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Serggg [28]
3 years ago
10

6CO2 + 6 H2O —> C6H12O6 + 6O2

Chemistry
1 answer:
PilotLPTM [1.2K]3 years ago
3 0

Answer:

The answer to your question is 0.5 moles

Explanation:

Data

moles of Glucose = ?

moles of carbon dioxide = 3

Balanced chemical reaction

                6CO₂  +  6H₂O   ⇒   C₆H₁₂O₆  +  6O₂

Process

To solve this problem, use proportions, and cross multiplication.

Use the coefficients of the balanced equation.

                6 moles of CO₂ ----------------- 1 mol of C₆H₁₂O₆

                3 moles of CO₂ ----------------    x

                    x = (3 x 1) / 6

-Simplification

                    x = 3/6

-Result

                   x = 0.5 moles of Glucose

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3 years ago
How many pints of a 30% sugar solution must be added to a 5 pint of a 5% sugar solution to obtain a 20% sugar solution?
ser-zykov [4K]

You must add 7.5 pt of the 30 % sugar to the 5 % sugar to get a 20 % solution.

You can use a modified dilution formula to calculate the volume of 30 % sugar.

<em>V</em>_1×<em>C</em>_1 + <em>V</em>_2×<em>C</em>_2 = <em>V</em>_3×<em>C</em>_3

Let the volume of 30 % sugar = <em>x</em> pt. Then the volume of the final 20 % sugar = (5 + <em>x</em> ) pt

(<em>x</em> pt×30 % sugar) + (5 pt×5 % sugar) = (<em>x</em> + 5) pt × 20 % sugar

30<em>x</em> + 25 = 20x + 100

10<em>x</em> = 75

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5 0
3 years ago
How would you convert 2930 kcal to kJ?
andrezito [222]

It is quite easy:

1 cal = 4,1868 J

Solution is:

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3 years ago
What is the difference between valence and valency
zhuklara [117]

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Explanation:

4 0
3 years ago
Calculate the molaity of a solution that is prepared by dissolving 50.4g sucrose (C12H22O11) in 0.332kg of water
maks197457 [2]

Answer:

0.443 mol/kg.

Explanation:

The formula for molal concentration (<em>b</em>) is

b = \frac{\text{moles of solute}}{\text{kilograms of solvent}}

\text{Moles of solute} = \text{50.4 g sucrose} \times \frac{ \text{1 mol sucrose}}{\text{342.30 g sucrose} } = \text{0.1472 mol sucrose}

b = \frac{ \text{0.1472 mol}}{ \text{0.332 kg}} = \text{0.443 mol}\cdot\text{kg}^{-1}

4 0
3 years ago
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