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joja [24]
3 years ago
6

A cook had a jar containing a sweet food and a jar containing a sour food. The image above shows the sweet and sour foods. At ro

om temperature, both foods are liquids. The same amount of energy was transferred into both substances. Later, one of the foods had changed phase while the other had not. Which food changed phase, and how did it change? PLEASE ANSWER ASAP
Chemistry
1 answer:
mamaluj [8]3 years ago
5 0

Answer: One of the foods was sour and one was sweet

Explanation:

The sweet one has changed because it has more sugar then the sour one, It either got moldy or melted.

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Assuming 100% dissociation, calculate the freezing point and boiling point (in °C) of a solution with 83.6 g of AgNO 3 in 1.00 k
Thepotemich [5.8K]

Answer:

  • <u>Freezing point: - 1.83ºC</u>
  • <u>Boiling point: 100.50ºC</u>

Explanation:

The <em>freezing point</em> and<em> boiling point</em> of solvents, when a solute is added, will change accordingly to the concentration of the solute particles.

The freezing point will decrease and the boiling point will increase. These are two colligative properties.

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3 years ago
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allochka39001 [22]
MgCl2 because it is the only option in which a metal appears with a nonmetal. In this case, the metal transfers electrons to the nonmental because the metal has a lower ionization energy.
7 0
3 years ago
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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}

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:

\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.

4 0
3 years ago
Show all work and units cancellation. 1000mL = 1 L, 1000 g = 1kg
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Answer:

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2 years ago
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Otrada [13]

Answer:

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