Chemistry, 25.09.2021 14:00 cjdieball2106

How many molecules are contained in 4.00g of CO2 if the Molar mass of CO2 is 44.00 g/mole?

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Answer from: reghanhoward3

5.47 × 10²² CO₂ molecules

Explanation:

We want to determine the amount of molecules of CO₂ in 4.00 grams of CO₂ given that its molar mass is 44.00 g/mol.

To convert from grams to molecules, we can first convert from grams to moles and then from moles to molecules using the following two ratios:

$\displaystyle \frac{1 \text{ mol CO$_2$}}{44.00 \text{ g CO$_2$}}\text{ and } \frac{6.02 \cdot 10^{23} \text{ molecules CO$_2$}}{1 \text{ mol CO$_2$}}$

Begin with the initial value and multiply:

$\displaystyle \begin{aligned} &4.00 \text{ g CO$_2$} \cdot \frac{1 \text{ mol CO$_2$}}{44.00 \text{ g CO$_2$}}\cdot \frac{6.02 \times 10^{23} \text{ molecules CO$_2$}}{1 \text{ mol CO$_2$}} \\ \\ &= 5.47 \times 10^{22} \text{ molecules CO$_2$}}\end{aligned}$

Note that the final result should have three significant digits since 4.00 has three sig. digs.

In conclusion, there are about 5.47 × 10²² CO₂ molecules in 4.00 grams of CO₂.

Answer from: 23gordns

$\boxed {\boxed {\sf 5.47 \times 10^{22} \ molecules \ CO_2}}$

Explanation:

We are asked to find how many molecules are contained in 4.00 grams of carbon dioxide.

1. Grams to Moles

Convert grams to moles using the molar mass (mass of 1 mole of a substance). The molar mass of carbon dioxide is given: 44.00 grams per mole.

We will convert using dimensional analysis. Set up a conversion factor using the molar mass.

$\frac {44.0 \ g \ CO_2}{ 1 \ mol \ CO_2}$

We are converting 4.00 grams of carbon dioxide to moles, so we multiply by this value.

$4.00 \ g \ CO_2 * \frac {44.0 \ g \ CO_2}{ 1 \ mol \ CO_2}$

Flip the conversion factor. The value remains the same, but it allows us to cancel the units of grams of carbon dioxide.

$4.00 \ g \ CO_2 * \frac { 1 \ mol \ CO_2}{44.0 \ g \ CO_2}$

$4.00 * \frac { 1 \ mol \ CO_2}{44.0 }$

$\frac {4.00}{44.00 } \ mol \ CO_2$

$0.09090909091 \ mol \ CO_2$

2. Moles to Molecules

Convert moles to molecules using Avogadro's Number or 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this case, the particles are molecules of carbon dioxide.

Set up another conversion factor, this time with Avogadro's Number.

$\frac {6.022 \times 10^{23} \ molecules \ CO_2}{ 1\ mol \ CO_2}$

We are converting 0.09090909091 moles of carbon dioxide to molecules, so we multiply by this value.

$0.09090909091 \ mol \ CO_2*\frac {6.022 \times 10^{23} \ molecules \ CO_2}{ 1\ mol \ CO_2}$

The units of moles of carbon dioxide cancel.

$0.09090909091 *\frac {6.022 \times 10^{23} \ molecules \ CO_2}{ 1}$

$0.09090909091 *{6.022 \times 10^{23} \ molecules \ CO_2}$

$5.47454545 \times 10^{22}} \ molecules \ CO_2$

3. Significant Figures

The measurement of molar mass (44.00) has 4 sig fig and the measurement of molecules (4.00) has 3 sig fig. We round to the least number of sig fig, or 3.

For the number we found, that is the hundredth place. The 4 in the thousandth place tells us to leave the 7.

$5.47 \times 10^{22} \ molecules \ CO_2$

4.00 grams of carbon dioxide contains 5.47 × 10²² molecules of carbon dioxide.

Answer from: Quest

2b2t

explanation:

2b2t

Answer from: Quest

Answer is: the equilibrium shifts toward the reactants, to the left side of chemical reaction.it is because lead(ii) nitrate dissociates in water: pb(no₃)₂ → pb²⁺(aq) + 2no₃⁻(aq).according to lechatelier's principle, the position of equilibrium moves to counteract thechange (more lead cations, common ion), the position of equilibrium will move so that the concentration ofproducts decrease and concentration of reactants of chemical reaction increase.read more on -

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