#1: Which element has the same number of valence electrons as hydrogen (H)?

1 answer:

6 0

Valence electrons are those in the outer most orbital shell (the valence shell). Elements in the same column (group) have the same number of valence electrons. Hydrogen is in Group 1. If you were to look at a periodic table of the elements, Lithium is right under Hydrogen. You are correct, the answer is D. Lithium.

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Complete the nuclear equation 42 K → 0 e 19____-1

kenny6666 [7]

Answer:

42 19 K→42 20 Ca+e−

Explanation:

Naturally-occurring potassium atoms have a weighted average atomic mass of 39.10 (as seen on most modern versions of the periodic table.) Each potassium atom contains 19 protons p+ and thus an average potassium atom contains about 39.10−19≈20 neutrons n0.

This particular isotope of potassium-42 contains 42 nucleons (i.e., protons and neutrons, combined;) Like other isotopes of potassium 19 out of these nucleons are protons; the rest 42−19=23 are therefore neutrons.

5 0

2 years ago

Calculate the volume in mL of a 0.708 M KOH solution containing 0.098 mol of solute

Olenka [21]

Molarity = mol/liter

0.708M = 0.098mol/L
Rearrange to find L:
0.098mol/0.708M = .138L

For every liter there is 1000 mL:
.138L • 1000mL =138mL KOH

7 0

3 years ago

Silicon has 2 isotopes, Silicon-28 with an abundance of 90% and Silicon-30 with an abundance of 10%. Find the AAM for Silicon.

JulsSmile [24]

Answer:

RAM

$RAM = (mass \: of \: ^{28} Si \: \times \%abundance) + (mass \: of \: ^{30} Si \: \times \%abundance) \\ RAM = (28 \times 90\%) + (30 \times 10\%) \\ RAM = 25.2 + 3 \\ RAM = 28.2$

5 0

2 years ago

8. What volume (ml) of a 5.45 M lead nitrate solution must be diluted to 820.7 ml to make a

Ganezh [65]

212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.

Answer:

Option A.

Explanation:

Similar to Avagadro's law, there is another law termed as dilution law. As the product of volume and normality of the reactant is equal to the product of volume and normality of the product from the Avagadro's law. In dilution law, it will be as product of volume and concentration of the solute of the reactant is equal to the product of volume and concentration of solution.

$C_{1} V_{1} =C_{2} V_{2}$

So, as per the given question C1 = 5.45 M of lead nitrate and V1 has to be found. While C2 is 1.41 M of lead nitrate and V2 is 820.7 ml.

Then, $(5.45*V_{1} ) = (820.7*1.41)$

$V_{1}=\frac{820.7*1.41}{5.45}= 212.33 ml$

So nearly 212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.

3 0

2 years ago

Predict the empirical formula of these compounds. a. C2N2. b. P4O10. c. N2O5. d. NaCl. e. C9H20. f. B2H6. g. K2Cr2O7. h. Al2Br6.

Stella [2.4K]

The empirical formula is the simplest form of the formula expressed in the lowest ratio. In this case, we just have to divide each subscript by the greatest common factor. Hence.
a. CN
b. P2O5
c.N2O5
d.NaCl
e. C9H20
f. BH3
g.K2Cr2O7
h.AlB3
i.CH
j.SiCl4

5 0

2 years ago