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2 years ago

Why is hydrogen (H) placed with the elements in Group 1A?

Chemistry

2 answers:

Alex787 [66]2 years ago

6 0

It has ns1 electron configuration like the alkali metals.

NISA [10]2 years ago

6 0

The answer is -

It has the same number of valence electrons

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Write a balanced nuclear equation for the β emission of the following isotopes

andrezito [222]

The balanced nuclear equation for the β emission of the following isotopes is seen below:

92 92 0

Sr ⇒ Y + e

38 39 -1

<h3>What is Beta emission?</h3>

This is also known as beta decay in which a beta ray is emitted from an atomic nucleus.

The element formed during the beta emission of strontium is referred to as Yttrium.

Read more about Beta emission here brainly.com/question/16334873

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6 0

1 year ago

How many ions are there in 0.187 mol of Na+ ions

kondaur [170]

Best Answer
1 mole of a substance contains 6.022x10^23 "units" of that substance.

So 0.187 mol of Na+ is 1.13x10^23 ions (6.022x10^23 x 0.187).

4 0

3 years ago

Read 2 more answers

What quality is attributed to water due to “capillary action”?

irakobra [83]

Answer:

I think it's B. The ability of water molecules to adhere to the surfaces of objects

4 0

2 years ago

D=___g/mL<br> v=100mL<br> m=1.5kg=___g

Bess [88]

The volume is 100mL.

The mass is 1.5kg which is equal to 1500g.

Thus, the density is 1500g / 100mL which is 15g/mL.

8 0

2 years ago

Two hydrogen atoms collide in a head on collision and end up with zero kinetic energy. Each then emits a photon of 121.6 nm (n=2

Zinaida [17]

Explanation:

Expression for the kinetic energy is as follows.

K.E = $\frac{1}{2}mv^{2}$

Now, total kinetic energy will be as follows.

K.E = $2 \times \frac{1}{2}mv^{2}$ = $m \times v^{2}$

Since, this energy converts into electromagnetic radiation of wavelength 121.6 nm.

Relation between energy and photon is as follows.

Energy of photon = $\frac{hc}{\lambda}$

= $\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{121.6 \times 10^{-9}}$

= $1.63 \times 10^{-18} J$

$m \times v^{2} = 1.63 \times 10^{-18}$

v = $\sqrt{\frac{1.63 \times 10^{-18}}{1.67 \times 10^{-27}}$

= $3.12 \times 10^{4}$ m/s

Thus, we can conclude that atoms were moving at a speed of $3.12 \times 10^{4}$ m/s before the collision.

8 0

2 years ago