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2 years ago

What characteristics are shared by elements in the same period row?

Chemistry

1 answer:

kodGreya [7K]2 years ago

5 0

Answer:

Elements in the same period have the same number of electron shells.

Explanation:

elements gain electrons and protons and become less metallic

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A sample of 28 Mg decays initially at a rate of 53500 disintegrations per minute, but the decay rate falls to 10980 disintegrati

prisoha [69]

Answer : The half life of 28-Mg in hours is, 6.94

Explanation :

First we have to calculate the rate constant.

Expression for rate law for first order kinetics is given by:

$k=\frac{2.303}{t}\log\frac{a}{a-x}$

where,

k = rate constant

t = time passed by the sample = 48.0 hr

a = initial amount of the reactant disintegrate = 53500

a - x = amount left after decay process disintegrate = 53500 - 10980 = 42520

Now put all the given values in above equation, we get

$k=\frac{2.303}{48.0}\log\frac{53500}{42520}$

$k=9.98\times 10^{-2}hr^{-1}$

Now we have to calculate the half-life.

$k=\frac{0.693}{t_{1/2}}$

$9.98\times 10^{-2}=\frac{0.693}{t_{1/2}}$

$t_{1/2}=6.94hr$

Therefore, the half life of 28-Mg in hours is, 6.94

8 0

3 years ago

Which element would you expect to have properties similar to Chlorine (Cl)

jeyben [28]

Answer:

Iodine

Explanation:

It's in the same group as chlorine.

5 0

3 years ago

Which statement is correct about the conclusions of historical events? A) Historical event conclusions can be tested using the s

BARSIC [14]

C is the correct answer I believe

4 0

3 years ago

Read 2 more answers

M/ABCI<br>m.CADC = 112<br>BK D 117

raketka [301]

Answer:

what do u mean????? i dont get the question

4 0

3 years ago

Purification of copper can be achieved by electrorefining copper from an impure copper anode onto a pure copper cathode in an el

Likurg_2 [28]

Answer: 281 hours

Explanation:-

1 electron carry charge= $1.6\times 10^{-19}C$

1 mole of electrons contain= $6.023\times 10^{23}$ electrons

Thus 1 mole of electrons carry charge= $\frac{1.6\times 10^{-19}}{1}\times 6.023\times 10^{23}=96500C$

$Cu^{2+}+2e^-\rightarrow Cu$

$96500\times 2=193000Coloumb$ of electricity deposits 1 mole or 63.5 g of copper

0.0635 kg of copper is deposited by 193000 Coloumb

11.5 kg of copper is deposited by= $\frac{193000}{0.0635}\times 11.5=34952756$ Coloumb

$Q=I\times t$

where Q= quantity of electricity in coloumbs = 34952756 C

I = current in amperes = 34.5 A

t= time in seconds = ?

$34952756 C=34.5A\times t$

$t=1013123sec=281hours$

Thus it will take 281 hours to plate 11.5 kg of copper onto the cathode if the current passed through the cell is held constant at 34.5 A.

8 0

3 years ago