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GrogVix [38]
2 years ago
15

What is the second quantum number of a 3p3 electron in phosphorus 1s22s22p63s23p3

Chemistry
1 answer:
slega [8]2 years ago
7 0

Answer:

l=1

Explanation:

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How does solar energy compare to wind energy and fossil fuels?
Helga [31]
All of them are a source of getting energy but solar and wind energy are a more beneficial way of producing the energy we need.
6 0
3 years ago
How many moles of carbon (C) are there in 92.5 g of C?
aksik [14]

Answer:

c are there 93.5 of c yan siguro

8 0
3 years ago
Which equation correctly relates the heat of reaction to the standard heats of formation?
VARVARA [1.3K]

Answer:

D

Explanation:

∆H° = ∆Hf ° (products) – ∆Hf ° (reactants)

4 0
3 years ago
How much energy is released by the decay of 3 grams of 230Th in the following reaction 230 Th - 226Ra + "He (230 Th = 229.9837 g
Serhud [2]

<u>Answer:</u> The energy released for the decay of 3 grams of 230-Thorium is 2.728\times 10^{-15}J

<u>Explanation:</u>

First we have to calculate the mass defect (\Delta m).

The equation for the alpha decay of thorium nucleus follows:

_{90}^{230}\textrm{Th}\rightarrow _{88}^{226}\textrm{Ra}+_2^{4}\textrm{He}

To calculate the mass defect, we use the equation:

Mass defect = Sum of mass of product - Sum of mass of reactant

\Delta m=(m_{Ra}+m_{He})-(m_{Th})

\Delta m=(225.9771+4.008)-(229.9837)=1.4\times 10^{-3}amu=2.324\times 10^{-30}kg

(Conversion factor: 1amu=1.66\times 10^{-27}kg )

To calculate the energy released, we use Einstein equation, which is:

E=\Delta mc^2

E=(2.324\times 10^{-30}kg)\times (3\times 10^8m/s)^2

E=2.0916\times 10^{-13}J

The energy released for 230 grams of decay of thorium is 2.0916\times 10^{-13}J

We need to calculate the energy released for the decay of 3 grams of thorium. By applying unitary method, we get:

As, 230 grams of Th release energy of = 2.0916\times 10^{-13}J

Then, 3 grams of Th will release energy of = \frac{2.0916\times 10^{-13}}{230}\times 3=2.728\times 10^{-15}J

Hence, the energy released for the decay of 3 grams of 230-Thorium is 2.728\times 10^{-15}J

5 0
3 years ago
Nitrogen gas reacts with hydrogen gas to produce ammonia. How many liters of hydrogen gas at 95kPa and 15∘C are required to prod
viktelen [127]

Answer:

222.30 L

Explanation:

We'll begin by calculating the number of mole in 100 g of ammonia (NH₃). This can be obtained as follow:

Mass of NH₃ = 100 g

Molar mass of NH₃ = 14 + (3×1)

= 14 + 3

= 17 g/mol

Mole of NH₃ =?

Mole = mass /molar mass

Mole of NH₃ = 100 / 17

Mole of NH₃ = 5.88 moles

Next, we shall determine the number of mole of Hydrogen needed to produce 5.88 moles of NH₃. This can be obtained as follow:

N₂ + 3H₂ —> 2NH₃

From the balanced equation above,

3 moles of H₂ reacted to produce 2 moles NH₃.

Therefore, Xmol of H₂ is required to p 5.88 moles of NH₃ i.e

Xmol of H₂ = (3 × 5.88)/2

Xmol of H₂ = 8.82 moles

Finally, we shall determine the volume (in litre) of Hydrogen needed to produce 100 g (i.e 5.88 moles) of NH₃. This can be obtained as follow:

Pressure (P) = 95 KPa

Temperature (T) = 15 °C = 15 + 273 = 288 K

Number of mole of H₂ (n) = 8.82 moles

Gas constant (R) = 8.314 KPa.L/Kmol

Volume (V) =?

PV = nRT

95 × V = 8.82 × 8.314 × 288

95 × V = 21118.89024

Divide both side by 95

V = 21118.89024 / 95

V = 222.30 L

Thus the volume of Hydrogen needed for the reaction is 222.30 L

8 0
3 years ago
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