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Lemur [1.5K]
2 years ago
10

A sample of hydrogen gas was collected over water at 21ºc and 685 mmhg. the volume of the container was 7.80 l. calculate the m

ass of h2(g) collected. (vapor pressure of water = 18.6 mmhg at 21c.)
Chemistry
1 answer:
murzikaleks [220]2 years ago
8 0
To determine the mass of the hydrogen gas that was collected, we calculate for the moles of hydrogen gas from the conditions given. In order to do this, we need an equation which would relate pressure, volume and temperature. For simplicity, we assume the gas is an ideal gas so we use the equation PV = nRT where P is the pressure, V is the volume, n is the number of moles of the gas, T is the temperature and R is the universal gas constant. We calculate as follows:

PV = nRT
n = PV / RT
n = (18.6/760) (7.80) / 0.08205 ( 21 + 273.15)
n = 0.0079 mol

Mass = 0.0079 mol ( 18.02 g / mol ) = 0.1425 g H2
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Answer:

9.2x10²g

Explanation:

Data obtained from the question include the following:

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Therefore, 1L of olive will weigh 9.2x10²g.

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in sample of elemental bromine, 55% or the atoms are Br-79, and the remainder are Br-81. if this sample is typical of naturally
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Answer:

The average atomic mass of bromine is 79.9 amu.

Explanation:

Given data:

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Average atomic mass of bromine = ?

Formula:

Average atomic mass = [mass of isotope× its abundance] + [mass of isotope× its abundance] +...[ ] / 100

Now we will put the values in formula.

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3 0
2 years ago
A 32.5 g piece of aluminum (which has a specific heat capacity of 0.921 J/g°C) is heated to 82.4°C and dropped into a calorimete
N76 [4]

Answer:

The mass of water = 219.1 grams

Explanation:

Step 1: Data given

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specific heat capacity aluminium = 0.921 J/g°C

Temperature = 82.4 °C

Temperature of water = 22.3 °C

The final temperature = 24.2 °C

Step 2: Calculate the mass of water

Heat lost = heat gained

Qlost = -Qgained

Qaluminium = -Qwater

Q = m*c*ΔT

m(aluminium)*c(aluminium)*ΔT(aluminium) = -m(water)*c(water)*ΔT(water)

⇒with m(aluminium) = the mass of aluminium = 32.5 grams

⇒with c(aluminium) = the specific heat of aluminium = 0.921 J/g°C

⇒with ΔT(aluminium) = the change of temperature of aluminium = 24.2 °C - 82.4 °C =  -58.2 °C

⇒with m(water) = the mass of water = TO BE DETERMINED

⇒with c(water) = 4.184 J/g°C

⇒with ΔT(water) = the change of temperature of water = 24.2 °C - 22.3 °C = 1.9 °C

32.5 * 0.921 * -58.2 = -m * 4.184 * 1.9

-1742.1 = -7.95m

m = 219.1 grams

The mass of water = 219.1 grams

8 0
2 years ago
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