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3 years ago

Which processes represent one chemical change in one physical change

Chemistry

2 answers:

prisoha [69]3 years ago

7 0

decomposition and melting

Alex Ar [27]3 years ago

4 0

Answer:

Decomposition and melting

Explanation:

Decomposition is the process of splitting two or more compounds apart from each other. Usually, the decomposed substance have different chemicals identities hence the reason why it is a chemical process since it causes chemical change. Example is the decomposition of hydrated salts of metals. CuSO4.5H20 is a penta-hydrate salt I.e contains 5 molecules of water. If this compound is heated, two different compounds are formed and changes the chemical identity of the compound. Melting is a process of liquidizing compounds or elements to the point it changes physical state from solid-liquid. Melting does not change the chemical identity of the compound but rather the physical state of the compound.

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A wooden tool is found at an archaeological site. Estimate the age of the tool using the following information: A 100 gram sampl

lord [1]

Answer:

2578.99 years

Explanation:

Given that:

100 g of the wood is emitting 1120 β-particles per minute

Also,

1 g of the wood is emitting 11.20 β-particles per minute

Given, Decay rate = 15.3 % per minute per gram

So,

Concentration left can be calculated as:-

C left = $[A_t]=\frac{11.20\ per\ minute}{15.3\ per\ minute\ per\ gram}\times [A_0]= 0.7320[A_0]$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Also, Half life of carbon-14 = 5730 years

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,

$k=\frac {ln\ 2}{t_{1/2}}$

$k=\frac {ln\ 2}{5730}\ years^{-1}$

The rate constant, k = 0.000120968 year⁻¹

Time =?

Using integrated rate law for first order kinetics as:

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

So,

$\frac {[A_t]}{[A_0]}=e^{-0.000120968\times t}$

$\frac {0.7320[A_0]}{[A_0]}=e^{-0.000120968\times t}$

$0.7320=e^{-0.000120968\times t}$

$ln\ 0.7320=-0.000120968\times t$

t = 2578.99 years

8 0

3 years ago

Calculate the volume in mL of 0.589 M NaOH needed to neutralize 52.1 mL of 0.821 M HCl in a titration.

Igoryamba

Answer:

72.6 mL

Explanation:

A quick way to solve this titration problem when you have a monoprotic acid is to use the Dilution equation, M1V1=M2V2.

.589(x)=.821(52.1)

X=72.6 mL

4 0

2 years ago

What is the pH and pOH of a 2.2 x 10^-3 HBr solution?

antiseptic1488 [7]

Answer:

The pH and pOH of a 2.2*10⁻³ HBr solution is 2.66 and 11.34 respectively.

Explanation:

pH - short for hydrogen potential - is a measure of the acidity or alkalinity of a solution. So the pH is a parameter that indicates the concentration of hydrogen ions [H]⁺ that exist in a solution.

The pH is expressed as the negative base 10 logarithm of the hydrogen ion concentration. This is represented by:

pH= - log [H⁺]

pH is measured on a scale of 0 to 14. On this scale, a pH value of 7 is neutral, which means that the substance or solution is neither acidic nor alkaline. A pH value of less than 7 means that it is more acidic, and a pH value of more than 7 means that it is more alkaline.

HBr is a strong acid. Then, in aqueous solution it will be totally dissociated. So the proton concentration is equal to the initial concentration of acid:

[H⁺]= [HBr]= 2.2*10⁻³ M

So:

pH= - log (2.2*10⁻³)

pH= 2.66

On the other hand, pOH is a measure of the concentration of hydroxyl ions in a solution. The sum of pH and pOH equals 14:

pH + pOH= 14

2.66 + pOH= 14

pOH= 14 - 2.66

pOH= 11.34

The pH and pOH of a 2.2*10⁻³ HBr solution is 2.66 and 11.34 respectively.

4 0

3 years ago

The picture is there

Lena [83]

Answer:

Explanation:

4 0

2 years ago

In the laboratory a student determines the specific heat of a metal. She heats 19.6 grams of zinc to 98.37°C and then drops it i

fgiga [73]

Answer:

The specific heat of zinc is 0.375 J/g°C

Explanation:

Step 1: Data given

Mass of zinc = 19.6 grams

Mass of water = 82.9 grams

Initial temperature of zinc = 98.37 °C

Initial temperature of water = 24.16 °C

Final temperature of water (and zinc) = 25.70 °C

Specific heat of water = 4.184 J/g°C

Step 2: Calculate Specific heat of zinc

Heat lost by zinc = heat won by water

Q=m*c*ΔT

Qzinc = -Qwater

m(zinc)*C(zinc)*ΔT(zinc) = -m(water)*C(water)*ΔT(water)

⇒ with mass of zinc = 19.6 grams

⇒ with C(zinc) = TO BE DETERMINED

⇒ with ΔT(zinc) = T2 -T1 = 25.70 - 98.37 = -72.67°C

⇒ with mass of water = 82.9 grams

⇒ with C(water) = 4.184 J/g°C

⇒ with ΔT(water) = T2 - T1 = 25.70 - 24.16 = 1.54

Qzinc = -Qwater

19.6g* C(zinc) * (-72.67°C) = - 82.9g* 4.184 J/g°C * 1.54 °C

-1424.332*C(zinc) = -534.155

C(zinc) = 0.375 J/g°C

The specific heat of zinc is 0.375 J/g°C

7 0

3 years ago