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USPshnik [31]
3 years ago
7

The Beryllium (Be) atom has 4 protons, 4 electrons (2 valence e-​ ​), and 5 neutrons. It is on the 2nd period, so it has two ele

ctron shells

Chemistry
1 answer:
Semmy [17]3 years ago
4 0

(diagram below)

1. has two electron shells, so draw two rings around the nucleus

2. has 4 electrons, and since the first shell always has two the second shell would have 2 as well (to add up to four). this makes sense, because the description says that Be has 2 valence electrons

3. Draw 4 protons inside the grey circle (the nucleus)

4. Draw 5 neutrons inside the grey circle too (the nucleus)

they don't have to be in any particular order, hope this helps!

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URGENT ! PLEASE ANSWER QUICKLY
Bumek [7]

Answer:If we dissolve NaF in water, we get the following equilibrium:

text{F}^-(aq)+text{H}_2text{O}(l) rightleftarrows text{HF}(aq)+text{OH}^-(aq)

The pH of the resulting solution can be determined if the  K_b of the fluoride ion is known.

20.0 g of sodium fluoride is dissolve in enough water to make 500.0 mL of solution. Calculate the pH of the solution. The  K_b of the fluoride ion is 1.4 × 10 −11 .

Step 1: List the known values and plan the problem.

Known

mass NaF = 20.0 g

molar mass NaF = 41.99 g/mol

volume solution = 0.500 L

K_b of F – = 1.4 × 10 −11

Unknown

pH of solution = ?

The molarity of the F − solution can be calculated from the mass, molar mass, and solution volume. Since NaF completely dissociates, the molarity of the NaF is equal to the molarity of the F − ion. An ICE Table (below) can be used to calculate the concentration of OH − produced and then the pH of the solution.

Explanation:

7 0
3 years ago
Read 2 more answers
NEED HELP ASAP NOT DIFFICULT
Burka [1]
Answer- 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.

Given - Number of moles of Al(NO3)3 - 4 moles
Number of moles of NaCl - 9 moles
Find - Maximum amount of AlCl3 produced during the reaction.
Solution - The complete reaction is - Al(NO3)3 + 3NaCl --> 3NaNO3 + AlCl3
To find the maximum amount of AlCl3 produced during the reaction, we need to find the limiting reagent.
Mole ratio Al(NO3)3 - 4/1 - 4
Mole ratio NaCl - 9/3 - 3
Thus, NaCl is the limiting reagent in the reaction.
Now, 3 moles of NaCl produces 1 mole of AlCl3
9 moles of NaCl will produce - 1/3*9 - 3 moles.
Weight of AlCl3 - 3*133.34 - 400 grams
Thus, 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.
5 0
3 years ago
Read 2 more answers
1. What kinds of metals are REALLY reactive? why?
Olin [163]

Answer:

1. Alkali metals (group 1)

2. halogens (Group 17)

3. noble gasses (group 18)

Explanation:

1. alkali metals only have one valence electron meaning that they really want to lose that one valence electron to get a full octet.

2. halogens have 7 valence electrons meaning that they just need to gain 1 to get a full octet.

3. Nobel gasses already have a full octet meaning that they don't want to react. (atoms only react to get a full octet)

I hope this helps.  Let me know if anything is unclear.

8 0
3 years ago
An atom has radius of 227 pm and crystallizes in a body-centered cubic unit cell. What is the volume of the unit cell
sdas [7]

We will see that the volume of the unit cell is 144,070,699.06 pm^3

<h3>How to get the volume of a body-centered cubic unit cell?</h3>

In a body-centered cubic unit cell, the side length of the cube is given as:

S = \frac{4}{\sqrt{3} } *R

Where R is the radius of the atom.

And the volume of a cube is the side length cubed, then we can see that the volume of our cube will be:

V = S^3 = (\frac{4}{\sqrt{3} }*227pm)^3

Solving that we get:

V  = (\frac{4}{\sqrt{3} }*227pm)^3 = 144,070,699.06pm^3

This is the approximated volume of the unit cell.

If you want to learn more about unit cell structures, you can read:

brainly.com/question/13110055

5 0
2 years ago
A compound decomposes by a first-order process. What is the half-life of the compound if 25.0% of the compound decomposes in 60.
amid [387]

Answer : The half-life of the compound is, 145 years.

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  = 60.0 min

a = let initial amount of the reactant  = 100 g

a - x = amount left after decay process = 100 - 25 = 75 g

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

k=\frac{2.303}{60.0}\log\frac{100g}{75g}

k=4.79\times 10^{-3}\text{ years}^{-1}

Now we have to calculate the half-life of the compound.

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

4.79\times 10^{-3}\text{ years}^{-1}=\frac{0.693}{t_{1/2}}

t_{1/2}=144.676\text{ years}\approx 145\text{ years}

Therefore, the half-life of the compound is, 145 years.

8 0
4 years ago
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