• Explain, in terms of energy, particles motion & bonds,

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QveST [7]

Answer:

hi

Explanation:

4 0

3 years ago

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Zinc metal is added to a flask containing aqueous hydrochloric acid. The flask contains 0.400 mole of HCl. How much hydrogen gas

dusya [7]

Answer:

0.4 g of hydrogen gas would be produced.

1. 0.48 mole of HCl is needed to react completely with 15.5 g of zinc.

2. HCl is the limiting reactant.

3. 2.61 g of zinc is in excess

Explanation:

From the balanced equation of reaction:

$Zn(s)+ 2HCl(aq) --> H_2(g) + ZnCl_2(aq)$

1 mole of Zn requires 2 moles of HCl to produce 1 mole of hydrogen gas.

15.5 g of zinc = 15.5/65.3 = 0.24 moles of zinc.

0.24 moles Zn is supposed to require 0.24 x 2 moles HCl which is equivalent to 0.48 moles HCl.

But only 0.400 mole of HCl is present. Hence, HCl is the limiting reagent. For complete reaction with 15.5 g of Zinc, 0.48 mole HCl would be needed.

0.400 mole of HCl will require 0.2 mole of Zn for complete reaction. This thus means that 0.24 - 02 = 0.04 mole of Zn is in excess.

0.04 mole Zn = 0.04 x 65.3 = 2.61 g excess Zn.

Now, since HCl is the limiting reagent;

2 moles of HCl is required to produce 1 mole of H2 according to the equation.

0.400 mole HCl will therefore yield 0.400 x 1/2 = 0.2 mole H2

0.2 mole H2 = 2 x 0.2 = 0.4 g H2

Hence, 0.4 g of hydrogen gas would be produced.

6 0

3 years ago

What is the value for the reaction: N2(g) + 2 O2(g) --> N2O4(g) in terms of K values from the reactions:

Valentin [98]

Answer : The correct expression will be:

$K=(K_1)^2\times K_2$

Explanation :

The chemical reactions are :

(1) $\frac{1}{2}N_2(g)+\frac{1}{2}O_2(g)\rightleftharpoons NO(g)$ $K_1$

(2) $2NO(g)+O_2(g)\rightleftharpoons N_2O_4(g)$ $K_2$

The final chemical reaction is :

$N_2(g)+2O_2(g)\rightleftharpoons N_2O_4$ $K=?$

Now we have to calculate the value of $K$ for the final reaction.

Now equation 1 is multiply by 2 and then add both the reaction we get the value of 'K'.

If the equation is multiplied by a factor of '2', the equilibrium constant will be the square of the equilibrium constant of initial reaction.

If the two equations are added then equilibrium constant will be multiplied.

Thus, the value of 'K' will be:

$K=(K_1)^2\times K_2$

6 0

3 years ago

Individual particles move with the wave.

MArishka [77]

Answer:

No.

Explanation:

No, individual particles do not move with the wave, it only oscillates back and forth its mean position. The particles in the medium transfer its energy to their neighboring particles and in that way the energy moves in the form of wave. The particles only vibrates on its means position instead of moving from one place to another. So we can conclude that Individual particles do not move with the wave.

3 0

3 years ago

Enter your answer in the provided box. Calculate the number of moles of CrCl, that could be produced from 49.4 g Cr202 according

Mrrafil [7]

Answer:

0.4694 moles of CrCl₃

Explanation:

The balanced equation is:

Cr₂O₃(s) + 3CCl₄(l) → 2CrCl₃(s) + 3COCl₂(aq)

The stoichiometry of the equation is how much moles of the substances must react to form the products, and it's represented by the coefficients of the balanced equation. So, 1 mol of Cr₂O₃ must react with 3 moles of CCl₄ to form 2 moles of CrCl₃ and 3 moles of COCl₂.

The stoichiometry calculus must be on a moles basis. The compounds of interest are Cr₂O₃ and CrCl₃. The molar masses of the elements are:

MCr = 52 g/mol

MCl = 35.5 g/mol

MO = 16 g/mol

So, the molar mass of the Cr₂O₃ is = 2x52 + 3x35.5 = 210.5 g/mol.

The number of moles is the mass divided by the molar mass, so:

n = 49.4/210.5 = 0.2347 mol of Cr₂O₃.

For the stoichiometry:

1 mol of Cr₂O₃ ------------------- 2 moles of CrCl₃

0.2347 mol of Cr₂O₃----------- x

By a simple direct three rule:

x = 0.4694 moles of CrCl₃

6 0

3 years ago