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Maru [420]
3 years ago
10

A candy bar has a total mass of 75.0 grams. In a calorimetry experiment, a 1.0-g sample of this candy bar was burned in a calori

meter surrounded by 1000g of water. The temperature of the water in contact with the burning candy bar was measured and found to increase from an initial temperature of 21.2C to a final temperature of 24.3C.
a. Calculate the amount of heat in calories release when the 1.0-g sample burned.
b. Convert the heat in calories to nutritional Calories and then calculate the energy content (fuel value) in Cal/g.
c. Calculate the total caloric content of the candy bar in Calories.
Chemistry
1 answer:
geniusboy [140]3 years ago
6 0
<span>a. 3.1 calories in heat were released by the burning candy bar sample b. The energy value of the sample was 3.1 Cal/g c.The total caloric content of the candy bar is 232.5 Calories, derived by multiplying the total mass of the candy bar by its fuel value per gram</span>
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A total of 25.0 mL of 0.150 M potassium hydroxide (KOH) was required to neutralize 15.0 mL of sulfuric acid (H2SO4) of unknown c
Kamila [148]
We are told that KOH is being used to completely neutral H₂SO₄ according to the following reaction:

KOH + H₂SO₄ → H₂O + KHSO₄

If KOH can completely neutralize H₂SO₄, then there must be an equal amount of moles of each as they are in a 1:1 ratio:

0.025 L x 0.150 mol/L = .00375 mol KOH

0.00375 mol KOH x 1 mole H₂SO₄/1 mole KOH = 0.00375 mol H₂SO₄

We are told we have 15 mL of H₂SO₄ initially, so now we can find the original concentration:

0.00375 mol / 0.015 L = 0.25 mol/L

The concentration of H₂SO₄ being neutralized is 0.25 M.
6 0
3 years ago
or a particular isomer of C 8 H 18 , the combustion reaction produces 5104.1 kJ of heat per mole of C 8 H 18 ( g ) consumed, und
beks73 [17]

Answer:

The standard enthalpy of formation of this isomer of C_{8}H_{18} is -220.1 kJ/mol.

Explanation:

The given chemical reaction is as follows.

C_{8}H_{18}(g)+ \frac{25}{2}O_{2}(g) \rightarrow 8CO_{2}(g)+9H_{2}O(g)

\Delta H^{o}_{rxn}= -5104.1kJ/mol

The expression for the entropy change for the reaction is as follows.

\Delta H^{o}_{rxn}=[8\Delta H^{o}_{f}(CO_{2}) +9\Delta H^{o}_{f}(H_{2}O)]-[\Delta H^{o}_{f}(C_{8}H_{18})+ \frac{25}{2}\Delta H^{o}_{f}(O_{2})]

\Delta H^{o}_{f}(H_{2}O)= -241.8kJ/mol

\Delta H^{o}_{f}(CO_{2})= -393.5kJ/mol

\Delta H^{o}_{f}(O_{2})= 0kJ/mol

Substitute the all values in the entropy change expression.

-5104.1kJ/mol=[8(-393.5)+9(-241.8)kJ/mol]-[\Delta H^{o}_{f}(C_{8}H_{18})+ \frac{25}{2}(0)kJ/mol]

-5104.1kJ/mol=-5324.2kJ/mol -\Delta H^{o}_{f}(C_{8}H_{18})

\Delta H^{o}_{f}(C_{8}H_{18}) =-5324.2kJ/mol +5104.1kJ/mol

=-220.1kJ/mol

Therefore, The standard enthalpy of formation of this isomer of C_{8}H_{18} is -220.1 kJ/mol.

4 0
3 years ago
What is k in the rate law equation?
REY [17]

Answer:

B. A rate constant

Explanation:

The mathematical expression of rate law is given below,

Rate = K[A]m[B]n

This rate law show the relationship between the rate of chemical reaction and concentration of reactants.

In given equation [A] and [B] are molar concentration of reactants while K represent rate constant.

The value of K is specific for particular reaction at particular temperature,

m and n are represent exponents and determine experimentally. The value of K is not depend upon the concentrations of reactant but depend upon the surface area and temperature

5 0
3 years ago
Read 2 more answers
When two atoms form a covalent bond, they share electrons from all of their orbitals. All of their orbitals, in turn, combine to
erma4kov [3.2K]

Answer:

It's false.

Explanation:

Molecular orbital theory states that the number of molecular orbitals is equal to the number of atomic orbitals that overlap. The lowest energy molecular orbital is formed when two atomic orbitals that are in phase overlap, forming a bonding molecular orbital. However, another molecular orbital is also formed, called an anti-binding orbital.

So if an "n" quantity of atomic orbitals is combined, an "n" quantity of molecular orbitals is formed.

Have a nice day!

8 0
4 years ago
What is the mass, in grams, of 8 mol of 12c?
Kitty [74]

The mass of 8 mol of ¹²C is 96 g.

Mass = 8 mol × (12 g/1 mol) = 96 g

5 0
3 years ago
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