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Nezavi
5 days ago
15

A sample of an unknown substance has a mass of 0.158 kg. If 2,510.0 J of heat is required to heat the substance from 32.0°C to 6

1.0°C, what is the specific heat of the substance?
Chemistry
2 answers:
alisha [964]5 days ago
8 0
Specific heat refers to the quantity of heat a material can absorb or release to alter its temperature by one degree Celsius. To calculate specific heat, we apply the equation for the heat absorbed by the system. The heat taken in or released by a system can be expressed by multiplying the mass of the substance by its specific heat capacity and the change in temperature. The formula is:
Heat = mC(T2-T1)
By substituting the provided values, we can find C, the specific heat of the substance.
2510 J = 0.158 kg (1000 g / 1 kg)(C)(61.0 - 32.0 °C) C = 0.5478 J/g°C
lions [985]5 days ago
6 0

Information:

Mass of the unidentified material, m = 0.158 kg = 158 g

Heat intake Q = 2510.0 J

Starting temperature T1 = 32 C

Ending temperature T2 = 61 C

Required:

The specific heat capacity of the material

Analysis:

The heat required for the temperature increase from T1 to T2 is calculated as-

Q = mcΔT

where c represents specific heat

c = Q/mΔT = Q/m(T2-T1)

c = 2510 J / 158 g * (61 - 32) C = 0.548 J/gC

Result: The specific heat of the substance is 0.548 J/gC

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The metallic radius of a potassium atom is 231 pm. What is the volume of a potassium atom in cubic meters?
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11 days ago
An unknown compound, X is thought to have a carboxyl group with a pKa of 2.0 and another ionizable group with a pKa between 5 an
alisha [964]

Answer:

7.3

Explanation:

Using the Henderson Hasselbalch equation, one can determine the pH or pOH of a solution via its pKa. Remember, pH = -log[H^{+}], and pKa = -logKa, where Ka denotes the acid's equilibrium constant.

The Henderson Hasselbalch formula:

pH = pKa - log \frac{[HA]}{[A^{-}]}

In this context, acid X possesses two ionic forms: the carboxyl group and an alternative form. Initially, we have 0.1 mol/L of acid in 100 mL, which gives:

n1 = (0.1 mol/L)×(0.1 L) = 0.01 mol

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Introducing NaOH at a concentration of 0.1 mol/L and 75 mL, the moles of OH^- become:

n2 = (0.1 mol/L)×(0.075 L) = 0.0075 mol

Thus, 0.0075 mol of OH^- reacts with 0.005 mol of the carboxyl form, leading to 0.0025 mol of OH^-, which in turn reacts with 0.005 mol of the alternating group, leaving 0.0025 mol of the latter.

The new solution’s volume is 175 mL, but the concentrations of both forms remain unchanged in volume, so we can utilize the moles in the equation.

<pNote, the moles of the acid form remain 0.01 mol as it doesn’t undergo reaction!

Thus, we arrive at:

6.72 = pKa - log \frac{0.01}{0.0025}

6.72 = pKa - log 4

pKa - log 4 = 6.72

pKa = 6.72 + log 4

pKa = 6.72 + 0.6

pKa = 7.3

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