How many valence electrons do the halogens have? A. 8 B. 6 C. 7 D. 2

The equation below shows lithium reacting with nitrogen to produce lithium nitride. 6Li + N2 2Li3N If 12 mol of lithium were rea

Andreyy89

Answer:

4 mol of lithium nitride would be produced

Explanation:

6Li + N2 → 2Li3N

This equation means that 6 mol of lithium are reacted with 1 mol of nitrogen to make, 2 mol of lithium nitride.

So this is the rule of three to find out, how many moles of lithium nitride are been made, from 12 mol of Li.

6 Li _____ 2 Li3N

12 Li ________ (12. 2) / 6 = 4 moles

5 0

4 years ago

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What method of cleaning tools and equipment do you follow or apply?

AleksAgata [21]

I clean with a rag wdym

5 0

3 years ago

What is the final acetate ion concentration when 69. g Ba(C2H3O2)2 is dissolved in enough water to make 970. mL of solution? The

mario62 [17]

Answer:

0.56M of acetate ions

Explanation:

Given parameters:

Mass of Ba(C₂H₃O₂)₂ = 69g

Volume of water = 970mL = 0.97dm³

Molar mass of Ba(C₂H₃O₂)₂ = 255.415g/mol

Unknown:

Concentration of acetate ion in the final solution = ?

Solution:

Let us represent the dissociation;

Ba(C₂H₃O₂)₂ = Ba²⁺ + 2C₂H₃OO⁻

We see that 1M of will produce 2M of acetate ions

Now, let us find the molarity of the barium acetate;

Molarity = $\frac{number of moles }{volume}$

Number of moles of Ba(C₂H₃O₂)₂ = $\frac{mass}{molar mass}$

Number of moles = $\frac{69}{255.415}$ = 0.27moles

Molarity of Ba(C₂H₃O₂)₂ = $\frac{0.27}{0.97}$ = 0.28M

since 1M of Ba(C₂H₃O₂)₂ will produce 2M of acetate ions

0.28M of Ba(C₂H₃O₂)₂ will produce 2 x 0.28 = 0.56M of acetate ions

3 0

4 years ago

What is the difference expressed in kilograms between the mass of proton and the mass of electon?

Mashutka [201]

Particle, Relative Mass, Actual Mass (g) Electron- 1 , 9.11*10 power -28 g. ... Calculate: What is the diff expressed in kg between the mass of a proton and the mass ... Best Answer: differencein mass = [1.673 x 10^(-24) - 9.11 x ...

4 0

4 years ago

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During the electrolysis of cryolite ,aluminium and fluorine are formed in.......molar ratio

PtichkaEL [24]

${\huge{\boxed{\sf{\pink{☛Answer࿐}}}}}$

From the first mechanism, it can be seen that the Aluminum is reduced at cathode and Florine is oxidized at anode.

The ratio of aluminium and fluorine is the ratio of their change in oxidation number, i.e. 2:3

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8 0

3 years ago