What happens when an electron moves from n=2 to n=5?

The oxidation numbers of nitrogen in nh3, hno3, and no2 are, respectively:

vovikov84 [41]

The oxidation numbers for Nitrogen are respectively -3, +5, +4

3 0

3 years ago

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A hydrogen halide diffuses 1.49 times faster than HBr. This hydrogen halide is

marysya [2.9K]

To solve this problem, we must assume ideal gas behaviour so that we can use Graham’s law:

vA / vB = sqrt (MW_B / MW_A)

where,

vA = speed of diffusion of A (HBR)

vB = speed of diffusion of B (unknown)

MW_B = molecular weight of B (unkown)

MW_A = molar weight of HBr = 80.91 amu

We know from the given that:

vA / vB = 1 / 1.49

So,

1/1.49 = sqrt (MW_B / 80.91)

MW_B = 36.44 g/mol

Since this unknown is also hydrogen halide, therefore this must be in the form of HX.

HX = 36.44 g/mol , therefore:

x = 35.44 g/mol

From the Periodic Table, Chlorine (Cl) has a molar mass of 35.44 g/mol. Therefore the hydrogen halide is:

HCl

6 0

3 years ago

Fill in the missing data point. Show all calculations leading to an answer.

Aleksandr-060686 [28]

Answer:

1090 mmHg

Explanation:

We know that with gases we must use a Kelvin temperatures, so let’s try a plot of pressure against the Kelvin temperature.

We can create a table as follows

t/°C T/K p/mmHg

10 283 726

20 293 750

40 313 800

70 343 880

100 373 960

150 423 ???

I plotted the data and got the graph in the figure below.

It appears that pressure is a linear function of the Kelvin temperature.

y = mx + b

where x is the slope and b is the y-intercept.

===============

Calculate the slope

I will use the points (275, 700) and (380, 975).

Slope = Δy/Δx = (y₂ - y₁)/(x₂ -x₁) = (975 -700)/(380 – 275) = 275/105 = 2.619

So,

y = 2.619x + b

===============

Calculate the intercept

When x = 275, y = 700.

700 = 2.619 × 275 + b

700 = 720 + b Subtract 720 from each side and transpose.

b = -20

So, the equation of the graph is

y = 2.619x -20

===============

Calculate the pressure at 423 K (150°C)

y = 2.619 × 423 - 20

y = 1110 - 20

y = 1090

At 150 °C, the pressure 1090 mmHg.

The point is approximately at the position of the black dot in the graph.

7 0

3 years ago

You find a little bit (0.150g) of a chemical marked Tri-Nitro-Toluene, and upon combusting it in oxygen, collect 0.204 g of CO2

avanturin [10]

1) Answer is: the formula is C₇N₃O₆H₅.

M(TNT) = 0.150 g; mass of the trinitrotoluene.

ω(N) = 18.5% ÷ 100%.

ω(N) = 0.185; mass percentage of the nitrogen.

m(N) = 0.150 g · 0.185.

m(N) = 0.02775 ·; mass of the nitrogen.

n(N) = 0.02775 g ÷ 14 g/mol.

n(N) = 0.002 mol; amount of the nitrogen.

n(CO₂) = 0.204 g ÷ 44 g/mol.

n(CO₂) = 0.0046 mol.

n(C) = n(CO₂) = 0.0046 mol; amount of the carbon.

m(C) = 0.0046 mol · 12 g/mol.

m(C) = 0.0552 g; mass of the carbon.

n(H₂O) = 0.030 g ÷ 18 g/mol.

n(H₂O) = 0.00166 mol.

n(H) = 2 · n(H₂O) = 0.0033 mol; amount of the hydrogen.

m(H) = 0.0033 mol · 1 g/mol.

m(H) = 0.0033 g; mass of the hydrogen.

2) m(O) = m(TNT) - m(N) - m(C) - m(H).

m(O) = 0.150 g - 0.02775 g - 0.0552 g - 0.0033 g.

m(O) = 0.06375 g.

n(O) = 0.06375 g ÷ 16 g/mol.

n(O) = 0.004 mol; amount of oxygen.

n(C) : n(N) : n(O) : n(H) = 0.0046 mol : 0.002 mol : 0.004 mol : 0.0033 mol.

n(C) : n(N) : n(O) : n(H) = 2.33 : 1 : 2 : 1.66 /×3.

n(C) : n(N) : n(O) : n(H) = 7 : 3 : 6 : 5.

8 0

3 years ago

Draw a resonance structure, complete with all formal charges and lone (unshared) electron pairs, that shows the resonance intera

icang [17]

Answer:see the picture attached

Explanation:

3 0

3 years ago