All categories

3 years ago

A small sack of sand has a density of 1.5 g/cm3 and a mass of 1500 g. How much space (volume) does the sand occupy?

Chemistry

2 answers:

d1i1m1o1n [39]3 years ago

8 0

The answer would be 1000 cm3

Keith_Richards [23]3 years ago

7 0

1000cm3 would be the answer

You might be interested in

Which of the following are considered renewable resources?

Anna007 [38]

Answer:

wind

Explanation:

you cant run out of it

6 0

3 years ago

Read 2 more answers

100 POINTS HELP WILL GIVE BRAINLIEST

vodomira [7]

b. testing the hypothesis

6 0

1 year ago

N sub 2 +3H sub 2 rightwards arrow 2NH sub If 6 liters of hydrogen gas are used, how many liters of nitrogen gas will be needed

astraxan [27]

Answer:

2L of nitrogen gas will be needed

Explanation:

Based on the following reaction:

N₂ + 3H₂ → 2NH₃

1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia.

If 6L of hydrogen (In a gas, the volume is directly proportional to the moles, Avogadro's law) react, the volume of nitrogen gas required will be:

6L H₂ * (1mol N₂ / 3 moles H₂) =

<h3>2L of nitrogen gas will be needed</h3>

7 0

3 years ago

mixture of N 2 And H2 Gases weighs 13.22 g and occupies a volume of 24.62 L at 300 K and 1.00 atm.Calculate the mass percent of

anygoal [31]

Answer: The mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

Explanation:

To calculate the number of moles, we use the equation given by ideal gas equation:

PV = nRT

where,

P = Pressure of the gaseous mixture = 1.00 atm

V = Volume of the gaseous mixture = 24.62 L

n = number of moles of the gaseous mixture = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = Temperature of the gaseous mixture = 300 K

Putting values in above equation, we get:

$1.00atm\times 24.62L=n_{mix}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 300K\\\\n_{mix}=\frac{1.00\times 24.62}{0.0821\times 300}=0.9996mol$

We are given:

Total mass of the mixture = 13.22 grams

Let the mass of nitrogen gas be 'x' grams and that of hydrogen gas be '(13.22 - x)' grams

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

For nitrogen gas:

Molar mass of nitrogen gas = 28 g/mol

$\text{Moles of nitrogen gas}=\frac{x}{28}mol$

For hydrogen gas:

Molar mass of hydrogen gas = 2 g/mol

$\text{Moles of hydrogen gas}=\frac{(13.22-x)}{2}mol$

Equating the moles of the individual gases to the moles of mixture:

$0.9996=\frac{x}{28}+\frac{(13.22-x)}{2}\\\\x=12.084g$

To calculate the mass percentage of substance in mixture we use the equation:

$\text{Mass percent of substance}=\frac{\text{Mass of substance}}{\text{Mass of mixture}}\times 100$

Mass of the mixture = 13.22 g

For nitrogen gas:

Mass of nitrogen gas = x = 12.084 g

Putting values in above equation, we get:

$\text{Mass percent of nitrogen gas}=\frac{12.084g}{13.22g}\times 100=91.41\%$

For hydrogen gas:

Mass of hydrogen gas = (13.22 - x) = (13.22 - 12.084) g = 1.136 g

Putting values in above equation, we get:

$\text{Mass percent of hydrogen gas}=\frac{1.136g}{13.22g}\times 100=8.59\%$

Hence, the mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

5 0

3 years ago

Find the partial pressure of helium gas in mmHg

mezya [45]

Answer:

214 mmhg

Explanation:

5 0

2 years ago