemical reaction: n(LiOH) : n(HCl) = 1 : 1.
c(HCl) · V(HCl) = c(LiOH) · V(LiOH).
c(HCl) = 8.55 mL · 0.140 M / 12.35 mL.
c(LiOH) = 0.097 M.
Answer:
I think u are supposted to find out the are the missing things
Explanation:
The theoretical mass of is 110.
<u>Explanation:</u>
Mass of one atom of calcium = 40 .
Mass of one atom of chlorine=35.5 .
Hence , two atoms would be
Therefore, mass of calcium chloride would be net sum both constituent atoms
=
So the theoretical mass of is 110.
The theoretical mass is the measure of item coming about because of an ideal chemical reaction, and in this manner not equivalent to the sum you'll really get from a response in the lab. It is the measure of item coming about because of an ideal chemical reaction, and therefore not equivalent to the sum you'll really get from a response in the lab.