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Vlad [161]
2 years ago
5

Calcium carbonate decomposes when heated to solid calcium oxide and carbon dioxide gas. The balanced equation is: CaCO3(s) → CaO

(s) + CO2(g) Before this reaction was run, the reaction container, including the CaCO3, had a mass of 24.20 g. After the reaction, the container with product (and any unreacted reactant) had a mass of only 22.00 g because the CO2 gas produced did not remain in the container. What mass of CaCO3 reacted?
Chemistry
1 answer:
wlad13 [49]2 years ago
8 0

Answer:

The mass of CaCO3 reacted  is 5.00 grams

Explanation:

Step 1 :Data given

Before the reaction, the container, including the CaCO3, had a mass of 24.20 g

After the reaction the container with product had a mass of only 22.00 g because the CO2 gas produced did not remain in the container.

Molar mass of CO2 = 44.01 g/mol

Molar mass CaCO3 = 100.09 g/mol

Step 2: The balanced equation

CaCO3 → CaO + CO2

Step 3: Calculate mass of CO2

Mass of CO2 = 24.20 grams - 22.00 grams

Mass of CO2 = 2.20 grams

Step 4: Calculate moles CO2

Moles CO2 = mass CO2 / molar mass CO2

Moles CO2 = 2.20 grams / 44.01 g/mol

Moles CO2 = 0.0500 moles

Step 5: Calculate moles CaCO3

For 1 mol CaCO3 we'll have 1 mol CaO and 1 mol CO2

For 0.0500 moles CO2 we need 0.0500 moles CaCO3

Step 6: Calculate mass CaCO3

Mass CaCO3 = moles CaCO3 * molar mass CaCO3

Mass CaCO3 = 0.0500 moles  * 100.09 g/mol

Mass CaCO3 = 5.00 grams

The mass of CaCO3 reacted  is 5.00 grams

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ipn [44]

Answer:

relative rate of diffusion is 1.05

Explanation:

According to Graham's law of difussion:

Rate of diffusion is inversely proportional to the square root of molecular weight of a molecule.

For two given molecules:

\frac{(rate)_{1}}{(rate)_{2}}=\frac{\sqrt{M_{2}} }{\sqrt{M_{1}}}

The given molecules are

Water = 18.01

Heavy water =20.03

Thus the relative rate of diffusion will be:

\frac{(rate)_{water}}{(rate)_{heavywater}}=\sqrt{ \frac{20.03}{18.01}}=1.05

8 0
2 years ago
How many molecules are in 3.6 grams of NaCl?
raketka [301]

Answer:

\boxed{\text{None}}

Explanation:

There are no molecules in NaCl, because it consists only of ions.

However, we can calculate the number of formula units (FU) of NaCl.

Step 1. Calculate the moles of NaCl

\text{No. of moles}=\text{3.6 g NaCl}\times \dfrac{\text{1 mol NaCl}}{\text{63.54 g NaCl}} = \text{0.0567 mol NaCl}

Step 2. Convert moles to formula units

\text{FU} = \text{0.0567 mol NaCl} \times \dfrac{6.022 \times 10^{23}\text{ FU NaCl}}{\text{1 mol NaCl}}\\\\= 3.4 \times 10^{22} \text{ FU NaCl}

There are \boxed{3.4 \times 10^{22} \text{ FU NaCl}} in 3.6 g of NaCl.

6 0
2 years ago
A sample of ammonia has a mass of 82.9 g. how many molecules are in this sample?
alina1380 [7]
The chemical formula for ammonia is NH3. So first, you need to find the molar mass of ammonia (how many grams in one mole).
N=14g
H3=3g
So one mole of NH3 is 17 grams, you can divide 82.9 grams by 17 grams to find the number of molecules. The answer should be 4.876 moles (molecules) of ammonia. Hope this helps!
8 0
2 years ago
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What is the density of a piece of metal in g/cm3 if its mass is determined to be 42.20 g and it is in the shape of a cube, with
maria [59]
It's a cube so the volume = edge^3
Volume = 2.5^3 cm^3 = 15.625 cm^3

density = mass / volume = 42.20 / 15.625 = 2.70 You have 3 places of accuracy.

density of object = 2.70 grams / cm^3 <<<<=== answer.
4 0
2 years ago
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How many grams of fluorine are contained in 8 molecules of boron trifluoride?
Lelu [443]
<h3>Answer:</h3>

             7.57 × 10⁻²² g of F

<h3>Solution:</h3>

Data Given:

                 Number of Molecules  =  8

                 M.Mass of BF₃ =  67.82 g.mol⁻¹

                 Mass of Fluorine atoms  =  ?

Step 1: Calculate Moles of BF₃

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

Putting value,

            Moles  =   8 Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹

            Moles  =  1.33 × 10⁻²³ mol

Step 2: Calculate Mass of BF₃:

                   Moles  =  Mass ÷ M.Mass

Solving for Mass,

                   Mass  =  Moles × M.Mass

Putting values,

                   Mass  =  1.33 × 10⁻²³ mol × 67.82 g.mol⁻¹

                   Mass  =  9.0 × 10⁻²² g

Step 3: Calculate Mass of Fluorine Atoms:

As,

                         67.82 g BF₃ contains  =  57 g of F

So,

                    9.0 × 10⁻²² g will contain  =  X g of F

Solving for X,

                       X =  (9.0 × 10⁻²² g × 57 g) ÷ 67.82 g

                        X  =  7.57 × 10⁻²² g of F

4 0
2 years ago
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