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natita [175]
3 years ago
10

What is the mass of 0.73 moles of AgNO3?

Chemistry
1 answer:
love history [14]3 years ago
6 0

Answer:

124 g (3 sig figs)

or

124.011 g (6 sig figs

Explanation:

Step 1: Calculate g/mol for AgNO₃

Ag - 107.868 g/mol

N - 14.01 g/mol

O - 16.00 g/mol

107.868 + 14.01 + 16.00(3) = 169.878 g/mol

Step 2: Multiply 0.73 moles by molar mass

0.73 mol (169.979 g/mol)

124 grams of AgNO₃

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All elements are made up of atoms. ➢ Atoms are made up of protons, neutrons, and electrons. Two different kinds of atoms can combine to form a compound. A molecule is a combination of atoms that cannot be broken apart while still retaining the same properties as the larger substance that it is a part of.
7 0
3 years ago
If 5.0 moles of C3H8 react, how many molecules of water are formed?
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Balance Chemical Equation for combustion of Propane is as follow,

                         C₃H₈  +  5 O₂    →    3 CO₂  +  4 H₂O

According to equation,

              1 mole of C₃H₈ on combustion gives  =  4 moles of H₂O
So,
        5 moles of C₃H₈ on combustion will give  =  X moles of H₂O

Solving for X,
                                    X  =  (5 mol × 4 mol) ÷ 1 mole

                                    X  =  20 moles of H₂O

Calculating number of molecules for 20 moles of H₂O,
As,
                        1 mole of H₂O contains  =  6.022 × 10²³ molecules
So,
                20 moles of H₂O will contain  =  X molecules

Solving for X,
                                 X  =  (20 mole × 6.022 × 10²³ molecules) ÷ 1 mol

                                 X  =  1.20 ×10²⁵ Molecules of H₂O
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3 years ago
A piece of metal with a mass of 611 g is placed into a graduated cylinder that contains 25.1 ml of water, raising the water leve
jeka94

Mass of metal piece is 611 g and volume of graduated cylinder is 25.1 mL. When metal piece is placed in the graduated cylinder water level increases to 56.7 mL. The increase in volume is due to volume of metal piece that gets added to the volume of water.

Thus, volume of metal piece can be calculated by subtracting initial volume from the final one.

V_{metal}=V_{final}-V_{initial}=(56.7-25.1)mL=31.6 mL

Thus, volume of metal piece will be 31.6 mL. The mass of metal piece is given 611 g, density of metal can be calculated as follows:

d=\frac{m}{V}=\frac{611 g}{31.6 mL}=19.33 g/mL

Therefore, density of metal is 19.33 g/mL.

3 0
3 years ago
When iron(III) oxide (Fe2O3) is exposed to carbon monoxide (CO) under high pressure, metallic iron (Fe) and carbon dioxide (CO2)
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Answer:

The equation is Fe₂O₃ + CO ⇒ Fe + CO₂.

The balanced reaction equation is Fe₂O₃ + 3CO ⇒ 2Fe + 3CO₂.

Explanation:

First, we have to write our equation. It's actually pretty straightforward - first we look for our reactants (looks like it's Fe₂O₃ and CO), then we look for our products (Fe and CO₂). Then, we have to balance it so that both sides have the same number of both element.

Currently, we have the equation Fe₂O₃ + CO ⇒ Fe + CO₂. There are 2 Fe atoms, 4 O atoms, and 1 C atom on the left side. There is 1 Fe atom, 2 O atoms, and 1 C atom on the right side.

First thing we can do is give our Fe on the right side a coefficient of 2. This will make it equivalent to the 2 Fe atoms on the left side:

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Fe₂O₃ + 3CO ⇒ 2Fe + 3CO₂

And that's how I balanced the equation. It can be confusing, but with enough practice, it will get easier and easier. :)

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