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yanalaym [24]
4 years ago
8

What is the symbol, atomic number, atomic mass, # of protons, # of neutrons, # of electrons, ion charge for the beryllium ion?

Chemistry
1 answer:
jolli1 [7]4 years ago
8 0

Answer:

Symbol Be

atomic number 4

protons 4

neutrons 5

electrons 4

ion charge +2

Explanation:

Look at the periodic table and all the answers can be found there. If you would like to learn how to find them please message me below.

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What is the molecular geometry of a molecule made of two atoms that share one pair of electrons and have no lone electrons pairs
morpeh [17]

Answer:

linear

Explanation:

Actually I consulted various sources too and that was the most sought answer,I just decided to answer you before studying it too

5 0
3 years ago
Teacher asks a group of students to measure The amplitude of the wave. Which labeled part on the image is the amplitude?
alexira [117]

Answer:

start from the crest down

Explanation:

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4 years ago
What is -2/3÷(1 1/4)
IRISSAK [1]

Answer:

to simplify, its -0.53

4 0
3 years ago
A sample consisting of 1.0 mol of perfect gas molecules with CV = 20.8 J K−1 is initially at 4.25 atm and 300 K. It undergoes re
Marat540 [252]

Answer : The value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

Explanation :

First we have to calculate the value of \gamma.

\gamma=\frac{C_p}{C_v}

As, C_p=R+C_v

So, \gamma=\frac{R+C_v}{C_v}

Given :

C_v=20.8J/K\\\\R=8.314J/K

\gamma=\frac{8.314+20.8}{20.8}=1.4

Now we have to calculate the initial volume of gas.

Formula used :

P_1V_1=nRT_1

where,

P_1 = initial pressure of gas = 4.25 atm

V_1 = initial volume of gas = ?

T_1 = initial temperature of gas = 300 K

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

(4.25atm)\times V_1=(1.0mol)\times (0.0821L.atm/mol.K)\times (300K)

V_1=5.80L

Now we have to calculate the final volume of gas by using reversible adiabatic expansion.

P_1V_1^{\gamma}=P_2V_2^{\gamma}

where,

P_1 = initial pressure of gas = 4.25 atm

P_2 = final pressure of gas = 2.50 atm

V_1 = initial volume of gas = 5.80 L

V_2 = final volume of gas = ?

\gamma = 1.4

Now put all the given values in above formula, we get:

(4.25atm)\times (5.80L)^{1.4}=(2.50atm)\times V_2^{1.4}

V_2=8.47L

Now we have to calculate the final temperature of gas.

Formula used :

P_2V_2=nRT_2

where,

P_2 = final pressure of gas = 2.50 atm

V_2 = final volume of gas = 8.47 L

T_2 = final temperature of gas = ?

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in above formula, we get:

(2.50atm)\times (8.47L)=(1.0mol)\times (0.0821L.atm/mol.K)\times T_2

T_2=257.9K\approx 258K

Now we have to calculate the work done.

w=nC_v(T_2-T_1)

where,

w = work done = ?

n = number of moles of gas =1.0 mol

T_1 = initial temperature of gas = 300 K

T_2 = final temperature of gas = 258 K

C_v=20.8J/K

Now put all the given values in above formula, we get:

w=(1.0mol)\times (20.8J/K)\times (258-300)K

w=-873.6J

Therefore, the value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

8 0
3 years ago
The table shows the total number of electrons in Atom A and Atom B.
alina1380 [7]

Answer:

Both A and B will be chemically unreactive

Explanation:

5 0
3 years ago
Read 2 more answers
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