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natita [175]
3 years ago
14

What is the oxidation number of an element?

Chemistry
1 answer:
tatiyna3 years ago
7 0

Answer:

Oxidation number is the total number of electrons that an atom either gains or loses in order to form a chemical bond with another atom.

Hope this helps

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What is the energy of a photon of green light with a frequency of 5.50 x 101^4 Hz?
Minchanka [31]
I didn't know if you meant to the power of 14 but if you did here your answer:
3.64 x 10^-19
(you just multiply the frequency by Planck's constant= 6.63 × 10^–34)
8 0
4 years ago
Be sure to use proper significant figures. The molecular weight of NaCl is 58.44 grams/mole. To prepare a 0.50 M solution: You w
Misha Larkins [42]
Your answer would be,
Molarity = moles of solute/volume of solution we needed, 29.22(g)(mol) of NaCI

= 29.22(g)/58.44(g)(mol^-1)(1)/1(L)
= 0.500(mol)(L^-1)



Hope that helps!!!
5 0
3 years ago
Read 2 more answers
What mass of water is produced by the combustion 1.2 x 10^26 molecules of butane ?
podryga [215]

Answer:

17.934 kg of water

Explanation:

If balanced equation is not given; this format can come in handy.

For any alkane of the type : CₙH₂ₙ₊₂ , it's combustion reaction will follow:

2CₙH₂ₙ₊₂ + (3n+1) O₂  →  (2n)CO₂ + 2(n+1) H₂O

For butane:

2C₄H₁₀(g) + 13O₂(g) →  8CO₂(g) + 10H₂O(l)

2 moles of butane gives 10 moles of water.

1 mol of any substance has Avogadro number(N) of molecules in it( 6.022 x 10²³)

Mass of 1 mole of any substance is equal to it's molar mass

So, if 2 x N molecules of butane gives 10 x 18 g of water.

Then  1.2 x 10²⁶ molecules will give:

\frac{1.2 \times 10^{26} \times 180}{2 \times 6.022 \times  10^{23}}

= 17.934 x 10³ g of water

= 17.934 kg of water

5 0
3 years ago
In the Haber process for ammonia synthesis, K " 0.036 for N 2 (g) ! 3 H 2 (g) ∆ 2 NH 3 (g) at 500. K. If a 2.0-L reactor is char
lisabon 2012 [21]

Answer : The partial pressure of N_2,H_2\text{ and }NH_3 at equilibrium are, 1.133, 2.009, 0.574 bar respectively. The total pressure at equilibrium is, 3.716 bar

Solution :  Given,

Initial pressure of N_2 = 1.42 bar

Initial pressure of H_2 = 2.87 bar

K_p = 0.036

The given equilibrium reaction is,

                              N_2(g)+H_2(g)\rightleftharpoons 2NH_3(g)

Initially                   1.42      2.87             0

At equilibrium    (1.42-x)  (2.87-3x)     2x

The expression of K_p will be,

K_p=\frac{(p_{NH_3})^2}{(p_{N_2})(p_{H_2})^3}

Now put all the values of partial pressure, we get

0.036=\frac{(2x)^2}{(1.42-x)\times (2.87-3x)^3}

By solving the term x, we get

x=0.287\text{ and }3.889

From the values of 'x' we conclude that, x = 3.889 can not more than initial partial pressures. So, the value of 'x' which is equal to 3.889 is not consider.

Thus, the partial pressure of NH_3 at equilibrium = 2x = 2 × 0.287 = 0.574 bar

The partial pressure of N_2 at equilibrium = (1.42-x) = (1.42-0.287) = 1.133 bar

The partial pressure of H_2 at equilibrium = (2.87-3x) = [2.87-3(0.287)] = 2.009 bar

The total pressure at equilibrium = Partial pressure of N_2 + Partial pressure of H_2 + Partial pressure of NH_3

The total pressure at equilibrium = 1.133 + 2.009 + 0.574 = 3.716 bar

6 0
3 years ago
I need help in science
astraxan [27]
The answer is D . I hope this help you :) .
7 0
2 years ago
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