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3 years ago

I WILL MARK AS BRAINLIEST IF YOU GIVE ME ANSWER

Chemistry

2 answers:

Maru [420]3 years ago

7 0

Answer:

Explanation:

olga2289 [7]3 years ago

3 0

Answer:

Explanation:

I think it is c because the tep. will change and the volume will change.

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A rigid vessel contains water at 0.15 MPa (at 25 °C). The vessel is heated to the critical point of water. Calculate the fractio

pishuonlain [190]

Answer:

0.8078 Kg

Explanation:

Pressure of water = 0.15 MPa = 1.5 bar

At critical point of water ,temperature = 647 K=374°C

From the ideal gas equation

P×V= m×R×T

Let us assume volume = 1 m^3

1.5 x 105 x 1 = m x 287 x 647

m= 0.8078 kg

the fraction of mass of liquid at 25°C.

4 0

4 years ago

If 11.5 ml of vinegar sample (d=1g/ml) is titrated with 18.5 ml of standardized Sodium hydroxide

insens350 [35]

Answer:

4.83% of acetic acid in the vinegar

Explanation:

3 0

4 years ago

What do we call the experimental apparatus that William crooks used in his experiments and what did he discover

34kurt

Answer: It is called a Crookes Tube, and he used it to discover cathode rays, which were later determined to be electrons.

4 0

3 years ago

A mixture of 15.0 g of the anesthetic halothane (C2HBrClF3 197.4 g/mol) and 22.6 g of oxygen gas has a total pressure of 862 tor

AlexFokin [52]

Answer : The partial pressure of $C_2HBrClF_3$ and $O_2$ are, 84 torr and 778 torr respectively.

Explanation : Given,

Mass of $C_2HBrClF_3$ = 15.0 g

Mass of $O_2$ = 22.6 g

Molar mass of $C_2HBrClF_3$ = 197.4 g/mole

Molar mass of $O_2$ = 32 g/mole

First we have to calculate the moles of $C_2HBrClF_3$ and $O_2$ .

$\text{Moles of }C_2HBrClF_3=\frac{\text{Mass of }C_2HBrClF_3}{\text{Molar mass of }C_2HBrClF_3}=\frac{15.0g}{197.4g/mole}=0.0759mole$

and,

$\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{22.6g}{32g/mole}=0.706mole$

Now we have to calculate the mole fraction of $C_2HBrClF_3$ and $O_2$ .

$\text{Mole fraction of }C_2HBrClF_3=\frac{\text{Moles of }C_2HBrClF_3}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.0759}{0.0759+0.706}=0.0971$

and,

$\text{Mole fraction of }O_2=\frac{\text{Moles of }O_2}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.706}{0.0759+0.706}=0.903$

Now we have to partial pressure of $C_2HBrClF_3$ and $O_2$ .

According to the Raoult's law,

$p^o=X\times p_T$

where,

$p^o$ = partial pressure of gas

$p_T$ = total pressure of gas

$X$ = mole fraction of gas

$p_{C_2HBrClF_3}=X_{C_2HBrClF_3}\times p_T$

$p_{C_2HBrClF_3}=0.0971\times 862torr=84torr$

and,

$p_{O_2}=X_{O_2}\times p_T$

$p_{O_2}=0.903\times 862torr=778torr$

Therefore, the partial pressure of $C_2HBrClF_3$ and $O_2$ are, 84 torr and 778 torr respectively.

6 0

3 years ago

Which of the following elements have properties similar to those of strontium (Sr)

dalvyx [7]

D
because they are elements in the same group as strontium

8 0

4 years ago

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