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2 years ago

Why a finely powdered sample should be used in a melting point measurement?

Chemistry

1 answer:

lapo4ka [179]2 years ago

6 0

because it a essential for good heat transfer

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How do I answer this???

Mumz [18]

Will you have to show me what you're talking

8 0

2 years ago

The half-life of cesium-137 is 30 years. Suppose we have a 180 mg sample. Find the mass that remains after t years. Let y(t) be

Stels [109]

Explanation:

1) Initial mass of the Cesium-137= $N_o$ = 180 mg

Mass of Cesium after time t = N

Formula used :

$N=N_o\times e^{-\lambda t}\\\\\lambda =\frac{0.693}{t_{\frac{1}{2}}}$

Half life of the cesium-137 = $t_{1/2}=30 years[p/tex]where,
[tex]N_o$ = initial mass of isotope

N = mass of the parent isotope left after the time, (t)

$t_{\frac{1}{2}}$ = half life of the isotope

$\lambda$ = rate constant

$N=N_o\times e^{-(\frac{0.693}{t_{1/2}})\times t}$

Now put all the given values in this formula, we get

$N=180mg\times e^{-\frac{0.693}{30 years}\times t}$

Mass that remains after t years.

$N=180 mg\times e^{0.0231 year^{-1}\times t}$

Therefore, the parent isotope remain after one half life will be, 100 grams.

t = 70 years

$N_o=180 mg$

$t_{1/2}= 30 yeras$

$N=180mg\times e^{-\frac{0.693}{30 years}\times 70 years}$

N = 35.73 mg

35.73 mg of cesium-137 will remain after 70 years.

$N_o=180 mg$

$t_{1/2}= 30 yeras$

N = 1 mg

t = ?

$1 mg =180mg\times e^{-\frac{0.693}{30 years}\times t}$

$\frac{-30 year}{0.693}\times \ln \frac{1 mg}{180 mg}=t$

t = 224.80 years ≈ 225 years

After 225 years only 1 mg of cesium-137 will remain.

7 0

3 years ago

I have 2 Valence electrons and 1 energy levels

natima [27]

Answer:

Hydrogen(H) and Heluim(He)

Explanation:

These are the only two valennce electrons and 1 energy levels.

3 0

2 years ago

Read 2 more answers

What is the pH of a 0.100 M HI solution? Is that neutral, acidic, or basic? acidic What is the pOH of a 0.100 M HI solution? Is

inessss [21]

<u>Answer:</u> The pH and pOH of the solution is 1 and 13 respectively and the solution is acidic in nature.

<u>Explanation:</u>

There are three types of solution: acidic, basic and neutral

To determine the type of solution, we look at the pH values.

The pH range of acidic solution is 0 to 6.9
The pH range of basic solution is 7.1 to 14
The pH of neutral solution is 7.

We are given:

Concentration of HI = 0.100 M

1 mole of HI produces 1 mole of hydrogen ions and 1 mole of iodide ions

To calculate the pH of the solution, we use the equation:

$pH=-\log[H^+]$

We are given:

$[H^+]=0.100M$

Putting values in above equation, we get:

$pH=-\log(0.100)\\\\pH=1$

To calculate the pOH of the solution, we use the equation:

pH + pOH = 14

$pOH=14-1=13$

Hence, the pH and pOH of the solution is 1 and 13 respectively and the solution is acidic in nature.

4 0

3 years ago

Which characteristic do electronegativity differences indicate about reactions between atoms

emmasim [6.3K]

O valence electron number is the answer

6 0

2 years ago