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slava [35]
3 years ago
11

En la ecuación NaOH (ac) + H2SO4 (ac) -- > Na2SO4 (S) + H2O (l) ¿Quienes son los reactivos?

Chemistry
2 answers:
ser-zykov [4K]3 years ago
8 0

Answer:

that's a lot

Explanation:

Drupady [299]3 years ago
4 0
Im sorry whats the question
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What is the energy of light with a wavelength of 468 nm? (The speed of light
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The answer for the problem is explained below.

The option for the answer is "D".

<u><em>Therefore the energy of the light is  4.25 × 10^-19 J</em></u>

Explanation:

Given:

wavelength (λ) = 468 nm = 468×10^-9 m

speed of light (c) = 3.00 x 10^8m/s

Planck's constant is 6.626 x 10^-34J·s

To solve:

energy of light (E)

We know,

E =(h×c) ÷ λ

E = ( 6.626 x 10^-34 ×  3.00 x 10^8) ÷ 468×10^-9

E = 4.25 × 10^-19 J

<u><em>Therefore the energy of the light is  4.25 × 10^-19 J</em></u>

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2 years ago
What is the mass, in grams, of one mole of any substance known as?
lbvjy [14]
It is known as its molar mass.
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3 years ago
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In atmospheric chemistry, the following chemical reaction converts SO2, the predominant oxide of sulfur that comes from combusti
Misha Larkins [42]

Answer:

Explanation:

From the given information;

The chemical reaction can be well presented as follows:

\mathtt{SO_{2(g)} + \dfrac{1}{2}O_{2(g)} }  ⇄ \mathtt{3SO_{2(l)}}

Now, K is known to be the equilibrium constant and it can be represented in terms of each constituent activity:

i.e

K = \dfrac{a_{so_3}}{a_{so_2} a_{o_2}^{\frac{1}{2}}}

However, since we are dealing with liquids solutions;

K = \dfrac{1}{\dfrac{Pso_2}{P^0}\Big ( \dfrac{Po_2}{P^0} \Big)^{1/2}}   since the activity of a_{so_3} is equivalent to 1

Hence, under standard conditions(i.e at a pressure of 1 bar)

K = \dfrac{1}{Pso_2Po_2^{1/2}}

(b)

From the CRC Handbook, we are meant to determine the value of the Gibb free energy by applying the formula:

\Delta _{rxn} G^o = \sum \Delta_f \ G^o (products) - \sum \Delta_fG^o (reactants) \\ \\ = (1) (-368 \ kJ/mol) - (\dfrac{1}{2}) (0) - ((1) (-300.13 \ kJ/mol)) \\ \\ = -368 \ kJ/mol + 300.13 \ kJ/mol \\ \\  \simeq -68 \ kJ/mol

Thus, for this reaction; the Gibbs frree energy = -68 kJ/mol

(c)

Le's recall that:

At equilibrium, the instantaneous free energy is usually zero &

Q(reaction quotient) is equivalent to K(equilibrium constant)

So;

\mathtt{\Delta _{rxn} G = \Delta _{rxn} G^o + RT In Q}

\mathtt{0- \Delta _{rxn} G^o = RTIn K } \\ \\ \mathtt{ \Delta _{rxn} G^o = -RTIn K }  \\ \\  K = e^{\dfrac{\Delta_{rxn} G^o}{RT}} \\ \\  K = e^{^{\dfrac{67900 \ J/mol}{8.314 \ J/mol \times 298 \ K}} }

K =7.98390356\times 10^{11} \\ \\  \mathbf{K = 7.98 \times 10^{11}}

(d)

The direction by which the reaction will proceed can be determined if we can know the value of Q(reaction quotient).

This is because;

If  Q < K, then the reaction will proceed in the right direction towards the products.

However, if Q > K , then the reaction goes to the left direction. i.e to the reactants.

So;

Q= \dfrac{1}{Pso_2Po_2^{1/2}}

Since we are dealing with liquids;

Q= \dfrac{1}{1 \times 1^{1/2}}

Q = 1

Since Q < K; Then, the reaction proceeds in the right direction.

Hence, SO2 as well O2 will combine to yield SO3, then condensation will take place to form liquid.

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Answer:

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