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IRISSAK [1]
3 years ago
10

A substance absorbed 895 J and caused its temperature to increase by 10 ºC. If the substance has a mass of 27.9 g, what is the s

ubstance's specific heat in J/(gºC) ?
1.) 3.21

2.) 4.62

3.) 1,406.25

4.) 249,705
Chemistry
2 answers:
Irina18 [472]3 years ago
5 0

Answer:

i dont get the equation sorry wish i can help

vova2212 [387]3 years ago
4 0
4.62 is the correct answer
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The dissociation of cadmium chloride is as follows: CdCl₂(s) → Cd⁺²(aq) + 2Cl⁻(aq)

<h3>What is dissociation?</h3>

Dissociation is the process by which a compound body breaks up into simpler constituents; said particularly of the action of heat on gaseous or volatile substances.

It is a chemical reaction in which a compound breaks apart into two or more components. The general formula for a dissociation reaction follows the form:

AB → A + B

According to this question, cadmium chloride undergoes dissociation into cadmium and chlorine ions as follows:

CdCl₂(s) → Cd⁺²(aq) + 2Cl⁻(aq)

Learn more about dissociation at: brainly.com/question/28952043

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8 0
1 year ago
What is a sign that a chemical change has occurred?
blondinia [14]

Answer:

The production of heat, light, or smoke is observed.

3 0
3 years ago
Read 2 more answers
Why series fatty acid carbons seen in double from?
BARSIC [14]
Stearic acid is a fully saturated fatty acid with no carbon-carbon double bonds. (Bottom) Oleic acid is an unsaturated fatty acid.
8 0
2 years ago
While exploring a coal mine, scientists found plant fossils in the ceiling of the mine which had been preserved by an earthquake
ale4655 [162]

Answer:

11552.45 years

Explanation:

Given that:

Half life = 5730 years

t_{1/2}=\frac{\ln2}{k}

Where, k is rate constant

So,  

k=\frac{\ln2}{t_{1/2}}

k=\frac{\ln2}{5730}\ years^{-1}

The rate constant, k = 0.00012 years⁻¹

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,  

[A_t] is the concentration at time t

[A_0] is the initial concentration

Given that:

The rate constant, k = 0.00012 years⁻¹

Initial concentration [A_0] = 160.0 counts/min

Final concentration [A_t] = 40.0 counts/min

Time = ?

Applying in the above equation, we get that:-

40.0=160.0e^{-0.00012\times t}

e^{-0.00012t}=\frac{1}{4}

-0.00012t=\ln \left(\frac{1}{4}\right)

t=11552.45\ years

7 0
3 years ago
A sample of methane gas, CH4, occupies 3.25 L at temperature of 19.0 o C. If the pressure is held constant, what will be the tem
Artemon [7]

Answer:

625.46 °C

Explanation:

We'll begin by converting 19 °C to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

T(°C) = 19 °C

T(K) = 19 °C + 273

T(K) = 292 K

Next, we shall determine the Final temperature. This can be obtained as follow:

Initial volume (V₁) = 3.25 L

Initial temperature (T₁) = 292 K

Final volume (V₂) = 10 L

Final temperature (T₂) =?

V₁/T₁ = V₂/T₂

3.25 / 292 = 10 / T₂

Cross multiply

3.25 × T₂ = 292 × 10

3.25 × T₂ = 2920

Divide both side by 3.25

T₂ = 2920 / 3.25

T₂ = 898.46 K

Finally, we shall convert 898.46 K to celsius temperature. This can be obtained as follow:

T(°C) = T(K) – 273

T(K) = 898.46 K

T(°C) = 898.46 – 273

T(°C) = 625.46 °C

Therefore the final temperature of the gas is 625.46 °C

4 0
3 years ago
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