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3 years ago

Emperical formula of carbon dioxide

Chemistry

2 answers:

Vladimir79 [104]3 years ago

8 0

Empirical formula of Carbon dioxide :

$\mathrm{CO_2}$

Stells [14]3 years ago

6 0

CO2 is the emperical formula of carbon dioxide

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How many atoms will there be in 5.00 g of gold. (Use ^ to show the exponent in scientific notation. For example, 4 X 1011 should

gayaneshka [121]

Answer:

1.53x10^22 atoms of Au

Explanation:

To find the atoms of gold we need first, to convert the mass of gold to moles using molar mass of gold (196.97g/mol). Then, these moles must be converted to number of atoms based on definition of moles (1 mole = 6.022x10²³ atoms).

<em>Moles Au:</em>

5.00g Au * (1mol / 196.97g) = 0.0254 moles of Au

<em>Atoms of Au:</em>

0.0254 moles * (6.022x10²³ atoms / 1 mole) =

<h3>1.53x10^22 atoms of Au</h3>

3 0

3 years ago

Suppose an aluminum- nuclide transforms into a phosphorus- nuclide by absorbing an alpha particle and emitting a neutron. Comple

AfilCa [17]

Explanation:

An alpha particles is basically a helium nucleus and it contains 2 protons and 2 neutrons.

Symbol of an alpha particle is $^{4}_{2}\alpha$ . Whereas a neutron is represented by a symbol $^{1}_{0}n$ , that is, it has zero protons and only 1 neutron.

Therefore, reaction equation when an aluminum- nuclide transforms into a phosphorus- nuclide by absorbing an alpha particle and emitting a neutron is as follows.

$^{27}_{13}Al + ^{1}_{0}n \rightarrow ^{30}_{15}P + ^{1}_{0}n$

5 0

3 years ago

It has more protons than Cl but less than K

timama [110]

Answer: Argon (Ar), which has 18 protons.

8 0

3 years ago

Nicotine, a component of tobacco, is composed of C, H, and N. A 7.875-mg sample of nicotine was combusted, producing 21.363 mg o

Gnom [1K]

Answer: The empirical formula for the given compound is $C_5H_7N$

Explanation:

The chemical equation for the combustion of compound having carbon, hydrogen, and nitrogen follows:

$C_xH_yN_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of carbon, hydrogen and nitrogen respectively.

We are given:

Mass of $CO_2=21.363mg=21.363\times 10^3g=21363g$

Mass of $H_2O=6.125g=6.125\times 10^3g=6125g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 21363 g of carbon dioxide, $\frac{12}{44}\times 21363=5826.27g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 6125 g of water, $\frac{2}{18}\times 6125=680.55$ of hydrogen will be contained.

Now we have to calculate the mass of nitrogen.

Mass of nitrogen in the compound = (7875) - (5826.27 + 680.55) = 1368.18 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{5826.27g}{12g/mole}=485.52moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{680.55g}{1g/mole}=680.55moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{1368.18g}{14g/mole}=97.73moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0154 moles.

For Carbon = $\frac{485.52}{97.73}=4.96\approx 5$

For Hydrogen = $\frac{680.55}{97.73}=6.96\approx 7$

For Nitrogen = $\frac{97.73}{97.73}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 5 : 7 : 1

Hence, the empirical formula for the given compound nicotine is $C_5H_7N_1=C_5H_7N$

7 0

3 years ago

Another gas was used for many years to float transportation balloons.

Alja [10]

Another gas that was used to fill balloons was hydrogen. The reason why hydrogen is not used today is because it use to cause many fires when it was used to fill balloons up.

Hope I helped you! :)

6 0

3 years ago