All categories

2 years ago

How do you calculate mass using density and volume??

Chemistry

1 answer:

LiRa [457]2 years ago

8 0

Just reorder the equation

Density= mass/volume

You might be interested in

I need help with 2, 3, and 4 please help i will give the brainliest

Sphinxa [80]

Answer:

2=2.28*10^5

3=3.98*10^3

4=5.76*10^3

8 0

2 years ago

During the combustion of a peanut that weighed 0.341 g, the temperature of the 100 mL of water in the calorimeter rose from 23.4

blondinia [14]

Answer:

119 kCal per serving.

Explanation:

The heat energy necessary to elevates water's temperature from 23.4°C to 37.9°C can be calculated by the equation below:

Q = mcΔT

Q: heat energy

m: mass in g

c: specific heat capacity in cal/g°C

ΔT = temperature variation in °C

m is the mass of water, considering the density of water to be 1g/mL, 100 mL of water weights 100g. Therefore:

Q = 100 g x 1.00 cal/g°C x (37.9 - 23.4)°C

Q = 1450 cal

1450 cal ____ 0.341 g peanuts

x ____ 28 g peanuts

x = 119061.58 cal

This means that the cal from fat per serving of peanuts is at least 119 kCal.

3 0

2 years ago

Write 151.5 cm to meters in the correct significant figures

iogann1982 [59]

Answer:

1.515m

hope that helps uh :)

3 0

2 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

2 years ago

52.0 g of LiCl represents ______ moles of LiCl. A) 0.88 Eliminate B) 1.23 C) 1.51 D) 2.45

xxMikexx [17]

The correct option is a

4 0

2 years ago