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3 years ago

Please only answer if you know the actual thing, don't search it up! Are these correct? :P

Chemistry

2 answers:

makkiz [27]3 years ago

4 0

Hey buddy I am here to help!

1. C

2. A

3. A & B

4. C

5. C

6. A

7. A

8. A & C

Hope it helps!

Plz mark brainlist!

lawyer [7]3 years ago

3 0

Answer:

1. C

2. A

3. A & B

4. C

5. C

6. A

7. A

8. A & C

Explanation:

Hope you have a fantastic day! <3

You might be interested in

What is the freezing point in °C) of a 0.743 m

lesantik [10]

Answer:

-0.276 Degrees C

Explanation:

kf of water is 1.86

Freezing point depression= m x kf x i

i= ions present ( K+ Cl-) 1=2

1.86x2X.743= .276

Since its freezinf point depression the freezing point will lower

0-.276= -.276

7 0

3 years ago

PLEASE HELP!! Thanks! How much heat (in kJ) is required to warm 13.0 g of ice, initially at -10.0 ∘C, to steam at 111.0 ∘C? The

ZanzabumX [31]

Answer:

Approximately 39.7 kJ.

Assumptions: the specific heat capacity of water is $\rm 4.182\; J \cdot mol^{-1}$ , the melting point of water is $\rm 0\, ^{\circ} C$ , and that the boiling point of water is $\rm 100 \,^{\circ} C$ .

Explanation:

It takes five steps to convert 13.0 grams of $\rm \text{-}10.0\, ^{\circ}C$ ice to steam at $\rm 111.0\,^{\circ}C$ .

Step one: heat the 13.0 gram of ice from $\rm \text{-}10.0\, ^{\circ}C$ to $\rm 0\,^{\circ}C$ . The change in temperature would be $\rm 10.0\,^{\circ}C$ .
Step two: supply the heat of fusion to convert that 13.0 gram of ice to water.
Step three: heat the 13.0 gram of water from $\rm 0\,^{\circ}C$ to $\rm 100\,^{\circ}C$ . The change in temperature would be $\rm 100\,^{\circ}C$ .
Step four: supply the heat of vaporization to convert that 13.0 gram of water to steam.
Step five: heat the 13.0 gram of steam from $\rm 100\,^{\circ}C$ to $\rm 111.0\,^{\circ}C$ . The change in temperature would be $\rm 11.0\,^{\circ}C$ .

<h3>Energy required for step one, three, and five</h3>

The following equation gives the amount of energy $Q$ required to raise the temperature of an object by a $\Delta T$ :

$Q = c \cdot m \cdot \Delta T$ .

In this equation,

$c$ is the specific heat of this substance,
$m$ is the mass of the substance, and
$\Delta T$ is the change in the temperature of the object.

Assume that there's no mass loss in this whole process. The value of $m$ would stay the same at $13.0\; \rm g$ .

$\begin{aligned}& &&\text{Energy required for raising temperature} \cr &=&& c(\text{Ice}) \cdot m \cdot \Delta(\text{Ice}) \cr & && + c(\text{Water}) \cdot m \cdot \Delta(\text{Water})\cr & && + c(\text{Steam}) \cdot m \cdot \Delta(\text{Steam}) \cr & = && (2.09 \times 13.0 \times 10) \cr & && + (4.182 \times 13.0 \times 100) \cr & &&+ ( 2.01 \times 13.0 \times 10) \cr & = && 5969.6\;\rm J \cr & = && 5.969\; \rm kJ\end{aligned}$ .

<h3>Energy required for step two and four</h3>

The equations for the energy of fusion and energy of vaporization are quite similar:

$E(\text{Fusion}) = n \cdot \Delta H_\text{Fusion}$ .

$E(\text{Vaporization}) = n \cdot \Delta H_\text{Vaporization}$ .

where $n$ is the number of moles of the substance.

Look up the relative atomic mass of oxygen and hydrogen from a modern periodic table:

H: 1.008,
O: 15.999.

Hence the molar mass of water:

$M(\rm H_2O) = 2\times 1.008 + 15.999 = 18.015\; g \cdot mol^{-1}$ .

Number of moles of $\rm H_2O$ molecules in $\rm 13.0\; g$ :

$\displaystyle n = \frac{m}{M} \approx 0.721621\; \rm mol$ .

$\begin{aligned}& &&\text{Energy required for phase changes} \cr &=&& n \cdot \Delta H_\text{Fusion} \cr & &&+n \cdot \Delta H_\text{Vaporization} \cr & = &&0.721621 \times 6.02 + 0.721621 \times 40.7 \cr & = &&33.7\; \rm kJ \end{aligned}$