A microscope containing two or more lenses is an

Which statement is true about a liquid but not a gas?

vovangra [49]

Answer:

I think its D

Explanation:

.........................

3 0

3 years ago

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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

Onsider the reaction below.

mamaluj [8]

Answer:

3.14 moles of hydrogen are produced

Explanation:

This is because for every 1 molesof hydrogen are produced 2 moles of oxygen are produced. So we take 6.28 divide it by 2 and we wend up with 3.14.

8 0

2 years ago

Can anyone help me in my chemistry homework?

sdas [7]

Answer:

what is in your chemistry home work I can try for...

7 0

3 years ago

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For an alloy that consists of 93.1 g copper, 111.7 g zinc, and 4.0 g lead, what are the concentrations of (a) Cu, (b) Zn, and (c

Andrew [12]

Answer:

(a) weight percent of Cu = 44.59%

(b) weight percent of Zn = 53.49%

(c) weight percent of Pb = 1.91%

Explanation:

Given: Alloy contains- Cu=93.1 g, Zn=111.7 g, Pb=4.0 g

Therefore, the total mass of the alloy (m) = mass of Cu (m₁) + mass of Zn (m₂)+ mass of Pb (m₃)

m= 93.1 g + 111.7 g + 4.0 g = 208.8 g

(a) weight percent of Cu = (m₁ ÷ m)× 100% = (93.1 g ÷ 208.8 g)× 100% =44.59%

(b) weight percent of Zn = (m₂ ÷ m)× 100% = (111.7 g ÷ 208.8 g)× 100% =53.49%

(c) weight percent of Pb = (m₃ ÷ m)× 100% = (4.0 g ÷ 208.8 g)× 100% =1.91%

5 0

3 years ago