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Rashid [163]
3 years ago
8

Calculate the approximate number of atoms in a bacterium, assuming the average mass of an atom is ten times the mass of a hydrog

en atom. The mass of a bacterium is 10−15 kg and the mass of a hydrogen atom is of the order of 10−27 kg.
Chemistry
1 answer:
mezya [45]3 years ago
8 0

Answer: -

The approximate number of atoms in a bacterium is 10¹¹

Explanation: -

We are given the mass of a bacterium is 10⁻¹⁵ kg.

We are told that the mass of a hydrogen atom is 10⁻²⁷ kg.

Finally we learn that the average mass of an atom of the bacterium is ten times the mass of a hydrogen atom.

Mass of an atom of bacterium = 10 x mass of hydrogen atom

                                                   = 10 x 10⁻²⁷ kg.

                                                   = 10⁻²⁶ kg.

Thus the number of atoms in a bacterium = \frac{Total mass}{mass of 1 bacterium atom}

                                                                      = \frac{10-15 kg}{10-26 kg}

                                                                      = 10¹¹


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3 years ago
Which sample contains more atoms: 3.89 g of nickel or 6.61 g of silver? defend your answer
Lubov Fominskaja [6]

Nickel contains more atoms.

<u>Explanation:</u>

Number of atoms is determined by the Avogadro number.

1 mole of substance contains 6.022 X 10²³ atoms.

Given:

Mass of Nickel, Ni = 3.89g

Molecular weight of Ni = 58.6934g

Number of moles of Ni = 3.89/58.6934g

Number of atoms of Ni = \frac{3.89 X 6.022 X 10^2^3}{58.6934}

Number of atoms of Ni = 0.39 X 10²³

Mass of silver, Ag = 6.61g

Molecular weight of Ag = 107.8682g

Number of moles of Ag = 6.61/107.8682

Number of atoms of Ag = \frac{6.61 X 6.022 X 10^2^3}{107.8682}

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How many more grams should be added​
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During the following chemical reaction, 46.3 grams of C3H6O react with 73.2 grams of O2
ra1l [238]

Answer:

a) O2 is the limiting reactant

b) 75.70 grams CO2 (theoretical yield)

c) There remains 12.81 grams of C3H6O

d) The actual yield CO2 is 34.29 grams

Explanation:

Step 1: Data given

Mass of C3H6O = 46.3 grams

Mass of O2 = 73.2 grams

Molar mass of C3H6O = 58.08 g/mol

Molar mass  of O2 = 32 g/mol

Step 2: The balanced equation

C3H6O + 4O2 → 3 CO2 + 3H2O

Step 3: Calculate moles C3H6O

Moles C3H6O = mass C3H6O / molar mass C3H6O

Moles C3H6O = 46.3 grams / 58.08 g/mol

Moles C3H6O = 0.793 moles

Step 4: Calculate moles O2

Moles O2 = 73.2 grams / 32 g/mol

Moles O2 = 2.29 moles

Step 5: Calculate limiting reactant

For 1 mol C3H6O we need 4 moles of O2 to produce 3 moles CO2 and 3 moles H2O

O2 is the limiting reactant. It will completely be consumed. (2.29 moles).

C3H6O is in excess. There will react 2.29/4 = 0.5725 moles C3H6O

There will remain 0.793 - 0.5725 = 0.2205 moles C3H6O

This is 0.2205 moles * 58.08 g/mol =<u> 12.81 grams</u>

Step 6:  Calculate moles of CO2

For 1 mol C3H6O we need 4 moles of O2 to produce 3 moles CO2 and 3 moles H2O

For 2.29 moles O2 we need 3/4 * 2.29 = 1.72 moles CO2

This is 1,72 moles * 44.01 g/mol = <u>75.70 grams CO2</u>

Step 7: Calculate actual yield

% yield = 45.3 % = 0.453 = (actual yield / theoretical yield)

actual yield = 0.453 * 75.70 = <u>34.29 grams</u>

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