Which came first M or O

he equation represents the combustion of sucrose. C12H22O11 + 12O2 12CO2 + 11H2O If there are 10.0 g of sucrose and 8.0 g of oxy

xeze [42]

Answer:

The moles of sucrose that are available for this reaction is 0.0292 moles

Explanation:

Combustion is an specifyc reaction where the reactants react with O₂ in order to produce CO₂ and H₂O

This combustion is: C₁₂H₂₂O₁₁ + 12O₂ → 12CO₂ + 11H₂O

We have to conver the mass to moles, to find out the limiting reactant

10 g . 1 mol / 342 g = 0.0292 moles of sucrose

8 g . 1mol / 32g = 0.250 moles of O₂

The moles of sucrose that are available for this reaction is 0.0292 moles

Before we start to work with the equation we must find the limiting reactant. When you find it, you can do all the calculations.

6 0

4 years ago

A 33.153 mg sample of a chemical known to contain only carbon, hydrogen, sulfur, and oxygen is put into a combustion analysis ap

elena-14-01-66 [18.8K]

Answer:

The empirical formula of the compound = $C_4H_8S_1O_1$

Explanation:

Mass of carbon dioxide gas = 59.060 mg = 0.059060 g

1 mg = 0.001 g

Moles of carbon dioxide = $\frac{0.059060 g}{44 g/mol}=0.0013 mol$

Moles of carbon in 0.0013 moles of carbon dioxide gas = 1 × 0.0013 mol = 0.0013 mol

Mass of 0.0013 moles of carbon = $12 g/mol\times 0.0013 mol=0.0156 g$

Mass of water = 24.176 mg = 0.024176

Moles of water = $\frac{0.024176 g}{18 g/mol}=0.0013 mol$

Moles of hydrogen in 0.0013 moles of water = 2 × 0.0013 mol = 0.0026 mol

Mass of 0.0013 moles of hydrogen= $1 g/mol\times 0.0013 mol=0.0013 g$

Mass of sulfur dioxide = 20.326 mg = 0.020326 g

Moles of sulfur dioxide = $\frac{0.020326 g}{64 g/mol}=0.00032 mol$

Moles of sulfur in 0.00032 moles of sulfur dioxide = 1 × 0.00032 mol = 0.00032 mol

Mass of 0.00032 moles of sulfur = $32 g/mol\times 0.00032 mol=0.01024 g$

Mass of oxygen in the sample = x

Mass of sample = 33.153 mg = 0.033153 g

0.033153 g = 0.0156 g + 0.0013 g + 0.01024 g + x

x = 0.006013 g

Moles of oxygen = $\frac{0.006013 g}{16 g/mol}=0.00038 mol$

For empirical formula divide the lowest number of moles of elemnt from all the moles of the all the elements:

Carbon : $\frac{0.0013 mol}{0.00032 mol}=4$

Hydrogen: $\frac{0.0026 mol}{0.00032 mol}=8$

Sulfur : $\frac{0.00032 mol}{0.00032 mol}=1$

Oxygen : $\frac{0.00038 mol}{0.00032 mol}=1$

The empirical formula of the compound = $C_4H_8S_1O_1$

8 0

4 years ago

Which is an example of a homogeneous mixture?

goldenfox [79]

Answer:

Clean Air So B

Explanation:

5 0

3 years ago

Rachel loves gardening. This summer, her 10 ft x 10 ft garden space is divided equally among tomatoes, lettuce, and strawberries

Ray Of Light [21]

Answer:

Rachel needs 0.52 kg of fertilizer.

Explanation:

Since the surface is a square of 10 ft of size its area is A = 10 ft x 10 ft = 1.0 x 10² ft². One third of that space is devoted to tomatoes ("space is divided equally among tomatoes, lettuce, and strawberries"), that is, 1/3 x 100 ft² = 33 ft².

We calculate the kilograms of fertilizer required we need to know 2 equivalences:

1 ac = 43,560 ft²

1 lb = 0.4535 kg

Then, we can use proportions to find out the amount of fertilizer used:

$33ft^{2} \frac{1ac}{43,560ft^{2} } .\frac{1,500lb}{1ac} .\frac{0.4535kg}{1lb} =0.52kg$

7 0

3 years ago

Calculate the density of SO3 gas at STP. Show all work.

Kipish [7]

When the density formula at STP is:
density = molar mass / molar volume

So we can get the molar mass of SO3 from the periodic table = 32 + (16x3) = 80g

and at STP each 1 mole of gas occupy 22.4 L so we have the molar volume at STP = 22.4.
So by substitution in the density formula:
∴ density of SO3 = 80 g / 22.4 L = 3.571 g /L

8 0

3 years ago