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3 years ago

Plz Answer Fam & Con Chemistry (8TH GRADE)

Chemistry

2 answers:

Fantom [35]3 years ago

5 0

Claw technique is what is represented

olga_2 [115]3 years ago

3 0

Answer:

claw techniques

Explanation:

because it is

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If 0.35 moles of kcl was dissolved in enough water to make 0.2 L of solution, what is molarity?

Mars2501 [29]

Answer:

1.75M

Explanation:

molarity = number of moles of solute/ number of L of solution =

=0.35 mol/0.2L = 1.75 mol/L = 1.75 M

8 0

3 years ago

Heat moves from what source by convection and radiation?

Alona [7]

Answer:

heat can move from any source but if we are being legitimate it moves from convection

Explanation:

8 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 conversion factor do you use first to calculate the number of grams of CO 2 produced by the reaction of 50.6 g of CH 4 wit

Brilliant_brown [7]

Answer:

Thus, first conversion of mass of methane into moles by dividing it with 16.04 g/mol

Mass = 138.63 g

Explanation:

The balanced chemical reaction is shown below:-

$CH_4+2O_2\rightarrow CO_2+2H_2O$

Firstly the moles of methane gas reacted must be calculate as:-

Given, mass of methane = 50.6 g

Molar mass of methane gas = 16.04 g/mol

The formula for the calculation of moles is:-

$Moles=\frac{Mass\ taken}{Molar\ mass}=\frac{50.6}{16.04}\ mol=3.15\ mol$

Thus, from the reaction stoichiometry,

1 mole of methane produces 1 mole of carbon dioxide

Also,

3.15 mole of methane produces 3.15 mole of carbon dioxide

Moles of $CO_2$ = 3.15 mole

Molar mass of $CO_2$ = 44.01 g/mol

Mass = Moles*Molar mass = $3.15\times 44.01$ g = 138.63 g

3 0

3 years ago

Why does the flame go out on the Bunsen burner when the gas is turned off? (use the collision theory in your answer)

Gre4nikov [31]

Answer:

Explanation:

Well the gas is the fuel for the flame of course. The collision theory comes into play when the gas turns on, chemicals collide with one another. Then reactions occur causing the flame. Then when you take away the fuel, the flame stops because there is no atoms or molecules to come together/collide with one another.

Sorry if its wrong or doesn't make sense... Wish you the best of luck on whatever your doing!

8 0

3 years ago