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3 years ago

7

If you have a 1.0 m aqueous solution of NaCl, by how much will it increase the water’s boiling point, if KB = 0.512 °C/m? In oth

er words, what is the boiling point elevation (increase)?

1 answer:

fgiga [73]3 years ago

5 0

<u>Answer:</u> The elevation in boiling point is 1.024°C.

<u>Explanation:</u>

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=ik_b\times m$

where,

i = Van't Hoff factor = 2 (for NaCl)

$\Delta T_b$ = change in boiling point = ?

$k_b$ = boiling point constant = $0.512^oC/m$

m = molality = 1.0 m

Putting values in above equation, we get:

$\Delta Tb=2\times 0.512^oC/m\times 1.0m\\\\\Delta Tb=1.024^oC$

Hence, the elevation in boiling point is 1.024°C.

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The membrane that surrounds a certain type of living cell has a surface area of 5.1 x 10-9 m2 and a thickness of 1.4 x 10-8 m.

ziro4ka [17]

a) The charge on the outer surface is $1.2\cdot 10^{-12} C$

b) The number of ions is $7.5\cdot 10^6$

Explanation:

a)

The membrane behaves as a parallel plate capacitor, whose capacity is given by the equation

$C=\frac{k\epsilon_0 A}{d}$

where

k = 4.3 is the dielectric constant

$\epsilon_0 =8.85\cdot 10^{-12} F/m$ is the vacuum permittivity

$A=5.1\cdot 10^{-9} m^2$ is the surface area

$d=1.4\cdot 10^{-8} m$ is the distance between the two plates

Substituting,

$C=\frac{(4.3)(8.85\cdot 10^{-12})(5.1\cdot 10^{-9})}{1.4\cdot 10^{-8}}=1.4\cdot 10^{-11} F$

The capacity of the membrane is related to the potential difference between the two surfaces by

$C=\frac{Q}{\Delta V}$

where here we have

Q = excess charge on one surface

$\Delta V = 85.5 mV = 0.0855 V$ is the potential difference between the two surfaces

Solving for Q, we find

$Q=C\Delta V=(1.4\cdot 10^{-11})(0.0855)=1.2\cdot 10^{-12} C$

b)

We said that the net charge on the outer surface is

$Q=1.2\cdot 10^{-12} C$

The charge of one K+ ions is equal to the electron charge

$+e=1.6\cdot 10^{-19} C$

Therefore, the number of ions on the outer surface can be found by dividing the total charge by the charge of a single ion:

$N=\frac{Q}{e}=\frac{1.2\cdot 10^{-12}}{1.6\cdot 10^{-19}}=7.5\cdot 10^6$

Learn more about capacity:

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3 years ago

A block and tackle is used to lift an automobile engine that weighs 1800 N. The person exerts a force of 300 N to lift the engin

Sidana [21]

Answer:

1800/300 = 6ropes

Explanation:

The engine weighs 1800N and the person exerts a force of 300N, so for him to lift the engine and exerting a force of 300N all through we divide the weight of the engine by the force exerted to know how many ropes are used. Which makes it 6 thereby each rope uses 300N to lift the engine.

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3 years ago

I need help with answer 51.

Troyanec [42]

Answer:

2

Explanation:

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2 years ago

4. How much force is required to stop a 60 kg person traveling at 30 m/s during a time of a)

11111nata11111 [884]

Explanation:

F = ma, and a = Δv / Δt.

F = m Δv / Δt

Given: m = 60 kg and Δv = -30 m/s.

a) Δt = 5.0 s

F = (60 kg) (-30 m/s) / (5.0 s)

F = -360 N

b) Δt = 0.50 s

F = (60 kg) (-30 m/s) / (0.50 s)

F = -3600 N

c) Δt = 0.05 s

F = (60 kg) (-30 m/s) / (0.05 s)

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3 0

3 years ago

Say I have a series circuit with 20v and four 65 ohm resistors, what is the current in each resistor?

Komok [63]

Data:

$E = 20 V$
$R_{1} = 65\Omega$
$R_{2} = 65\Omega$
$R_{3} = 65\Omega$
$R_{4} = 65\Omega$

<span>Now that we have all the values we need properly identified, simply calculate the equivalent total resistance of the circuit and the intensity of the total electric current using the Ohm's Law:

</span> $R_{T} = R_{1} + R_{2} + R_{3} + R_{4}$
$R_{T} = 65 + 65 + 65+ 65$
$R_{T} = 260\Omega$

<span>Like this:
</span>
$I_{T} = \frac{E}{ R_{T} }$

$I_{T} = \frac{20}{ 260 }$
$I_{T} = 0,076923076...$

$\boxed{\boxed{I_{T} \approx 0,07A}}$
Answer:
<span>The intensity of the total electric current
</span> $\boxed{\boxed{I_{T} \approx 0,07A}}$

P.S:. Since the association is in series, the current of 0.07A is the same for all resistors.

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3 years ago