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Kobotan [32]
2 years ago
15

Which has a higher ionization energy magnesium or calcium?

Chemistry
1 answer:
LenKa [72]2 years ago
4 0

Answer:

magnesium has a higher ionization energy because its radius is smaller. calcium has a higher ionization energy because it outermost sub-energy level is full. they have the same ionization energy because they have the same number of valence electrons.

Explanation:

hope this helps

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Periodic table the first row of elements fits in the period blank after the element blank the second row of elements fits in per
Bezzdna [24]

Answer:

The first row fits in period 2

The second row fits in period 3

Explanation:

Elements arranged vertically in the periodic table in groups that share similar chemical properties.

Elements are also organized horizontally in rows or periods

The length of each period is determined by the number of electrons that can occupy the sublevels being filled in that period.

7 0
3 years ago
What happens if you were to swallow a small amount if wet nail polish? Answer truthfully in your own words.
Scilla [17]

Answer:

I'm pretty sure you'll be fine

Explanation:

I've swallowed more than a small amount and I'm ok sort of

5 0
2 years ago
Please answer!! I WILL GIVE EXTRA POINTS AND BRAINLIEST
sveta [45]
The answer is D I believe
5 0
3 years ago
Read 2 more answers
How many sulfur atoms are generated when 9.42 moles of H2S react according to the following equation: 2H2S+SO2→3S+2H2O
8_murik_8 [283]

Answer:

A) 8.51 × 10²⁴  

Explanation:

1. Gather all the information

            2H₂S + SO₂ ⟶ 3S + 2H₂O

n/mol:   9.42

2. Calculate the moles of S atoms

The molar ratio is 3 mol S:2 mol H₂S

\text{Moles of S} = \text{9.42 mol H$_{2}$S} \times \dfrac{\text{3 mol S }}{\text{2 mol H$_{2}$S }} = \text{14.13 mol S}

3. Calculate the atoms of S

\text{Atoms of S } = \text{14.13 mol S} \times \dfrac{6.022 \times 10^{23}\text{ S atoms}}{\text{1 mol S}} = \mathbf{8.51 \times 10^{24}}\textbf{ S atoms}

 

6 0
3 years ago
Read 2 more answers
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
3 years ago
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