What is the number of proton in this atom

Nitrogen forms a surprising number of compounds with oxygen. A number of these, often given the collective symbol NOx (for "nitr

kvv77 [185]

Answer:

9.2

Explanation:

Let's do an equilibrium chart of this reaction:

2NO(g) + O₂(g) ⇄ 2NO₂(g)

4.9 atm 5.1 atm 0 Initial

-2x -x +2x Reacts (stoichiometry is 2:1:2)

4.9-2x 5.1-x 2x Equilibrium

The mole fraction of NO₂ (y) can be calculated by the Raoult's law, that states that the mole fraction is the partial pressure divided by the total pressure:

y = 2x/(4.9 - 2x + 5.1 -x + 2x)

0.52 = 2x/(10 - x)

2x = 5.2 -0.52x

2.52x = 5.2

x = 2.06 atm

Thus, the partial pressure at equilibrium are:

pNO = 4.9 -2*2.06 = 0.78 atm

pO₂ = 5.1 - 2.06 = 3.04 atm

pNO₂ = 2*2.06 = 4.12 atm

Thus, the pressure equilibrium constant Kp is:

Kp = [(pNO₂)²]/[(pNO)²*(pO₂)]

Kp = [(4.12)²]/[(0.78)²*3.04]

Kp = [16.9744]/[1.849536]

Kp = 9.2

4 0

3 years ago

A sample compound contains 5.723g Ag, 0.852g S and 1.695g O. Determine its empirical formula.

Lubov Fominskaja [6]

Answer:

$Ag_2SO_4$

Explanation:

Formula for the calculation of no. of Mol is as follows:

$mol=\frac{mass\ (g)}{molecular\ mass}$

Molecular mass of Ag = 107.87 g/mol

Amount of Ag = 5.723 g

$mol\ of\ Ag=\frac{5.723\ g}{107.87\ g/mol} =0.05305\ mol$

Molecular mass of S = 32 g/mol

Amount of S = 0.852 g

$mol\ of\ S=\frac{0.852\ g}{32\ g/mol} =0.02657\ mol$

Molecular mass of O = 16 g/mol

Amount of O = 1.695 g

$mol\ of\ O=\frac{1.695\ g}{16\ g/mol} =0.10594\ mol$

In order to get integer value, divide mol by smallest no.

Therefore, divide by 0.02657

$Ag, \frac{0.05305}{0.02657} \approx 2$

$S, \frac{0.02657}{0.02657} \approx 1$

$O, \frac{0.10594}{0.02657} \approx 4$

Therefore, empirical formula of the compound = $Ag_2SO_4$

7 0

4 years ago

Read 2 more answers

For each of the following cases, identify the order with respect to the reactant, A. Case (A ----> product)

damaskus [11]

Answer:

The answers to the question are

1. 2nd and above order order

2. 2nd order

3. 1/2 order

4. 1st order

5. 0 order

Explanation:

We have $\frac{N}{N_{0} } = e^{-\lambda t}$

1. For nth order reaction half life $t_{\frac{1}{2} }$ ∝ $\frac{1}{[A_{0} ]^{n-1} }$

Therefore for a 0 order reaction increasing concentration of the reactant there will increase $t_{\frac{1}{2} }$

First order reaction is independent [A₀].

Second order reaction [A₀] decrease, increase.

Similarly for a third order reaction

1. 2nd order

2. 2nd order reaction

3. Order of reaction is 1/2.

4. 1st order reaction.

5. Zero order reaction.

5 0

3 years ago

Determine the bonding type for boron trihydride given the electronegativity on the Pauling scale for boron is 2.0 and for hydrog

Anika [276]

Answer: The bond between boron and hydrogen in boron trihydride is covalent bond.

Explanation:

The type of bonding between the atoms forming a compound is determined by using the electronegativity difference between the atoms. According to the pauling's electronegativity rule: