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ANTONII [103]
2 years ago
11

What is the value of freezing point depression of a solution of 15.0g of NaCl in 250g of water?

Chemistry
1 answer:
kobusy [5.1K]2 years ago
7 0

Answer:

-3.82ºC is the freezing point of solution

Explanation:

We work with the Freezing point depression to solve the problem

ΔT = m . Kf . i

ΔT = Freezing point of pure solvent - freezing point of solution

Let's find out m, molality (moles of solute in 1kg of solvent)

15 g / 58.45 g/mol = 0.257 moles of NaCl

NaCl(s) → Na⁺ (aq) + Cl⁻(aq)

i = 2 (Van't Hoff factor, numbers of ions dissolved)

m = mol /kg → 0.257 mol / 0.250kg = 1.03 m

Kf = Cryoscopic constant → 1.86 ºC/m (pure, for water)

0ºC - Tºf = 1.03m . 1.86ºC/m . 2

Tºf  = -3.82ºC

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A fossil was analyzed and determined to have a carbon-14 level that is 70 % that of living organisms. The half-life of C-14 is 5
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Answer: 2948

Explanation:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

t_{\frac{1}{2}}=\frac{0.693}{k}

k=\frac{0.69}{t_{\frac{1}{2}}}=\frac{0.693}{5730}=1.21\times 10^{-4}years^{-1}

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = 1.21\times 10^{-4}years^{-1}

t = age of sample  = ?

a = let initial amount of the reactant  = 100

a - x = amount left after decay process = \frac{70}{100}\times 100=70

t=\frac{2.303}{1.21\times 10^{-4}}\log\frac{100}{70}

t=2948years

Thus the fossil is 2948 years old.

5 0
2 years ago
How many atoms of both elements are there in this molecule: 3N205; O 3 Nitrogen and 15 Oxygen 6 Nitrogen and 5 Oxygen KD 2 Nitro
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6 0
3 years ago
Lead−206 is the end product of 238u decay. one 206pb atom has a mass of 205.974440 amu. (a) calculate the binding energy per nuc
Dovator [93]
The atomic number for Pb is 82
∴ Pb has 82 protons and 206-82 = 14 protons
The actual mass of Pb nuclei is
=(82 × mass of the proton) + (124 × mass of neutron)
=(82× 1.00728) + (124 × 1.008664) amu
= 207.6713 amu
The mass of lead which is given is 205.9744 amu
∴mass defect is
m = 207.6713 - 205.9744 = 1.6969 amu
=1.6969 × 1.66054 × 10⁻²⁷kg
=2.818 × 10⁻²⁷kg
The binding energy is E = mc²
C is the speed of light in vacuum = 2.9979 × 10⁸m/s
∴ E = 2.532 × 10×⁻¹⁰ J/mol
= 2.532 × 10⁻¹⁰ × 6.023 × 10²³ J/mol 
= 1.53811 × 10¹⁴ J/mol

8 0
3 years ago
Which of the following is the correct isotope notation for the element above?
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7 becuse it splits in half
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2 years ago
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