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3 years ago

Select all statements that are correct:____

Chemistry

1 answer:

topjm [15]3 years ago

4 0

Answer:

A, C and D are correct.

Explanation:

Hello.

In this case, since the relationship between the vapor pressure of a solution is directly proportional to the mole fraction of the solvent and the vapor pressure of the pure solvent as stated by the Raoult's law:

$P_{vap}^{solution}=x_{solvent}P_{solvent}$

Since the solute is not volatile, the mole fraction of the solute is not taken into account for vapor pressure of the solution, therefore A is correct whereas B is incorrect.

Moreover, since the higher the vapor pressure, the weaker the intermolecular forces due to the fact that less more molecules are like to change from liquid to vapor and therefore more energy is required for such change, we can evidence that both C and D are correct.

Best regards.

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PLEASE HELP ASAP<br> Label the correct descriptors (volume, temperature, pressure, moles) *

Zanzabum

Explanation:

1. volume

2. pressure

3. moles

4. temperature

5. temperature

6. moles

7. pressure

8. volume

9. pressure

10. volume

5 0

2 years ago

The electron pair in a C-F bond could be considered Question 3 options: closer to C because carbon has a larger radius and thus

Leokris [45]

Answer:

closer to F because fluorine has a higher electronegativity than carbon

Explanation:

Electronegativity refers to the ability of an atom in a bonding situation to draw the shared electrons of the bond closer to itself.

Electronegativity increases across the period and decreases down the group. A highly electronegative atom draws the shared electron pair of a bond towards itself.

When two atoms are bonded together, the electron pair is always drawn closer to the atom that has a higher electronegativity.

Hence, the electron pair in a C-F bond could be considered closer to F because fluorine has a higher electronegativity than carbon.

4 0

3 years ago

The cost of painting the circular traffic sign shown below is $3.50 per square foot. How much, to the nearest dollar, will it co

KiRa [710]

The cost of painting the traffic sign of 7.065 square feet is approximately equal to 25 dollar.

<u>Explanation:</u>

The cost of painting the circular traffic sign is given as 3.50 dollar per square feet. The diameter of the traffic sign is 36 inch, then its radius will be $\frac{36}{2} = 18$ inches.

But as the cost is given in unit of feet, we have to convert the radius from inches to feet.

1 inches = 0.0833 feet

18 inches = 18 × 0.0833 feet

So, the radius of the traffic sign will be approximately equal to 1.5 feet.

The area of the traffic sign will be $\pi r^{2} = 3.14 \times 1.5 \times 1.5 = 7.065$ square feet.

So, if the cost of painting 1 square feet of traffic sign is 3.50 dollar, then

cost of painting 7.065 square feet of traffic sign = 3.50 × 7.065 = 24.7 dollar.

Thus, the cost of painting the traffic sign of 7.065 square feet is approximately equal to 25 dollar.

8 0

3 years ago

Using at least two properties of ionic compounds, explain why cookware is not made from ionic compounds.

muminat

Answer:

They mostly dissolve in water, and they are poor conductors of heat.

7 0

3 years ago

A sample of CaCO3 (molar mass 100. g) was reported as being 30. percent Ca. Assuming no calcium was present in any impurities, c

natka813 [3]

Answer:

Approximately 75%.

Explanation:

Look up the relative atomic mass of Ca on a modern periodic table:

Ca: 40.078.

There are one mole of Ca atoms in each mole of CaCO₃ formula unit.

The mass of one mole of CaCO₃ is the same as the molar mass of this compound: $\rm 100\; g$ .
The mass of one mole of Ca atoms is (numerically) the same as the relative atomic mass of this element: $\rm 40.078\; g$ .

Calculate the mass ratio of Ca in a pure sample of CaCO₃:

$\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} = \frac{40.078}{100} \approx \frac{2}{5}$ .

Let the mass of the sample be 100 g. This sample of CaCO₃ contains 30% Ca by mass. In that 100 grams of this sample, there would be $\rm 30 \% \times 100\; g = 30\; g$ of Ca atoms. Assuming that the impurity does not contain any Ca. In other words, all these Ca atoms belong to CaCO₃. Apply the ratio $\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} \approx \frac{2}{5}$ :

$\begin{aligned} m\left(\mathrm{CaCO_3}\right) &= m(\mathrm{Ca})\left/\frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)}\right. \cr &\approx 30\; \rm g \left/ \frac{2}{5}\right. \cr &= 75\; \rm g \end{aligned}$ .

In other words, by these assumptions, 100 grams of this sample would contain 75 grams of CaCO₃. The percentage mass of CaCO₃ in this sample would thus be equal to:

$\displaystyle 100\%\times \frac{m\left(\mathrm{CaCO_3}\right)}{m(\text{sample})} = \frac{75}{100} = 75\%$ .

3 0

3 years ago