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3 years ago

What is the molecular geometry, or shape, of phosphorus trichloride (PC13)?

Chemistry

2 answers:

tankabanditka [31]3 years ago

8 0

The answer is C. Trigonal pyramidal.

Vedmedyk [2.9K]3 years ago

7 0

Answer:

the answer is c

Explanation:

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A 57.07 g sample of a substance is initially at 24.3°C. After absorbing of 2911 J of heat, the temperature of the substance is 1

icang [17]

Answer:

Approximately $0.551\; \rm J\cdot kg^{-1} \cdot \left(^\circ\! C \right)^{-1}$ .

Explanation:

The specific heat of a material is the amount of energy required to increase unit mass (one gram) of this material by unit temperature (one degree Celsius.)

Calculate the increase in the temperature of this sample:

$\Delta T = (116.9 - 24.3)\; \rm ^\circ\! C= 92.6\; \rm ^\circ\! C$ .

The energy that this sample absorbed should be proportional the increase in its temperature (assuming that no phase change is involved.)

It took $2911\; \rm J$ of energy to raise the temperature of this sample by $\Delta T = 92.6\; \rm ^\circ\! C$ . Therefore, raising the temperature of this sample by $1\; \rm ^\circ\! C$ (unit temperature) would take only $\displaystyle \frac{1}{92.6}$ as much energy. That corresponds to approximately $31.436\; \rm J$ of energy.

On the other hand, the energy required to raise the temperature of this material by $1\; \rm ^\circ\! C$ is proportional to the mass of the sample (also assuming no phase change.)

It took approximately $31.436\; \rm J$ of energy to raise the temperature of $57.07\; \rm g$ of this material by $1\; \rm ^\circ C$ . Therefore, it would take only $\displaystyle \frac{1}{57.07}$ as much energy to raise the temperature of $1\; \rm g$ (unit mass) of this material by $1\; \rm ^\circ \! C\!$ . That corresponds to approximately $0.551\; \rm J$ of energy.

In other words, it takes approximately $0.551\; \rm J$ to raise $1\; \rm g$ (unit mass) of this material by $1\; \rm ^\circ \! C$ . Therefore, by definition, the specific heat of this material would be approximately $0.551\; \rm J\cdot kg^{-1} \cdot \left(^\circ\! C \right)^{-1}$ .

8 0

3 years ago

Caroline has a condition where the amount of oxygen her blood can hold is greatly reduced. This leads to a decrease in oxygen su

Sphinxa [80]

The correct answer is : Circulatory and respiratory.

This condition primarily affects Caroline's circulatory system. Since the main purpose of the circulatory system is to supply the rest of the body with oxygen, this condition is greatly reducing the effectiveness of the circulatory system.
The decrease in oxygen supply to the lungs causes the reduction of the respiratory system effectivness as well, since the lung cells need oxygen for their metabolic processes.

5 0

3 years ago

Read 2 more answers

An automobile antifreeze mixture is made by mixing equal volumes of ethylene glycol (d = 1.114 g/mL; M = 62.07 g/mol) and water

maks197457 [2]

Answer :

(a) The volume percent is, 50.63 %

(b) The mass percent is, 52.69 %

(c) Molarity is, 9.087 mole/L

(d) Molality is, 17.947 mole/L

(e) Moles fraction of ethylene glycol is, 0.244

Explanation : Given,

Density of ethylene glycol = 1.114 g/mL

Molar mass of ethylene glycol = 62.07 g/mole

Density of water = 1.00 g/mL

Density of solution or mixture = 1.070 g/mL

According to the question, the mixture is made by mixing equal volumes of ethylene glycol and water.

Suppose the volume of each component in the mixture is, 1 mL

First we have to calculate the mass of ethylene glycol.

$\text{Mass of ethylene glycol}=\text{Density of ethylene glycol}\times \text{Volume of ethylene glycol}=1.114g/mL\times 1mL=1.114g$

Now we have to calculate the mass of water.

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1.00g/mL\times 1mL=1.00g$

Now we have to calculate the mass of solution.

Mass of solution = Mass of ethylene glycol + Mass of water

Mass of solution = 1.114 + 1.00 = 2.114 g

Now we have to calculate the volume of solution.

$\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}=\frac{2.114g}{1.070g/mL}=1.975mL$

(a) Now we have to calculate the volume percent.

$\text{Volume percent}=\frac{\text{Volume of ethylene glycol}}{\text{Volume of solution}}\times 100=\frac{1mL}{1.975mL}\times 100=50.63\%$

(b) Now we have to calculate the mass percent.

$\text{Mass percent}=\frac{\text{Mass of ethylene glycol}}{\text{Mass of solution}}\times 100=\frac{1.114g}{2.114g}\times 100=52.69\%$

(c) Now we have to calculate the molarity.

$\text{Molarity}=\frac{\text{Mass of ethylene glycol}\times 1000}{\text{Molar mass of ethylene glycol}\times \text{Volume of solution (in mL)}}$

$\text{Molarity}=\frac{1.114g\times 1000}{62.07g/mole\times 1.975L}=9.087mole/L$

(d) Now we have to calculate the molality.

$\text{Molality}=\frac{\text{Mass of ethylene glycol}\times 1000}{\text{Molar mass of ethylene glycol}\times \text{Mass of water (in g)}}$

$\text{Molality}=\frac{1.114g\times 1000}{62.07g/mole\times 1kg}=17.947mole/kg$

(e) Now we have to calculate the mole fraction of ethylene glycol.

$\text{Mole fraction of ethylene glycol}=\frac{\text{Moles of ethylene glycol}}{\text{Moles of ethylene glycol}+\text{Moles of water}}$

$\text{Moles of ethylene glycol}=\frac{\text{Mass of ethylene glycol}}{\text{Molar of ethylene glycol}}=\frac{1.114g}{62.07g/mole}=0.01795mole$

$\text{Moles of water}=\frac{\text{Mass of water}}{\text{Molar of water}}=\frac{1g}{18g/mole}=0.0555mole$

$\text{Mole fraction of ethylene glycol}=\frac{0.01795mole}{0.01795mole+0.0555mole}=0.244$

6 0

2 years ago

What is the [H+] of a solution with a pH of 5.6

lisabon 2012 [21]

Answer:

Hi there!

The answer to this question is 2.5118X10^-6

Explanation:

The formula to solve this is 10^-pH

5 0

3 years ago

Please help me fast, I will give brainliest if correct!

GrogVix [38]

Answer:

C. 24 mol

Explanation:

To say it quickly, 6 moles of C3H8 times 4 moles of HCl is 24. 32 moles of Cl2 times 4/4 moles HCl is 32. Therefore 24 is the correct answer.

7 0

3 years ago