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ruslelena [56]
3 years ago
13

Ethanol has a Kb of 1.22 °C/m and usually boils at 78.4 °C. How many mol of an nonionizing solute would need to be added to 48.8

0 g ethanol in order to raise the boiling point to 85.2 °C.
Chemistry
1 answer:
Galina-37 [17]3 years ago
6 0

Answer:

0.272 mol

Explanation:

∆Tb = m × Kb

∆Tb = 85.2°C - 78.4°C = 6.8°C

Kb = 1.22°C/m

mass of ethanol = 48.80 g = 48.80/1000 = 0.0488 kg

Let the moles of non-ionizing solute be y

m (molality) = y/0.0488

6.8 = y/0.0488 × 1.22

y = 6.8×0.0488/1.22 = 0.272 mol

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Answer:

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Explanation:

The symbol for TIN is Sn.

When Sn loses 2 electrons, it gets a double positive charge ( +2 ) and becomes \sf Sn ^{+2}.

It becomes a cation.

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The students in the picture below are using a globe and a lamp to model the Sun and the Earth. If the model Earth acts the same
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Answer:

The only one that makes sense IF the model behaves as the Earth is D.

Explanation:

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3 years ago
Water is poured into a conical container at the rate of 10 cm3/sec. The cone points directly down, and it has a height of 20 cm
8090 [49]

Answer:

\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{2}

Explanation:

Hello,

The suitable differential equation for this case is:

\frac{dV}{dt}=10\frac{cm^3}{s}

As we're looking for the change in height with respect to the time, we need a relationship to achieve such as:

\frac{dh}{dt} = ?*\frac{dV}{dt}

Of course, ?=\frac{dh}{dV}.

Now, since the volume of a cone is V=\pi r^2h/3 and the ratio r/h=15/20=3/4 or r=3/4h, the volume becomes:

V=\pi (\frac{3}{4} h)^2h/3= \frac{3}{16}\pi h^3

We proceed to its differentiation:

\frac{dV}{dh} =\frac{9}{16} \pi h^2\\\frac{dh}{dV} =\frac{16}{9 \pi h^2}

Then, we compute \frac{dh}{dt}

\frac{dh}{dt} = \frac{16}{9 \pi h^2}*\frac{dV}{dt}\\\frac{dh}{dt} = \frac{16}{9\pi h^2}*10\frac{cm^3}{s} =\frac{160}{9 \pi h^2}

Finally, at h=2:

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Best regards.

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What describes a change in velocity?
siniylev [52]
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The heat capacity of water is 1cal degree'g1 (1 calorie per degree centigrade, per gram). You are given 1 gallon of water at 25
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Answer:

The heat needed to boil 1 gallon of water is 81,490.62 Joules.

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Q= 4546.09 g\times 1 Cal/g^oC\times 75^oC=340,956.75 Cal

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Q=\frac{340,956.75}{4.184} J = 81,490.62 J

The heat needed to boil 1 gallon of water is 81,490.62 Joules.

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