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lorasvet [3.4K]
3 years ago
12

Write the symbols of following elements. a. Phosphorus b. Bromine c. Silver d. Sodium

Chemistry
1 answer:
Vikki [24]3 years ago
7 0

Answer:

a. P

b. Br

c. Ag

d. Na

Explanation:

The Periodic Table says so

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How many liters of water vapor are in 36.21 g?
Elden [556K]

Answer:

45.02 L.

Explanation:  

  • Firstly, we need to calculate the no. of moles of water vapor.
  • n = mass / molar mass = (36.21 g) / (18.0 g/mol) = 2.01 mol.
  • We can calculate the volume of knowing that 1.0 mole of a gas at STP occupies 22.4 L.

<em><u>Using cross multiplication:</u></em>

1.0 mole of CO occupies → 22.4 L.

2.01 mole of CO occupies → ??? L.

∴ The volume of water vapor in 36.21 g = (22.4 L)(2.01 mole) / (1.0 mole) = 45.02 L.

4 0
3 years ago
127) Thirty-six colonies grew in nutrient agar from 1.0 ml of undiluted sample in a standard plate count. How many cells were in
SIZIF [17.4K]

Answer:

36

Explanation:

Since the sample was undiluted the number of colonies is the number that grew on the nutrient agar which is 36 colonies. If it was diluted for example let say 0.1 ml from a dilution in which 1 ml of the sample was added to 9 ml of water, and it grew  colonies then  0.1 ml  yielded  6 colonies, 1 ml of the diluted sample will yield 60 colonies and 10 ml will have 600 colonies and therefore the 1 ml undiluted sample will have 600 colonies.

6 0
3 years ago
30. The density of an unknown gas at 27°C and 2 atm pressure is equal with density of N2 gas at
Zanzabum

Answer:

Molar mass of the unknown gas is 64.6 g/mol

Explanation:

Let's think this excersise with the Ideal Gases Law.

We start from the N₂. At STP conditions we know that 1 mol of anything occupies 22.4L.

We apply: P . V = n . R . T

5 atm . V = 1 mol . 0.082 . 325K

V = (1 mol . 0.082 . 325K) / 5 atm = 5.33 L

It is reasonable to say that, if we have more pressure, we may have less volume.

As this is the volume for 1 mol of N₂, our mass is 28 g. Then, the density of the nitrogen and the unknown gas is 28 g/5.33L = 5.25 g/L

Our unknown gas has, this density at 27°C and 2 atm.

If we star from this, again: 1 mol of any gas occupy 22.4L at STP, we can calculate the volume for 1 mol at those conditions:

P₁ . V₁ / T₁ = P₂ . V₂ / T₂

1 atm . 22,4L / 273K = 2 atm . V₂ / 300K

Remember that the value for T° is Absolute (T°C + 273)

[ (1 atm . 22.4L / 273K) . 300K] / 2 atm = V₂ → 12.3L

This is the volume for 1 mol of the unknown gas at 2 atm and 27°C

We use density to determine the mass: 12.3 L . 5.25 g/L = 64.6 g

That's the molar mass: 64.6 g/mol

6 0
2 years ago
Arrange these elements according to atomic Pb Si C Sn Ge
choli [55]
........C<Si<Ge<Sn<Pb
5 0
3 years ago
You have 1.6 x10 21 atoms of oxygen, what is the mass of that number of atoms
iren [92.7K]

Answer: 166.023×1023

Explanation:

Mass of 6.023×1023 atoms of oxygen = gram atomic mass of oxygen.

Mass of 6.023×1023 atoms of oxygen = 16g

∴ Mass of 1 atom of oxygen =166.023×1023

6 0
3 years ago
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