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alexandr1967 [171]
3 years ago
13

Calculate the specific heat of a substance if a 35g sample absorbs 48 j as the temperature is raised from 293 k to 313 k

Chemistry
1 answer:
bulgar [2K]3 years ago
8 0

Answer:

c = 0.07 j/g.k

Explanation:

Given data:

Mass of sample = 35 g

Heat absorbed = 48 j

Initial temperature = 293 K

Final temperature = 313 K

Specific heat of substance = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature - initial temperature

ΔT = 313 k - 293 K

ΔT = 20 k

Now we will put the values in formula.

48 j = 35 g × c× 20 k

48 j = 700 g.k ×c

c = 48 j/700 g.k

c = 0.07 j/g.k

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How many digits to the right of the decimal point should be used to report the result?
aalyn [17]

Answer:

In engineering and science the common stand is two places.

For example if you get a calculation of 4.567 round up and give the result of 4.57

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3 years ago
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What is the mass, in grams, of a liquid having a density of 1.50g/mL and a volume of 3500mL
mojhsa [17]

Answer:

5,250g

Explanation:

Density = Mass / Volume

*Note: mass = x

1.50g/mL = x / 3500mL

multiply 3500mL on both sides

1.50g/mL * 3500mL = 3500mL(x) / 3500mL

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5 0
3 years ago
On the reaction below, label the BSA, BSB, CA, and CB. CH3COOH + H2O → CH3COO– + H3O+
Ostrovityanka [42]

Answer:

Acid(BSA) = CH₃COOH

Base (BSB) = H₂O

Conjugate base (CB) = CH₃COO⁻

Conjugate acid (CA) = H₃O⁺

Explanation:

Equation of reaction;

CH₃COOH + H₂O → CH₃COO⁻ + H₃O⁺

Hello,

From my understanding of the question, we are required to identify the

1) Acid

2) Base

3) conjugate acid

4) conjugate base in the reaction

Acid (BSA) = CH₃COOH

Base (BSB) = H₂O

CA = conjugate acid = H₃O⁺

CB = conjugate base = CH₃COO⁻

7 0
3 years ago
100 ml of a 0.300 m solution of agno3 reacts with 100 ml of a 0.300 m solution of hcl in a coffee-cup calorimeter and the temper
Olin [163]

Answer:

100 ml of a 0.300 m solution of agno3 reacts with 100 ml of a 0.300 m solution of hcl in a coffee-cup calorimeter and the temperature rises from 21.80 °c to 23.20 °c. Assuming the density and specific heat of the resulting solution is 1.00 g/ml and 4.18 j/g ∙ °c respectfully, what is the ΔH°rxn?

39.013 kJ/mol.

Explanation:

AgNO3(aq) + HCl(aq) --------------> AgCl(s) + HNO3(aq)

We can calculate the amount of heat (Q) released from the solution using the relation:

Q = m.c.ΔT,

Where, Q is the amount of heat released from the solution (Q = ??? J).

m is the mass of the solution (m of the solution = density of the solution x volume of the solution = (1.0 g/mL)(200 mL) = 200 g.

c is the specific heat capacity of the solution (c = 4.18 J/g∙°C).

ΔT is the difference in the T (ΔT = final temperature - initial temperature = 23.20 °C - 21.80 °C = 1.4 °C).

∴ Q = m.c.ΔT = (200 g)(4.18 J/g∙°C)(1.4 °C) = 1170.4 J.

∵ ΔH°rxn = Qrxn/(no. of moles of AgNO₃).

Molarity (M) is defined as the no. of moles of solute dissolved in a 1.0 L of the solution.

M = (no. of moles of AgNO₃)/(Volume of the solution (L)).

∴ no. of moles of AgNO₃

               = (M)(Volume of the solution (L))

               = (0.3 M)(0.1 L) = 0.03 mol.

∴ ΔH°rxn

           = Qrxn/(no. of moles of AgNO₃)

            = (1170.4 J)/(0.03 mol)

            = 39013.33 J/mol

           = 39.013 kJ/mol.

7 0
3 years ago
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