A solution is prepared by adding 1.43 mol of potassium chloride (kcl) to 889 g of water. The concentration of kcl is 1.61 molal.
mol of Kcl (potassium chloride)= 1.43
water = 889 g
the formula for calculating molality is:
molality = moles of solute/kilograms of solvent
1kg = 1000g so, 889g = 0.889kg
m = 1.43/0.889 = 1.61 molal
Answer:
Mass of solution=100g
mass of salt=20g
so; mass of solute=80g
percentage composition =(mass of salt/total
mass) ×100
= \frac{20}{100} \times 100 \\ = 20\%
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Answer:
Energy lost is 7.63×10⁻²⁰J
Explanation:
Hello,
I think what the question is requesting is to calculate the energy difference when an excited electron drops from N = 15 to N = 5
E = hc/λ(1/n₂² - 1/n₁²)
n₁ = 15
n₂ = 5
hc/λ = 2.18×10⁻¹⁸J (according to the data)
E = 2.18×10⁻¹⁸ (1/n₂² - 1/n₁²)
E = 2.18×10⁻¹⁸ (1/15² - 1/5²)
E = 2.18×10⁻¹⁸ ×(-0.035)
E = -7.63×10⁻²⁰J
The energy lost is 7.63×10⁻²⁰J
Note : energy is lost / given off when the excited electron jumps from a higher energy level to a lower energy level
Answer:
9.
a. NH2CH2COOH
b. The function group is what I put in bold.
c. carboxylic acid and amine
10.
a. NH2CH(CH3)COOH
b. The functional group is in bold.
c. carboxylic acid and amine
Explanation:
NH2 is amine (amino acid)
COOH is the carboxylic acid (acetic acid)
Answer:
Hydrogen.
Explanation:
We see in the nucleus that we have 1 proton and 1 neutron.
The number of protons in the nucleus determines which element it is.
Element with 1 proton is the 1st element on the Periodic Table, Hydrogen.