As per as the Multiplication rules of the significant figures, whenever any numbers in the decimals forms are multiplied or divided then result in mentioned in such a way so that the significant figures after the decimal will be same as that in the given least condition.

_______________________________

102900/12 = 8575

170 × 1.27 = 215.9

∴ (102,900 ÷ 12) + (170 × 1.27) = 8575 + 215.9

= 8790.9

Now, As per as Above rules, answer in correct significant figures will be = 8791.

8 0

4 years ago

On combustion, 1.0 L of a gaseous compound of hydrogen, carbon, and sulfur gives 2.0 L of CO2, 3.0 L of H2O vapor, and 1.0 L of

Murrr4er [49]

Answer:

The empirical formula of the organic compound is = $C_2H_6S_1$

Explanation:

At STP, 1 mole of gas occupies 22.4 L of volume.

Moles of $CO_2$ gas at STP occupying 2.0 L = n

$n\times 22.4L=2.0L$

$n=\frac{2.0 L}{22.4 L}=0.08929 mol$

Moles of carbon in 0.08920 mol = 1 × 0.08920 mol = 0.08920 mol

Moles of $H_2O$ gas at STP occupying 3.0 L = n'

$n'\times 22.4L=3.0L$

$n'=\frac{3.0 L}{22.4 L}=0.1339 mol$

Moles of hydrogen in 0.1339 moles of water vapor = 2 × 0.1339 mol = 0.2678 mol

Moles of $SO_2$ gas at STP occupying 1.0 L = n''

$n''\times 22.4L=1.0L$

$n''=\frac{1.0 L}{22.4 L}=0.04464 mol$

Moles of sulfur in 0.04464 mol = 1 × 0.04464 mol = 0.04464 mol

Moles of carbon , hydrogen and sulfur constituent of that organic compound .

Moles of carbon in 0.08920 mol = 1 × 0.08920 mol = 0.08920 mol

Moles of hydrogen in 0.1339 moles of water vapor = 2 × 0.1339 mol = 0.2678 mol

Moles of sulfur in 0.04464 mol = 1 × 0.04464 mol = 0.04464 mol

For empirical; formula divide the least number of moles from all the moles of elements.

carbon = $\frac{0.08920 mol}{0.04464 mol}=2$

Hydrogen = $\frac{0.2678 mol}{0.04464 mol}=6$

Sulfur = $\frac{0.04464 mol}{0.04464 mol}=1$

The empirical formula of the organic compound is = $C_2H_6S_1$

3 0

3 years ago

Rusting of iron is a very common chemical reaction. It results in one form from Fe reacting with oxygen gas to produce iron (III

Vlada [557]

Answer: The given amount of iron reacts with 9.0 moles of $O_2$ and produce 6.0 moles of $Fe_2O_3$

Explanation:

We are given:

Moles of iron = 12.0 moles

The chemical equation for the rusting of iron follows:

$4Fe+3O_2\rightarrow 2Fe_2O_3$

For oxygen gas:

By Stoichiometry of the reaction:

4 moles of iron reacts with 3 moles of oxygen gas

So, 12.0 moles of iron will react with = $\frac{3}{4}\times 12.0=9.0mol$ of oxygen gas

For iron (III) oxide:

By Stoichiometry of the reaction:

4 moles of iron produces 2 moles of iron (III) oxide

So, 12.0 moles of iron will produce = $\frac{2}{4}\times 12.0=6.0mol$ of iron (III) oxide

Hence, the given amount of iron reacts with 9.0 moles of $O_2$ and produce 6.0 moles of $Fe_2O_3$

5 0

3 years ago