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My name is Ann [436]
3 years ago
5

What subscripts are needed to complete the formula unit Li_N_?

Chemistry
2 answers:
MAXImum [283]3 years ago
5 0

Answer:

3,1

Explanation:

Lithium nitride is a ionic compound. The lithium cation is Li⁺ and the nitride anion is N³⁻

When writing the ionic compounds formed by the respective ions the general rule is to write the charge of the anion as the subscript of the cation and to write the charge of the cation as the subscript of the anion.  

Li⁺ N³⁻  

Li₃N₁

Li₃N

sleet_krkn [62]3 years ago
3 0
The Answer would be B.)3,1 
<span>Lithium nitride is the chemical compound </span>
The formula would look like this: Li3N
You might be interested in
Why does nuclear fusion occur in stars but not on Earth?
disa [49]

Answer:

C.

Fusion reactions require a lot of heat and pressure

Explanation:

nuclear fusion takes place only at extremely high temperatures. That's because a great deal of energy is needed to overcome the force of repulsion between the positively charged nuclei. ... A: Nuclear fusion doesn't occur naturally on Earth because it requires temperatures far higher than Earth temperatures.

4 0
3 years ago
Read 2 more answers
How many grams of NH3 can be produced from 2.51 mil of N2 and excess H2 ?
salantis [7]

Answer:

85.34g of NH3

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

N2 + 3H2 —> 2NH3

Step 2:

Determination of the number of moles of NH3 produced by the reaction of 2.51 moles of N2. This is illustrated below:

From the balanced equation above,

1 mole of N2 reacted to produce 2 moles of NH3.

Therefore, 2.51 moles of N2 will react to produce = (2.51 x 2)/1 = 5.02 moles of NH3.

Therefore, 5.02 moles of NH3 is produced from the reaction.

Step 3:

Conversion of 5.02 moles of NH3 to grams. This is illustrated below:

Molar mass of NH3 = 14 + (3x1) = 17g/mol

Number of mole of NH3 = 5.02 moles

Mass of NH3 =..?

Mass = mole x molar Mass

Mass of NH3 = 5.02 x 17

Mass of NH3 = 85.34g

Therefore, 85.34g of NH3 is produced.

3 0
2 years ago
4. Consider a 33.0 mL solution containing 0.0758 M NaF and 0.0955 M HF. What is the total number of moles of HF after the additi
Ostrovityanka [42]
Buffer solution resist the change in pH upon addition of small amount of strong acid or strong base.
Buffer consists of weak acid as HF / and its conjugate base NaF
When strong acid as HCl is added to buffer, it respond with its conjugate base to convert the strong acid to weak acid like this:
HCl (S.A) + NaF → NaCl + HF (W.A)
moles of HF we already have = M * V(in liters)
                                                = 0.0955 M * 0.033 L = 3.15 x 10⁻³ mole
moles of HCl added = 8.00 x 10⁻⁵ mole
one mole HCl reacts with 1 mole NaF to give 1 mole HF
so the amount added to HF = 8.00 x 10⁻⁵
Total moles of HF present = (3.15 x 10⁻³) + (8.00 x 10⁻⁵) = 3.23 x 10⁻³ mole

4 0
3 years ago
Calculate the number of sodium ions, perchlorate ions, Cl atoms and O atoms in 17.8 g of sodium perchlorate. Enter your answers
Degger [83]

17.8 g of sodium perchlorate contains 8.73 × 10²² Na⁺ ions, 8.73 × 10²² ClO₄⁻ ions, 8.73 × 10²² Cl atoms and 3.49 × 10²³ O atoms.

First, we will convert 17.8 g of NaClO₄ to moles using its molar mass (122.44 g/mol).

17.8 g \times \frac{1mol}{122.44g} = 0.145 mol

Next, we will convert 0.145 moles to molecules of NaClO₄ using Avogadro's number; there are 6.02 × 10²³ molecules in 1 mole of molecules.

0.145 mol \times \frac{6.02 \times 10^{23}molecules  }{mol} = 8.73 \times 10^{22}molecules

NaClO₄ is a strong electrolyte that dissociates according to the following equation.

NaClO₄ ⇒ Na⁺ + ClO₄⁻

The molar ratio of NaClO₄ to Na⁺ is 1:1. The number of Na⁺ in 8.73 × 10²² molecules of NaClO₄ is:

8.73 \times 10^{22}moleculeNaClO_4 \times \frac{1 ion Na^{+} }{1moleculeNaClO_4} = 8.73 \times 10^{22}ion Na^{+}

The molar ratio of NaClO₄ to ClO₄⁻ is 1:1. The number of ClO₄⁻ in 8.73 × 10²² molecules of NaClO₄ is:

8.73 \times 10^{22}moleculeNaClO_4 \times \frac{1 ion ClO_4^{-} }{1moleculeNaClO_4} = 8.73 \times 10^{22}ion ClO_4^{-}

The molar ratio of ClO₄⁻ to Cl is 1:1. The number of Cl in 8.73 × 10²² ions of ClO₄⁻ is:

8.73 \times 10^{22}ion ClO_4^{-} \times \frac{1 atomCl }{1ion ClO_4^{-}} = 8.73 \times 10^{22}atom Cl

The molar ratio of ClO₄⁻ to O is 1:1. The number of O in 8.73 × 10²² ions of ClO₄⁻ is:

8.73 \times 10^{22}ion ClO_4^{-} \times \frac{4 atomO }{1ion ClO_4^{-}} = 3.49 \times 10^{23}atom O

17.8 g of sodium perchlorate contains 8.73 × 10²² Na⁺ ions, 8.73 × 10²² ClO₄⁻ ions, 8.73 × 10²² Cl atoms and 3.49 × 10²³ O atoms.

You can learn more Avogadro's number here: brainly.com/question/13302703

8 0
2 years ago
Oxidation state of N in n2(g)
Lady bird [3.3K]

Answer:

0

Explanation:

did research

4 0
3 years ago
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