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2 years ago

Calculate the specific heat capacity of aluminum if 14,200 J of heat is released in cooling a 350.0

Chemistry

1 answer:

andrew-mc [135]2 years ago

3 0

Answer:

Specific Heat Capacity = 0.901 J.g⁻¹.°C⁻¹

Heat is Exothermic

Explanation:

Specific heat capacity is the amount of heat required to raise the temperature of a given amount of substance by one degree.

Also, Exothermic reactions are those reaction in which the heat is lost from the system to surrounding while, endothermic reactions are those in which the system gains heat from the surroundings.

The equation used for this problem is as follow,

Q = m Cp ΔT ----- (1)

Where;

Q = Heat = 14200 J

m = mass = 350 g

Cp = Specific Heat Capacity = ??

ΔT = Change in Temperature = 70 °C - 25 °C = 45 °C

Solving eq. 1 for Cp,

Cp = Q / m ΔT

Putting values,

Cp = 14200 J / (350 g × 45 °C)

Cp = 0.901 J.g⁻¹.°C⁻¹

As the heat is lost by the metal therefore, the heat is exothermic.

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How would the freezing point of seawater compare with that of pure water?

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Pure water has a freezing point of 0 degrees centigrade but seawater has a lower freezing point because of the salt (NaCl)

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3 years ago

A 50.0 mL sample of 12.0 M HCl is diluted to 200 mL. What is true about the diluted solution?

kobusy [5.1K]

The concentration of the solution reduces and the number of moles of solute isn't affected.

Data;

V1 = 50mL
C1 = 12.0M
V2 = 200mL
C2 = ?

<h3>Facts about the diluted solution</h3>

1. When the solution is diluted, the concentration changes and this time, the concentration reduces.

Using dilution formula

$c_1 v_1 = c_2 v_2\\12 * 50 = c_2 * 200\\c_2 = \frac{600}{200} \\c_2 = 3M$

The concentration of the solution reduces.

2. The number of moles remains the same.

When a solution is diluted, the number of moles remains the same because there's no change in the mass of the solute.

Learn more on concentration of a solution here;

brainly.com/question/2201903

7 0

2 years ago

A cylinder with a moving piston expands from an initial volume of 0.350 L against an external pressure of 1.90 atm. The expansio

myrzilka [38]

Answer:

1.84 L

Explanation:

Using the equation for reversible work:

$W = -P*(V_{2} - V_{1})$

Where:

W is the work done (J) = -287 J.

Since the gas did work, therefore W is negative.

P is the pressure in atm = 1.90 atm.

However, work done is in joules and pressure is in atm. We can use the values of universal gas constant as a convenient conversion unit. R = 8.314 J/(mol*K); R = 0.0821 (L*atm)/(mol*K)

Therefore, the conversion unit is 0.0821/8.314 = 0.00987 (L*atm)/J

$V_{1}$ is the initial volume = 0.350 L