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Maru [420]
2 years ago
13

Ordinary table sugar is primarily sucrose, which has the chemical formula c12h22o11. calculate the mass percentages of carbon, h

ydrogen, and oxygen in sucrose.
Chemistry
1 answer:
tester [92]2 years ago
6 0
<span>carbon = 42.1% hydrogen = 6.5% oxygen = 51.4% First lookup the molar mass of carbon, hydrogen, and oxygen. mass of carbon = 12.0107 mass of hydrogen = 1.00794 mass of oxygen = 15.999 Now calculate the molar mass of each element in sucrose multiplying the atomic weight of each element by the number of times the element is used. carbon = 12 * 12.0107 = 144.1284 hydrogen = 22 * 1.00794 = 22.17468 oxygen = 11 * 15.999 = 175.989 Calculate the molar mass of sucrose by adding the mass of each element used. 144.1284 + 22.17468 + 175.989 = 342.2921 Finally, calculate the percentage by mass of each element by dividing the mass used for each element by the total mass of sucrose. carbon = 144.1284 / 342.2921 = 0.421068 = 42.1% hydrogen = 22.17468 / 342.2921 = 0.064783 = 6.5% oxygen = 175.989 / 342.2921 = 0.514149 = 51.4%</span>
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When 13.6 g of calcium chloride, CaCl2, was dissolved in 100.0 mL of water in a coffee cup calorimeter, the temperature rose fro
DanielleElmas [232]

Answer:

THE ENTHALPY OF SOLUTION IS 3153.43 J/MOL OR 3.15 KJ/MOL.

Explanation:

1. write out the variables given:

Mass of Calcium chloride = 13.6 g

Change in temperature = 31.75°C - 25.00°C = 6.75 °C

Density of the solution = 1.000 g/mL

Volume = 100.0 mL = 100.0 mL

Specific heat of water = 4.184 J/g °C

Mass of the water = unknown

2. calculate the mass of waterinvolved:

We must first calculate the mass of water in the bomb calorimeter

Mass = density  * volume

Mass = 1.000 * 100

Mass = 0.01 g

3. calculate the quantity of heat evolved:

Next is to calculate the quantity of heat evolved from the reaction

Heat = mass * specific heat of water * change in temperature

Heat = mass of water * specific heat *change in temperature

Heat = 13.6 g * 4.184 * 6.75

Heat = 13.6 g * 4.184 J/g °C * 6.75 °C

Heat = 384.09 J

Hence, 384.09J is the quantity of heat involved in the reaction of 13.6 g of calcium chloride in the calorimeter.

4. calculate the molar mass of CaCl2:

Next is to calculate the molar mas of CaCl2

Molar mass = ( 40 + 35.5 *2) = 111 g/mol

The number of moles of 13.6 g of CaCl2 is then:

Number of moles of CaCl2 = mass / molar mass

Number of moles = 13.6 g / 111 g/mol

Number of moles = 0.1225 mol

So 384.09 J of heat was involved in the reaction of 1.6 g of CaCl2 in a calorimter which translates to 0.1225 mol of CaCl2..

5. Calculate the enthalpy of solution in kJ/mol:

If 1 mole of CaCl2 is involved, the heat evolved is therefore:

Heat per mole = 384.09 J / 0.1225 mol

Heat = 3 135.43 J/mol

The enthalpy of solution is therefore 3153.43 J/mol or 3.15 kJ/mol.

5 0
2 years ago
a gas sample is heated from -20.0 C to 57.0 C and the volume is increased from 2.00L to 4.50L. If the initial pressure is 0.109a
Andreas93 [3]
Idk what this is im not good at this 
7 0
2 years ago
BH+ClO4- is a salt formed from the base B (Kb = 1.00e-4) and perchloric acid. It dissociates into BH+, a weak acid, and ClO4-, w
Len [333]

Answer:

The pH of 0.1 M BH⁺ClO₄⁻ solution is <u>5.44</u>

Explanation:

Given: The base dissociation constant: K_{b} = 1 × 10⁻⁴, Concentration of salt: BH⁺ClO₄⁻ = 0.1 M

Also, water dissociation constant: K_{w} = 1 × 10⁻¹⁴

<em><u>The acid dissociation constant </u></em>(K_{a})<em><u> for the weak acid (BH⁺) can be calculated by the equation:</u></em>

K_{a}. K_{b} = K_{w}    

\Rightarrow K_{a} = \frac{K_{w}}{K_{b}}

\Rightarrow K_{a} = \frac{1\times 10^{-14}}{1\times 10^{-4}} = 1\times 10^{-10}

<em><u>Now, the acid dissociation reaction for the weak acid (BH⁺) and the initial concentration and concentration at equilibrium is given as:</u></em>

Reaction involved: BH⁺  +  H₂O  ⇌  B  +  H₃O+

Initial:                     0.1 M                    x         x            

Change:                   -x                      +x       +x

Equilibrium:           0.1 - x                    x         x

<u>The acid dissociation constant: </u>K_{a} = \frac{\left [B \right ] \left [H_{3}O^{+}\right ]}{\left [BH^{+} \right ]} = \frac{(x)(x)}{(0.1 - x)} = \frac{x^{2}}{0.1 - x}

\Rightarrow K_{a} = \frac{x^{2}}{0.1 - x}

\Rightarrow 1\times 10^{-10} = \frac{x^{2}}{0.1 - x}

As, x

\Rightarrow 0.1 - x = 0.1

\therefore 1\times 10^{-10} = \frac{x^{2}}{0.1 }

\Rightarrow x^{2} = (1\times 10^{-10})\times 0.1 = 1\times 10^{-11}

\Rightarrow x = \sqrt{1\times 10^{-11}} = 3.16 \times 10^{-6}

<u>Therefore, the concentration of hydrogen ion: x = 3.6 × 10⁻⁶ M</u>

Now, pH = - ㏒ [H⁺] = - ㏒ (3.6 × 10⁻⁶ M) = 5.44

<u>Therefore, the pH of 0.1 M BH⁺ClO₄⁻ solution is 5.44</u>

5 0
2 years ago
Find the molarity of 186.55 g of sugar (C12H22O11) in 250. mL of water.
Anna [14]

Answer:

The molarity of this sugar solution in water is 2.18 M

Explanation:

Step 1: Data given

Mass of sugar (C12H22O11) = 186.55 grams

Molar mass of C12H22O11 = 342.3 g/mol

Volume of water = 250.0 mL = 0.250 L

Step 2: Calculate moles sugar

Moles sugar = mass sugar / molar mass sugar

Moles sugar = 186.55 grams / 342.3 g/mol

Moles sugar = 0.545 moles

Step 3: Calculate molarity of the sugar solution

Molarity = moles sugar / volume of water

Molarity = 0.545 moles / 0.250 L

Molarity = 2.18 MThe molarity of this sugar solution in water is 2.18 M

6 0
2 years ago
Consider the acid H3PO4. This acid will react with water by the following equation. H3PO4+H2O↽−−⇀H2PO−4+H3O+ What will be true o
Sonbull [250]

Answer:

H+/PO-4^-2

Explanation:

hydrogen has dissolved completely

7 0
2 years ago
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