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pickupchik [31]
3 years ago
13

How much ice at 0°C could be melted by the addition of 15 kJ of heat? (ΔHfus = 6.01 kJ/mol)

Chemistry
2 answers:
BARSIC [14]3 years ago
7 0

Answer:

d. 45 g

Explanation:

The amount of ice melted is:

\Delta n = \frac{15\,kJ}{6.01\,\frac{kJ}{mol} }

\Delta n = 2.496\,moles

\Delta m = \Delta n \cdot M_w

\Delta m = (2.496\,moles)\cdot \left(18\,\frac{g}{mol} \right)

\Delta m = 44.928\,g

anzhelika [568]3 years ago
5 0

Answer: d.45 g

Explanation:see attached photo

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Which of the following pairs consists of a weak acid and a strong base? (1 point)
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Answer:

acetic acid, sodium hydroxide

Explanation:

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CH3COOH <=> H^+ + CH3COO^-

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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}

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Hello,

The suitable differential equation for this case is:

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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:

\frac{dh}{dt}_{h=2cm} =\frac{160}{9\pi 2^2}\\\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{s}

Best regards.

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