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Arisa [49]
3 years ago
10

A piece of silver releases 202.8 J of heat while cooling 65.0 °C. What is the mass of the sample? Silver has a specific heat of

0.240 J/g °C.
Chemistry
1 answer:
KiRa [710]3 years ago
7 0

Answer: 13 grams

Explanation:

The quantity of heat energy (Q) released from a heated substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Since,

Q = 202.8 Joules

Mass of silver = ?

C = 0.240 J/g °C.

Φ = 65°C

Then, Q = MCΦ

202.8J = M x 0.240 J/g °C x 65°C

202.8J = M x 15.6 J/g

M = (202.8J / 15.6 J/g)

M = 13 g

Thus, the mass of silver is 13 grams

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Answer:

see explanation

Explanation:

Ice usually has tightly packed molecules at a low temperature. When it comes in contact with a higher temperature or room temperature, the ice molecules gain energy and the molecular tension increases which causes the state to change to liquid. Therefore, a <u>high temperature</u> causes an ice block to melt.

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Scientists use models to
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B- Predict the damages due to an earthquake
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The heat energy needed to raise the temperature of vapor above its saturation point is called _____________________. A. sensible
lana [24]

Answer:

Option (D)

Explanation:

The super-heating is usually defined as a phenomenon where a certain amount of energy is needed to raise the temperature of the water vapor beyond its normal saturation point. This is also known as the boiling delay.

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Super-heat = Current temperature - Boiling point of the liquid.

Thus, super-heat refers to the amount of energy that is required to increase the temperature of vapor beyond its point of saturation.

This super-heat is essential as it helps in preventing the damages of machines like air conditioner, fridge and also helps in their soft running.

Hence, the correct answer is option (D).

7 0
3 years ago
A student dissolves 6.2g of aniline c6h5nh2 in 350.ml of a solvent with a density of 1.04/gml . the student notices that the vol
Troyanec [42]

Answer : The correct answer for molarity = 0.19 \frac{mol}{L} and Molality = 0.18 \frac{mol}{Kg}.

Given : Mass of aniline = 6.2 g

Volume of solvent = 350 mL Density of solvent = 1.04 g/mL

1) Molarity : It is defined as number of moles of solute present in Litre of solution . It is expressed as :

Molarity(M) = \frac{moles of solute(mole)}{volume of solution (L)}\

Molairty can be found in following steps :

Step 1 : To calculate mole :

Mole of solute can be calculate using mole formula :

Mole of solute = \frac{given mass of solute (g)}{molar mass of solute\frac{g}{1 mol}}

Molar mass of aniline (C₆H₅NH₂) = 93.13 \frac{g}{1 mol}

Mass of aniline = 6.2 g

Plugging values in mole formula :

Mole  =  \frac{6.2 g}{93.13 \frac{g}{1 mol}}

Mole = 0.0665 mol

Step 2 : To find volume of solution

Volume of solution = volume of solute + volume of solution

Since addition of aniline does not change final volume , so volume of solvent = volume of solution. Since volume is given in mL , so it need to be converted to L .

1 L = 1000mL

Volume of solution = \frac{350 mL}{1000mL}  * 1 L

Volume of solution = 0.350 L

Step 3: Plug value of mole and volume in molarity formula :

Molarity = \frac{0.0665 mol}{0.350 L }

Molarity = 0.19 M or 0.19 \frac{mol}{L}

------------------------------------------------------------------------------------------

2) Molality : It is defined as mole of solute present in Kilogram of solvent . It can be expressed as :

Molality (m) = \frac{mole of solute (mol)}{Kilogram of solvent (Kg)}

Following are the steps to calculate molality :

Step 1: To find mole of Solute :

Mole of solute can be found out using mole formula . It is same as done for molarity .

Mole = 0.0665 mol

Step 2 : To find kilogram of solvent :

Mass of solvent can be calculated using density formula as :

Density \frac{g}{mL} = \frac{mass (g) }{volume (mL)}

Plugging value in density formula :

1.04 \frac{g}{mL}  =  \frac{ mass }{350 mL}

Multiplying both side by 350 mL

1.04 \frac{g}{mL} * 350 mL = \frac{x}{350 mL } * 350 mL

Mass of solvent = 364 g

Since mass is in g, it need to be converted to Kg . ( 1 Kg = 1000 g )

Mass of solvent = \frac{364 g}{1000g} * 1 Kg

Mass of solvent = 0.364 Kg

Step 3: Plug values of mole and Kg in molality formula :

Molality = \frac{0.0665 mol}{0.364 Kg}

Molality = 0.18 m or 0.18 \frac{mol}{Kg}

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a_sh-v [17]
2Kl+HgCl2 --> 2KCl + HgI2 

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