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4 years ago

1. Balance this reaction: Fe2O3(s) + C(s) → Fe(s) + CO2(g)

1 answer:

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2. How many moles of sodium Na will react with 6 moles of sulfur S?

fredd [130]

Answer:

12 moles of Na

Explanation:

5 0

3 years ago

An oxide of nitrogen contains 30.45 mass % N.

N76 [4]

An oxide of nitrogen contains 30.45 mass % N, if the molar mass is 90± 5 g/mol the molecular formula is N₂O₄.

<h3>What is molar mass?</h3>

The molar mass of a chemical compound is determined by dividing its mass by the quantity of that compound, expressed as the number of moles in the sample, measured in moles. A substance's molar mass is one of its properties. The compound's molar mass is an average over numerous samples, which frequently have different masses because of isotopes.

<h3>How to find the molecular formula?</h3>

The whole-number multiple is defined as follows.

Whole-number multiple = $\frac{molar mass (g/mol)}{empirical formula mass (g/mol)}$

The empirical formula mass is shown below.

Mw of empirical formula = Mw of N+ 2 x (Mw of O)

= 14.01 g/mol + 2 x (16.00 g/mol)

= 46.01 g/mol

With the given molar mass or the molecular formula mass, we can get the whole-number multiple for the compound.

Whole-number multiple = $\frac{90 g/mol }{46.01 g/mol}$ ≈ 2

Multiplying the subscripts of NO2 by 2, the molecular formula is N(1x2)O(2x2)= N2O4.

To learn more about molar mass visit:

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2 years ago

Where do the lemons go after harvested? What gets done with lemons?

Temka [501]

Answer:

Lemon juice,lemonade

Explanation:

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3 years ago

How many total atoms are in 0.290 g of P2O5?

Sliva [168]

There are 8.61 × 10²⁰ atoms in 0.290 g P₂O₅.

Step 1. Convert <em>grams of P₂O₅ to moles of P₂O₅</em>.

$\text{Moles of P}_{2}\text{O}_{5} = \text{0.290 g } \text{P}_{2}\text{O}_{5} \times \frac{\text{1 mol }\text{P}_{2}\text{O}_{5}}{\text{141.94 g }\text{P}_{2}\text{O}_{5}} = \text{0.002 043 mol } \text{P}_{2}\text{O}_{5} \\$

Step 2. Convert <em>moles of P₂O₅ to molecules of P₂O₅</em>.

$\text{Molecules of } \text{P}_{2}\text{O}_{5} = \text{0.002 043 mol } \text{P}_{2}\text{O}_{5} \times \frac{6.022 \times 10^{23}\text{ molecules }\text{P}_{2}\text{O}_{5}}{\text{1 mol } \text{P}_{2}\text{O}_{5}}\\$

$= 1.23 \times10^{21}\text{ molecules } \text{P}_{2}\text{O}_{5}\\$

Step 3. Convert <em>molecules of P₂O₅ to atoms</em>.

There are seven atoms in 1 mol P₂O₅.

∴ $\text{Total atoms} = 1.23\times 10^{21 }\text{ molecules }\text{P}_{2}\text{O}_{5} \times\frac{\text{7 atoms}}{\text{1 molecule }\text{P}_{2}\text{O}_{5}} = 8.61 \times 10^{21}\text{ atoms}\\$

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4 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