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1 year ago

2.00 g of an unknown gas at STP fills a 500. mL flask. What is the molar mass of the gas?

Chemistry

1 answer:

otez555 [7]1 year ago

4 0

Answer:

100g/mol

Explanation:

Given parameters:

Mass of unknown gas = 2g

Volume of gas in flask = 500mL = 0.5dm³

Unknown:

Molar mass of gas = ?

Solution:

Since we know the gas is at STP;

1 mole of substance occupies 22.4dm³ of space at STP

Therefore,

0.5dm³ will have 0.02mole at STP

Now;

Number of moles = $\frac{mass}{molar mass}$

Molar mass = $\frac{mass}{number of moles}$ = $\frac{2}{0.02}$ = 100g/mol

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A 0.380 kg sample of aluminum (with a specific heat of 910.0 J/(kg x K)) is heated to 378 K and then placed in 2.40 kg of water

lbvjy [14]

Answer:

The equilibrium temperature of the system is 276.494 Kelvin.

Explanation:

Let consider the system formed by the sample of aluminium and water as a control mass, in which the sample is cooled and water is heated until thermal equilibrium is reached. The energy process is represented by First Law of Thermodynamics:

$Q_{water} -Q_{sample} = 0$

$Q_{water} = Q_{sample}$

Where:

$Q_{water}$ - Heat received by water, measured in joules.

$Q_{sample}$ - Heat released by the sample of aluminium, measured in joules.

Given that no mass is evaporated, the previous expression is expanded to:

$m_{w}\cdot c_{p,w}\cdot (T-T_{w}) = m_{s}\cdot c_{p,s}\cdot (T_{s}-T)$

Where:

$m_{s}$ , $m_{w}$ - Mass of water and the sample of aluminium, measured in kilograms.

$c_{p,s}$ , $c_{p,w}$ - Specific heats of the sample of aluminium and water, measured in joules per kilogram-Kelvin.

$T_{s}$ , $T_{w}$ - Initial temperatures of the sample of aluminium and water, measured in Kelvin.

$T$ - Temperature which system reaches thermal equilibrium, measured in Kelvin.

The final temperature is now cleared:

$(m_{w}\cdot c_{p,w}+m_{s}\cdot c_{p,s})\cdot T = m_{s}\cdot c_{p,s}\cdot T_{s}+m_{w}\cdot c_{p,w}\cdot T_{w}$

$T = \frac{m_{s}\cdot c_{p,s}\cdot T_{s}+m_{w}\cdot c_{p,w}\cdot T_{w}}{m_{w}\cdot c_{p,w}+m_{s}\cdot c_{p,s}}$

Given that $m_{s} = 0.380\,kg$ , $m_{w} = 2.40\,kg$ , $c_{p,s} = 910\,\frac{J}{kg\cdot K}$ , $c_{p,w} = 4186\,\frac{J}{kg\cdot K}$ , $T_{s} = 378\,K$ and $T_{w} = 273\,K$ , the final temperature of the system is:

$T = \frac{(0.380\,kg)\cdot \left(910\,\frac{J}{kg\cdot K} \right)\cdot (378\,K)+(2.40\,kg)\cdot \left(4186\,\frac{J}{kg\cdot K} \right)\cdot (273\,K)}{(2.40\,kg)\cdot \left(4186\,\frac{J}{kg\cdot K} \right)+(0.380\,kg)\cdot \left(910\,\frac{J}{kg\cdot K} \right)}$

$T = 276.494\,K$

The equilibrium temperature of the system is 276.494 Kelvin.

7 0

2 years ago

A scientist is wondering why a certain region in the ocean doesn't have maximum phytoplankton growth, despite having plenty of n

Advocard [28]

Answer:

C The water had adequate nitrogen and phosphorus, so it is likely iron limited.

Explanation:

Phytoplankton are single- cell organisms that live in oceans.

They require nitrogen, phosphorus and trace amount of iron to survive.

From the scientist's results after testing the water for nitrogen and phosphorus,there are reasonable amount of these elements.

Therefore insufficient iron in the water is the reason why he could find plenty phytoplankton in the ocean.

8 0

2 years ago

Categorize the following materials: bronze alloy, mouse growing an ear on its back, porcelain dentures

masha68 [24]

Bronze alloy and porcelain dentures

5 0

1 year ago

How many moles of nitrogen are in 3.7 moles of C8H11NO2?

Phoenix [80]

<h3>Answer:</h3>

3.7 Moles of Nitrogen

<h3>Explanation:</h3>

On observing the chemical formula C₈H₁₁NO₂ (might be formula of Dopamine) it is found that one mole of this compound contains;

8 Moles of Carbon

11 Moles of hydrogen

1 Mole of Nitrogen and

2 Moles of Oxygen respectively.

<u>Calculate Number of Moles of Nitrogen:</u>

As,

1 Mole of C₈H₁₁NO₂ contains = 1 Mole of Nitrogen

So,

3.7 Moles of C₈H₁₁NO₂ will contain = X Moles of Nitrogen

Solving for X,

X = (3.7 Moles × 1 Mole) ÷ 1 Mole

X = 3.7 Moles of Nitrogen

4 0

2 years ago

If a large marshmallow has a volume of 2.75 in3 and density of 0.242 g/cm3, how much would it weigh in grams? 1 in3=16.39 cm3.

Anestetic [448]

<span>
• </span>Volume of the marshmallow:

V = 2.75 in^3 (but, 1 in^3 = 16.39 cm^3)

V = 2.75 × 16.39 cm^3

V = 2.75 × 16.39 cm^3

V = 45.0725 cm^3

• Density:

d = 0.242 g/cm^3

<span>• </span>Mass:

m = d × V

m = (0.242 g/cm^3) × (45.0725 cm^3)

m = (0.242 g/cm^3) × (45.0725 cm^3)

m = 10.907545 g

m ≈ 10.9 g <——<span>— this is the answer.

I hope this helps. =)
</span>

4 0

2 years ago