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IrinaVladis [17]
2 years ago
14

A 135-g sample of a metal requires 2.50kj to change its temperature from 19.5 degrees Celsius to 100.0 degrees Celsius. What is

the specific heat of this metal?
Chemistry
2 answers:
PolarNik [594]2 years ago
7 0
Q = m x c x ΔT

2500 = 0.135 x C x 80.5

2500 = 10.8765 x C

C = 230.043 J/Kg.K

hope this helps
mars1129 [50]2 years ago
5 0

Answer: The specific heat of the metal is 0.23J/g^0C

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

Q=m\times c\times \Delta T

Q = Heat absorbed=2.50kJ=2500J

m= mass of substance = 135 g

c = specific heat capacity = ?

Initial temperature = T_i = 19.5°C

Final temperature = T_f  =100.0°C

Change in temperature ,\Delta T=T_f-T_i=(100-19.5)^0C=80.5^0C

Putting in the values, we get:

2500=135\times c\times 80.5^0C

c=0.23J/g^0C

The specific heat of the metal is 0.23J/g^0C

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Answer:

Ok:

Explanation:

So, you can use the Henderson-Hasselbalch equation for this:

pH = pKa + log(A^-/HA) where A- is the conjugate base of the acid. In other words, A- is the deprotonated form and HA is the protonated.

We can solve that

1 = log(A^-/HA\\) and so 10 = A^-/HA or 10HA = A-.  For every 1 protonated form of adenosine (HA), there are 10 A-. So, the percent in the protonated form will be 1(1+10) or 1/11 which is close to 9 percent.

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The table lists the chemical formulas or symbols of a few common substances. Which two substances are compounds? A) aluminum and
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A block of aluminum weighing 140 g is cooled from 98.4°C to 62.2°C with the release of 1080 joules of heat. From this data, calc
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<u>Answer:</u>

Specific heat of a substance is the value that describe how the added heat energy of substance has the impact on its temperature.

Unit is <em>(\frac {J}{Kg.K})</em>

<em>C = Q/m. ∆T</em>

<em>C – Specific heat (\frac {J}{Kg.K})</em>

<em>Q- heat energy (J)</em>

<em>M – Mass (Kg)</em>

<em>∆T- change in temperature (K) </em>

<u>Explanation:</u>

<em>Given data:</em>

<em>M= 140 g = 0.14 Kg</em>

<em>Q – 1080 Joules.</em>

<em>∆T – 98.4 – 62.2 = 36.2</em>

Substituting  the given data in Equation

<em>Specific heat of Aluminium  = \frac {1080}{(0.14 \times 36.2)} = 213.10 (\frac {J}{Kg.K})</em>

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<em>The correct option would be that the average kinetic energy of the gas particles is greater in container B because it has a higher temperature.</em>

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In the illustration, the gas in container B showed a higher temperature than that of container A as indicated on the thermometer, it thus means that the average kinetic energy of the molecules of gas B is higher than those of gas A.

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