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BlackZzzverrR [31]
2 years ago
12

The total number of electron pairs in a nitrogen molecules.is: a) 2 b) 3 c) 5 d) 7​

Chemistry
1 answer:
Mekhanik [1.2K]2 years ago
3 0

7

nitrogen's atomic no. is 7 which is the amount of protons and the amount of protons is equal to the amount of electrons

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Which type of atom has the strongest attraction for electrons in bond formation? barium (Ba) chlorine (Cl) iodine (I) strontium
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3 years ago
The molecular weight of NaCl is 58.44 grams/moles. How many grams of NaCl are found in a beaket with 100 ml of a 0.0050 M soluti
Viktor [21]

Answer:

The mass of NaCl is 0.029 grams

Explanation:

Step 1: Data given

Molecular weight of NaCl = 58.44 g/mol

Volume of solution = 100 mL = 0.100 L

Molarity = 0.0050 M

Step 2: Calculate moles NaCl

Moles NaCl = molarity * volume

Moles NaCl = 0.0050 M * 0.100 L

Moles NaCl = 0.00050 moles

Step 3: Calculate mass NaCl

Mass NaCl = moles NaCl * molar mass NaCl

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Mass NaCl = 0.029 grams

The mass of NaCl is 0.029 grams

4 0
3 years ago
For the equilibrium
Mamont248 [21]

Answer:

Equilibrium concentrations of the gases are

H_2S=0.596M

H_2=0.004 M

S_2=0.002 M

Explanation:

We are given that  for the equilibrium

2H_2S\rightleftharpoons 2H_2(g)+S_2(g)

k_c=9.0\times 10^{-8}

Temperature, T=700^{\circ}C

Initial concentration of

H_2S=0.30M

H_2=0.30 M

S_2=0.150 M

We have to find the equilibrium concentration of gases.

After certain time

2x number of moles  of reactant reduced and form product

Concentration of

H_2S=0.30+2x

H_2=0.30-2x

S_2=0.150-x

At equilibrium

Equilibrium constant

K_c=\frac{product}{Reactant}=\frac{[H_2]^2[S_2]}{[H_2S]^2}

Substitute the values

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

By solving we get

x\approx 0.148

Now, equilibrium concentration  of gases

H_2S=0.30+2(0.148)=0.596M

H_2=0.30-2(0.148)=0.004 M

S_2=0.150-0.148=0.002 M

3 0
3 years ago
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