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Neko [114]
2 years ago
13

A substance with a mass of 50.0 g has a temperature change of 40.0

Chemistry
2 answers:
grigory [225]2 years ago
7 0

Answer:

Mass=50.0g

H=670J

change in temperature=40

using. c=h÷m×change in temperature

c=670÷50×40

C=670÷2000

C=0.335jkg-1k-1

Alex2 years ago
6 0

Answer: The specific heat of a substance with a mass of 50. 0g and a temperature change of 40. 0 degrees whne it absorbs 670 J of heat is 337 J/kg. K

Explanation:

Specific heat of a substance is said to be the energy required for an increase in temperature by 1°C in the unit of J/Kg. K

Formula of Specific heat:

Q = mcΔθ

Q represents the quantity of heat

m represents the mass of the substance in Kilograms = 50. 0g÷ 1000 = 0.05kg

c represents the specific heat capacity of the substance = ?

θ represents the temperature change = 40° C

Specific heat, c = Q ÷ mΔθ

                           = 670 ÷ 0.05 × 40

                           = 670 ÷ 2

                           = 335 J/ Kg. K

Therefore, the specific heat of the substance is 335 J/kg. K

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mezya [45]

Answer:

See explanation

Explanation:

The law of conservation of mass states that mass can neither be created nor destroyed. This implies that in a chemical reaction, we can only have the same number of atoms of each element on both sides of the reaction equation.

If we write  4Fe2S3 it means that we have;

4 * 2 = 8 atoms of Fe

4 * 3 = 12 atoms of S

8 + 12 = 20 atoms in all

5 0
3 years ago
Choose two of the following scientists: Anton Lavoisier, John Dalton, JJ Thomson, Robert Millikan, Ernest Rutherford, James Chad
OverLord2011 [107]

Ernest Rutherford

J. J Thomson

Explanation:

<u>Ernest Rutherford</u>

In 1911, Ernest Rutherford, a New Zealand chemist performed the gold foil experiment where he gave the modelling of the atom a boost.

 Experiment

 In his experiment, he bombarded a thin gold foil with alpha particles generated from a radioactive source. He found that most of the alpha particles passed through the gold foil while a few of them were deflected back.

 Discovery and reflection on the atomic theory

To account for his observation, Rutherford suggested an atomic model in which an atom has small positively charged center where nearly all the mass is concentrated.

<u>J. J Thomson</u>

Experiment

In 1897 J.J Thomson performed experiments using the gas discharge tube that led to the discovery of the electrons. He called them cathode rays because they originate from the cathode and exits at the anode.

Discovery and reflection on the atomic theory

From his experiment on the gas discharge tube, Thomson was able determine the properties of cathode rays some of which are:

  • they move in a straight line
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Using his observation, he proposed the plum pudding model of the atom where it is made up of entirely electrons.

learn more:

Rutherford brainly.com/question/1859083

#learnwithBrainly

6 0
3 years ago
indicate whether the entropy of the system increases or decreases. Mixing 10 mL of 90.0 °C water with 10 mL of 10 °C water. The
Ivanshal [37]

Answer:

When the water is mixed with water at lower temperature the effective temperature of the system (i.e the water at lower temperature) will increase, thereby increasing it's entropy

Explanation:

The answer that "the entropy will is increases" is correct as:

The water at 90° C i.e at higher temperature is mixed with the water at 10° C i.e the water at the lower temperature.

The water at lower temperature will have molecules with lower energy while the water with higher temperature will have molecules undergoing high thermal collisions. Thereby, when the water is mixed with water at lower temperature the effective temperature of the system (i.e the water at lower temperature) will increase, thereby increasing it's entropy.

Therefore, the answer is correct with respect to the water at lower temperature.

Meanwhile, for the water at higher temperature , the temperature of the system will decrease. Thus, the entropy of the water at higher level will decrease.

5 0
4 years ago
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KengaRu [80]

Answer: The density of the object will be 0.573g/cm^3

Explanation:

Density is defined as the mass contained per unit volume.

Density=\frac{mass}{Volume}

Given : Mass of object = 19.6 grams

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Putting in the values we get:

Density=\frac{19.6g}{34.2cm^3}=0.573g/cm^3

Thus density of the object will be 0.573g/cm^3

8 0
4 years ago
What is 25 °C in °F?
kifflom [539]

Answer:

77°F

Explanation:

(25°C × 9/5) + 32 = 77°F

7 0
3 years ago
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