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2 years ago

Pcl3+H2O=H3PO3+Hcl=?

Chemistry

2 answers:

beks73 [17]2 years ago

7 0

Answer:

PCl3 + 3H2O → HPO(OH)2 + 3HCl. Phosphorus(III) chloride react with water to produce phosphorous acid and hydrogen chloride.

Explanation:

snow_lady [41]2 years ago

4 0

Answer:

the answer is stop studying,enjoy your life,party all night,bite like a dog and purr like a cat thats it

Explanation:

i dunno

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A parachute on a racing dragster opens and changes the speed of the car from 85 m/s to 45 m/s in a period of 4.5 seconds. What i

LiRa [457]

Answer:

- 8.89 $m/s^{2}$

Explanation:

$\frac{(Vf-Vo)}{time}$ = $\frac{(45-85 m/s)}{4.5s}$ = - 8.89 $m/s^{2}$

7 0

2 years ago

Read 2 more answers

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.

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2 years ago

Balance the following equation: __Ch4+__O2→__CO2+__H2O

konstantin123 [22]

Answer:

1,2,1,2

Explanation:

You would need only one of the CH4 but 2 of the O2 then 1 CO2 and 2 H2O on each side of the equation you now have 1 carbon, 4 hydrogen, and 4 oxygen.

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2 years ago

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igor_vitrenko [27]

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2 years ago

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vladimir1956 [14]

Answer:

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Explanation:

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2 years ago

Pcl3+H2O=H3PO3+Hcl=?​

Pcl3+H2O=H3PO3+Hcl=?