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3 years ago

Which relationship can be used to aid in the determination of the heat absorbed by bomb calorimeter?

Chemistry

1 answer:

NARA [144]3 years ago

5 0

Answer:

ΔH = $q_{p}$

Explanation:

In a calorimeter, when there is a complete combustion within the calorimeter, the heat given off in the combustion is used to raise the thermal energy of the water and the calorimeter.

The heat transfer is represented by

$q_{com}$ = $q_{p}$

where

$q_{p}$ = the internal heat gained by the whole calorimeter mass system, which is the water, as well as the calorimeter itself.

$q_{com}$ = the heat of combustion

Also, we know that the total heat change of the any system is

ΔH = ΔQ + ΔW

where

ΔH = the total heat absorbed by the system

ΔQ = the internal heat absorbed by the system which in this case is $q_{p}$

ΔW = work done on the system due to a change in volume. Since the volume of the calorimeter system does not change, then ΔW = 0

substituting into the heat change equation

ΔH = $q_{p}$ + 0

==> ΔH = $q_{p}$

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Nucleic acids are made of which of the following?

abruzzese [7]

Answer: I believe the correct answer would be A.

5 0

2 years ago

Read 2 more answers

Given the following balanced equation, if the rate of O2 loss is 3.64 × 10-3 M/s, what is the rate of formation of SO3? 2 SO2(g)

Fynjy0 [20]

Answer:

Rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

Explanation:

According to equation 2 SO₂(g) + O₂(g) → 2 SO₃(g)

Rate of disappearance of reactants = rate of appearance of products

⇒ $-\frac{1}{2} \frac{d[SO_{2} ]}{dt} = -\frac{d[O_{2} ]}{dt}=\frac{1}{2} \frac{d[SO_{3} ]}{dt}$ -----------------------------(1)

Given that the rate of disappearance of oxygen = $-\frac{d[O_{2} ]}{dt}$ = 3.64 x 10⁻³ M/s

So the rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = ?

from equation (1) we can write

$\frac{d[SO_{3}] }{dt} = 2 [-\frac{d[O_{2}] }{dt} ]$

⇒ $\frac{d[SO_{3}] }{dt}$ = 2 x 3.64 x 10⁻³ M/s

⇒ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

∴ So the rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

7 0

3 years ago

Certain hydrocarbon is 92.3 % carbon and 7.7 % hydrogen by mass.If the molar mass of the hydrocarbon is approximately 40 g/mol,

zimovet [89]

Answer:

The hydrocarbon has a molecular formula of C3H3

Explanation:

Step 1: Data given

Suppose the mass of the compound is 100 grams

A hydrocarbon contains:

⇒ 92.3 % carbon = 92.3 grams

⇒ 7.7 % hydrogen = 7.7 grams

Molar mass of the compound is 40 g/mol

Molar mass of carbon = 12.01 g/mol

Molar mass of hydrogen = 1.01 g/mol

Step 2: Calculate moles

Moles = mass / molar mass

Moles carbon = 92.3 grams / 12.01 g/mol

Moles carbon = 7.69 moles

Moles hydrogen = 7.70 grams / 1.01 g/mol

Moles hydrogen = 7.62

Step 3: Calculate mol ratio

We divide by the smallest amount of moles

Carbon: 7.69/7.62 = 1

Hydrogen: 7.62/7.62 = 1

The empirical formula is CH

The molar mass is 13.02 g/mol

Step 4: Calculate the molecular formula

n = 40 g/mol / 13.02 g/mol

n ≈ 3

We need to multiply the empirical formula by 3 to get the molecular formula

Molecular formula = 3*(CH) = C3H3

The hydrocarbon is C3H3

5 0

2 years ago

Which of the following statements best describes why a chemical change is different from a physical change.

Butoxors [25]

Answer:

C) Chemical changes involve formation of a new substance while physical change only impacts appearance or form.

Explanation:

The obvious difference between a chemical change and physical change is that in a chemical change new substances are produced while in a physical change, the form of the compounds are altered.

Most phase changes are products of physical changes. Such changes are easily reversible.
In a chemical change, a new kind of matter is formed. It is accompanied by energy changes.
The process is not easily reversible.

3 0

3 years ago

Manganese sulfate forms a pale pink hydrate with the formula MnSO 4 ⋅ n H 2 O ( s ) . If this hydrate is heated to a high enough

Mama L [17]

Answer:

Value of n in MnSO₄.nH₂O is one.

Explanation:

The n represents the number of moles of water attached to the formula unit manganese sulfate. These moles (n) can be determined by taking the ratio of the moles of anhydrous salt and the moles of water. The moles of water can be determined by taking the difference of final and initial mass of the salt. This difference is equal to the mass of the water, mathematically it can be represented as,

Mass of H₂O = initial mass of the salt (g) - final mass of the salt (g)

Mass of H₂O = 16.260 g - 14.527 g

Mass of H₂O = 1.733 g

moles of H₂O = (1.733 g) ÷ (18.015 g/mole)

moles of H₂O = 0.0962

For the moles of anhydrous salt:

moles of MnSO₄ = mass of MnSO₄ ÷ molar mass of MnSO₄

moles of MnSO₄ = 14.5277 ÷ 151.001

moles of MnSO₄= 0.0962

Now for n:

n = moles of water ÷ moles of MnSO₄

n = 0.0962 ÷ 0.0962

n = 1

The above calculations show that one mole of H₂O is attached to the one formula unit of MnSO₄

8 0

3 years ago

Which relationship can be used to aid in the determination of the heat absorbed by bomb calorimeter? ￼

Which relationship can be used to aid in the determination of the heat absorbed by bomb calorimeter?