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Otrada [13]
3 years ago
9

A chemist dissolves 1.3 mol NaCl in a 2.0-kg sample of water. What is the boiling point of this solution? (Assume that pure wate

r boils at 100°C and that Kb for water = 0.512°C/m.)
Chemistry
2 answers:
mafiozo [28]3 years ago
8 0
For this problem, the solution is exhibiting some colligative properties since the solute in the solution interferes with some of the properties of the solvent. We use equation for the boiling point elevation for this problem. We do as follows:
<span>
ΔT(boiling point)  = (Kb)mi
</span>ΔT(boiling point)  = (0.512)(1.3/2.0)(2)
ΔT(boiling point)  = 0.67 degrees Celsius
<span>
T(boiling point) = 100 + 0.67 = 100.67 degrees Celsius</span>
lara31 [8.8K]3 years ago
7 0

Answer:

100.67°C

Explanation:

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What is a common land form that is formed when chemical weathering, specifically carbonation, is taking place?
skad [1K]

Answer:

plateaus

Explanation:

6 0
3 years ago
If a proton and an electron are released when they are 5.50×10−10 mm apart (typical atomic distances), find the initial accelera
Vlada [557]

Answer: The initial acceleration of the proton = (4.56 × 10^23) m/s2

The initial acceleration of the electron = (8.36 × 10^26) m/s2

Explanation: The force of attraction between the proton and electron can be computed using the statements of Coulomb's law which state that the force of attraction between two charged particles is directly proportional to the product of the two charges and inversely proportional to the square of their distances apart.

F = (Kq1q2)/(r^2) where K = (9 × (10^9) Nm(C^-2))

But q1 is the charge on a proton = (1.6 × (10^-19)) C

q2 is charge on an electron = -(1.6 × (10^-19)) C

r = (5.50 × (10^-10))mm = (5.50 × (10^-13))m

Computing all that, F = 0.0007616529 N = (7.62 × 10^-4) N

But the force of attraction is converted to that required for motion when they're released.

F = ma.

For proton, m = (1.67 × 10^-27) kg

a = F/m = 0.000762/(1.67 × 10^-27) = (4.56 × 10^23) m/s2

For electron, m = (9.11 × 10^-31) kg

a = F/m = 0.000762/(9.11 × 10^-31) = (8.36 × 10^26) m/s2

QED!

7 0
3 years ago
A sample of gas has a volume of 215 cm3 at 23.5 degrees Celsius and 84.6kPa what volume will the gas occupy at stp
miss Akunina [59]

The volume of the gas that occupy at STP is 165. 28 cm^3

calculation

by use of combined gas law that is P1V1/T1=P2V2/T2, where

P1=84.6 kpa

T1=23.5 +273=296.5 K

V1=215 cm^3

At STP T= 273 K and P= 101.325 Kpa

therefore p2 = 101.325 Kpa and T2 = 272 K V2=?

by making V2 the subject of the formula V2 =T2P1V1/P2T1

V2 = 273 K x 84.6 Kpa x 215 cm^3/ 101,.325 Kpa x296.5 K =165.28 cm^3

5 0
3 years ago
Read 2 more answers
Weight is measured as the downward force of __________. a. gravity c. acceleration b. friction d. motion
Murljashka [212]
A. Gravity, hope this helps
6 0
3 years ago
Read 2 more answers
What does a bond do?
frozen [14]

Answer:

chemical substances being held together by attraction of atoms to each other through sharing, as well as exchanging, of electrons -or electrostatic forces.

Explanation:

I hope this helps

5 0
3 years ago
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