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Marina CMI [18]

2 years ago

9

When an electron is removed the atoms charge changes to a what?

1 answer:

inna [77]2 years ago

6 0

Answer:

It make the atom octet from i.e more stable form.

Explanation:

A neutral sodium atom is likely to achieve an octet in its outermost shell by losing its one valence electron. Na -----> Na+ e-

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Drawing Conclusions

Dmitrij [34]

<u>Answer:</u>

<em>The situation given here is imaginary such that the life of Rock has to be found using the half-life of the element lokium that has been found inside the rock. </em>

<u>Explanation:</u>

Half-life of any material is the amount of time taken by that particular material to decay. Now the amount of lokium found in rock can show after how many half-lives this amount has been left out.

The time elapsed will be log (L) atoms X half-life.

4 0

3 years ago

The total bonding energy for the products of a reaction is 2535 kJ/mol and the bonding energy of the reactants is 1375 kJ/mol. C

Dmitry [639]

Answer:

-1160kj/mol

Explanation:

the reaction is exothermic because heat is released to the environment

7 0

3 years ago

What important result did Becquerel observe when he placed the uranium salt crystals and unexposed photographic film in a drawer

nexus9112 [7]

<span>B) The crystals did not phosphoresce within the drawer but did expose the film</span>

4 0

3 years ago

Read 2 more answers

Water is poured into a conical container at the rate of 10 cm3/sec. The cone points directly down, and it has a height of 20 cm

8090 [49]

Answer:

\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{2}

Explanation:

Hello,

The suitable differential equation for this case is:

$\frac{dV}{dt}=10\frac{cm^3}{s}$

As we're looking for the change in height with respect to the time, we need a relationship to achieve such as:

$\frac{dh}{dt} = ?*\frac{dV}{dt}$

Of course, $?=\frac{dh}{dV}$ .

Now, since the volume of a cone is $V=\pi r^2h/3$ and the ratio $r/h=15/20=3/4$ or $r=3/4h$ , the volume becomes:

$V=\pi (\frac{3}{4} h)^2h/3= \frac{3}{16}\pi h^3$

We proceed to its differentiation:

$\frac{dV}{dh} =\frac{9}{16} \pi h^2\\\frac{dh}{dV} =\frac{16}{9 \pi h^2}$

Then, we compute $\frac{dh}{dt}$

$\frac{dh}{dt} = \frac{16}{9 \pi h^2}*\frac{dV}{dt}\\\frac{dh}{dt} = \frac{16}{9\pi h^2}*10\frac{cm^3}{s} =\frac{160}{9 \pi h^2}$

Finally, at h=2:

$\frac{dh}{dt}_{h=2cm} =\frac{160}{9\pi 2^2}\\\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{s}$

Best regards.

4 0

3 years ago

In the iodoform test why is it necessary to add sodium hypochlorite

Kobotan [32]

Sodium hypochlorite is used in iodoform test for the oxidation of alcohol to aldehyde as shown in the image attached:

Here sodium hypochlorite is formed by the reaction of NaOH with I2 that further oxides alcohols to aldehydes.

8 0

2 years ago