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3 years ago

You have 55 grams of glucose(C6H12O6). How many grams of oxygen do you have?

Chemistry

1 answer:

zubka84 [21]3 years ago

8 0

Answer:

Mass of oxygen in glucose = 29.3g

Explanation:

Mass of glucose given is 55grams.

We are to find the mass of oxygen in this compound.

In the compound we have 6 atoms of oxygen.

Solution

To find the mass of oxygen in glucose, we calculate the formula mass of glucose. We now divide the formula mass of the oxygen atom with that of the glucose and multiply by the given mass to find the unkown mass.

Atomic mass of C = 12g

H = 1g

O = 16g

Formula mass of C₆H₁₂O₆ = {(12x6) + (1x12) + (16x6)} = 180

Mass of O in glucose = $\frac{6x16}{180}$ x 55

= $\frac{96}{180}$ x 55

= 0.53 x 55

Mass of oxygen in glucose = 29.3g

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A student determines the molar mass of acetone, CH3COCH3, by the method used in this experiment. She found that the equilibrium

Alexandra [31]

Answer:

(a). 4°C, (b). 2.4M, (c). 11.1 g, (d). 89.01 g, (e). 139.2 g and (f). 58 g/mol.

Explanation:

Without mincing words let's dive straight into the solution to the question.

(a). The freezing point depression can be Determine by subtracting the value of the initial temperature from the final temperature. Therefore;

The freezing point depression = [ 1 - (-3)]° C = 4°C.

(b). The molality can be Determine by using the formula below;

Molality = the number of moles found in the solute/ solvent's weight(kg).

Molality = ( 11.1 / 58) × (1000)/ ( 90.4 - 11.1) = 2.4 M.

(c). The mass of acetone that was in the decanted solution = 11.1 g.

(d). The mass of water that was in the decanted solution = 89.01 g.

(e). 2.4 = x/ 58 × (1000/1000).

x = 2.4 × 58 = 139.2 g.

(f). The molar mass of acetone = (12) + (1 × 3) + 12 + 16 + 12 + (1 x 3) = 58 g/mol.

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3 years ago

What is the molarity of a solution containing 6.0 moles of solute in 549 mL of solution?

Alexus [3.1K]

b because of the solution

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3 years ago

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The fuel used in many disposable lighters is liquid butane, C4H10. Butane has a molecular weight of 58.1 grams in one mole. How

BARSIC [14]

<h3>Answer:</h3>

6.21 × 10²² Carbon Atoms

<h3>Solution:</h3>

Data Given:

Mass of Butane (C₄H₁₀) = 1.50 g

M.Mass of Butane = 58.1 g.mol⁻¹

Step 1: Calculate Moles of Butane as,

Moles = Mass ÷ M.Mass

Putting values,

Moles = 1.50 g ÷ 58.1 g.mol⁻¹

Moles = 0.0258 mol

Step 2: Calculate number of Butane Molecules;

As 1 mole of any substance contains 6.022 × 10²³ particles (Avogadro's Number) then the relation for Moles and Number of Butane Molecules can be written as,

Moles = Number of C₄H₁₀ Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹

Solving for Number of Butane molecules,

Number of C₄H₁₀ Molecules = Moles × 6.022 × 10²³ Molecules.mol⁻¹

Putting value of moles,

Number of C₄H₁₀ Molecules = 0.0258 mol × 6.022 × 10²³ Molecules.mol⁻¹

Number of C₄H₁₀ Molecules = 1.55 × 10²² C₄H₁₀ Molecules

Step 3: Calculate Number of Carbon Atoms:

As,

1 Molecule of C₄H₁₀ contains = 4 Atoms of Carbon

So,

1.55 × 10²² C₄H₁₀ Molecules will contain = X Atoms of Carbon

Solving for X,

X = (1.55 × 10²² C₄H₁₀ Molecules × 4 Atoms of Carbon) ÷ 1 Molecule of C₄H₁₀

X = 6.21 × 10²² Atoms of Carbon

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3 years ago

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irinina [24]

Answer:

The DNA region is not easily accesible.

Explanation:

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2 years ago

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Lelu [443]

Answer:

Pascal's principle : Pressure applied to an enclosed fluid is transmitted undiminished to every part of the fluid, as well as to the walls of the container.

A common application of this is a hydraulic lift used to raise a car off the ground so it can be repaired at a garage. A small force applied to a small-area piston is transformed to a large force at a large-area piston. If a car sits on top of the large piston, it can be lifted by applying a relatively small force to the smaller piston, the ratio of the forces being equal to the ratio of the areas of the pistons.

Explanation:

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3 years ago

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