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vladimir1956 [14]
3 years ago
13

How many molecules in 4NH3

Chemistry
1 answer:
Snezhnost [94]3 years ago
8 0

Answer:

16

Explanation:

4 nitrogen (N) atoms and 12 (3×4) atoms in Hydrogen (H)

Total = 4 + 12 = 16 molecules

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In the equation. C + O2 --------> CO2; CO2 is the
lubasha [3.4K]

Answer: Product

Explanation:

A Chemical reaction shows how substances, written in symbols  known as  reactants, are converted to different substances called products. An arrow from the left hand side (reactants) points to right hand side ( products )shows the direction of  flow of reaction.

In the equation. C + O2 --------> CO2

                             reactant------> product

CO2  is the product

7 0
3 years ago
Name the region of the atom where protons and neutrons are located.
inn [45]

Answer:

Nucleus

Explanation:

5 0
3 years ago
What minimum volume of 0.200 m potassium iodide solution is required to completely precipitate all of the lead in 155.0 ml of a
Blababa [14]
First, we write the balanced equation for this reaction:

2KI + Pb(NO₃)₂ → 2KNO₃ + PbI₂

From this equation, we see that there are 2 moles of potassium iodide required for each mole of lead (II) nitrate. Moreover, we may use the formula:

Moles = volume (in L) * molarity

We find the molar relation ship for KI : Pb(NO₃)₂ to be 2 : 1. So:

M₁V₁ = 2M₂V₂

V₁ = 2M₂V₂/M₁
V₁ = 2 * 0.112 * 0.155 / 0.2
V₁ = 0.1736 L


The volume required is 173.6 mL
5 0
3 years ago
How many atoms does 2NaOH have
saul85 [17]
The substance of 2 Naoh contains 6 atoms
8 0
3 years ago
Read 2 more answers
The NaCl crystal structure consists of alternating Na⁺ and Cl⁻ ions lying next to each other in three dimensions. If the Na⁺ rad
Yuliya22 [10]

Answer : The radii of the two ions Cl⁻ ion and Na⁺ ion is, 181 and 102 pm respectively.

Explanation :

As we are given that the Na⁺ radius is 56.4% of the Cl⁻ radius.

Let us assume that the radius of Cl⁻ be, (x) pm

So, the radius of Na⁺ = x\times \frac{56.4}{100}=(0.564x)pm

In the crystal structure of NaCl, 2 Cl⁻ ions present at the corner and 1 Na⁺ ion present at the edge of lattice.

Thus, the edge length is equal to the sum of 2 radius of Cl⁻ ion and 2 radius of Na⁺ ion.

Given:

Distance between Na⁺ nuclei = 566 pm

Thus, the relation will be:

2\times \text{Radius of }Cl^-+2\times \text{Radius of }Na^+=\text{Distance between }Na^+\text{ nuclei}

2\times x+2\times 0.564x=566

2x+1.128x=566

3.128x=566

x=180.9\approx 181pm

The radius of Cl⁻ ion = (x) pm = 181 pm

The radius of Na⁺ ion = (0.564x) pm = (0.564 × 181) pm =102.084 pm ≈ 102 pm

Thus, the radii of the two ions Cl⁻ ion and Na⁺ ion is, 181 and 102 pm respectively.

4 0
3 years ago
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