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garri49 [273]
3 years ago
12

A flask contains 0.280 mol of liquid bromine, br2. Determine the number of bromine molecules present in the flask.

Chemistry
1 answer:
Nat2105 [25]3 years ago
4 0

Answer:

              1.68 × 10²³ Molecules

Explanation:

                     As we know that 1 mole of any substance contains exactly 6.022 × 10²³ particles which is also called as Avogadro's Number. So in order to calculate the number of particles (molecules) contained by 0.280 moles of Br₂, we will use following relation,

           Moles  =  Number of Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹

Solving for Number of Molecules,

           Number of Molecules  =  Moles × 6.022 × 10²³ Molecules.mol⁻¹

Putting values,

           Number of Molecules  =  0.280 mol × 6.022 × 10²³ Molecules.mol⁻¹

           Number of Molecules  = 1.68 × 10²³ Molecules

Hence,

           There are 1.68 × 10²³ Molecules present in 0.280 moles of Br₂.

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Answer:

D

Explanation:

D. V1P1 / T1=V2P2 / T2 is correct

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2 years ago
By definition, any substance that donates a proton is best defined as _________.
BabaBlast [244]
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How many liters of CO2 are in 4.76 moles? (at STP)
dem82 [27]

<u>Answer:</u> The volume of carbon dioxide gas at STP for given amount is 106.624 L

<u>Explanation:</u>

We are given:

Moles of carbon dioxide = 4.76 moles

<u>At STP:</u>

1 mole of a gas occupies a volume of 22.4 Liters

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Hence, the volume of carbon dioxide gas at STP for given amount is 106.624 L

6 0
3 years ago
Written in this form, carbon-13, what does the 13 represent and what is the atomic number?
tigry1 [53]
Explanation:
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Carbon-12:
Atomic number: 6
Mass number: 12
protons: 6
electrons: 6
neutrons: 6
Carbon-13:
Atomic number: 6
Mass number: 13
protons: 6
electrons: 6
neutrons: 7
Carbon-14:
Atomic number: 6
Mass number: 14
protons: 6
electrons: 6
neutrons: 8
8 0
3 years ago
Aluminum has a density of 2.70 g/cm3. what would be the mass of a sample whose volume is 10.0 cm3?
anastassius [24]
Hey there!

Density = 2.70 g/cm³

Volume = 10.0 cm³

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3 0
3 years ago
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