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Elza [17]
3 years ago
12

Why did scientists think that Rutherford’s model was incomplete?

Chemistry
1 answer:
kow [346]3 years ago
3 0
It did not explain how the atoms electrons are arranged in the space around the nucleus it did not explain whole the negative charged electrons are not pulled into the positive nucleus.
You might be interested in
Methane gas and oxygen gas react to form water vapor and carbon dioxide gas. What volume of carbon dioxide would be produced by
Bas_tet [7]

Answer:

V_{CO_2}=9.23m^3

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)

Now, as the stoichiometrical factors are in terms of mole but no information about neither the temperature nor the pressure is given, by means of the Avogadro's law, one could perform the stoichiometric calculations with the given volume as both the pressure and temperature remain the same, that is:

V_{CO_2}=9.23m^3CH_4*\frac{1m^3CO_2}{1m^3CH_4} \\\\V_{CO_2}=9.23m^3

Such 1:1 volume relationship equals the 1:1 molar relationship given in the chemical reaction in terms of their stoichiometric coefficients, therefore, the yielded volume of carbon dioxide is also 9.23m³

Best regards.

4 0
4 years ago
Natural gas or methane is a fossil fuel. In a plentiful supply of air, it burns to give carbon dioxide and water. Write a balanc
Alex17521 [72]

Answer:

CH_4 + 2O_2 --> CO_2 + 2H_2O

Explanation:

The chemical formula for methane is CH_4

The chemical formula for oxygen is O_2

The chemical formula for carbon dioxide is CO_2

The chemical formula for water is H_2O

Methane burns in oxygen to give carbon dioxide and water:

CH_4 + O_2 --> CO_2 + H_2O

The balanced equation that will ensure an equal number of each atom on the left and right-hand sides of the equation would be:

CH_4 + 2O_2 --> CO_2 + 2H_2O

3 0
3 years ago
An aqueous solution is prepared by dissolving 15.6 g of Cu(NO3)2 ⋅ 6 H2O in water and diluting to 345 mL of solution. What is th
Vladimir [108]

<u>Answer:</u> The molarity of NO_3^- ions in the solution is 0.306 M

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

We are given:

Mass of solute (Cu(NO_3)_2.6H_2O) = 15.6 g

Molar mass of (Cu(NO_3)_2.6H_2O) = 295.6 g/mol

Volume of solution = 345 mL

Putting values in above equation, we get:

\text{Molarity of }Cu(NO_3)_2.6H_2O=\frac{15.6g\times 1000}{295.6g/mol\times 345mL}\\\\\text{Molarity of }Cu(NO_3)_.6H_2O=0.153M

As, 1 mole of (Cu(NO_3)_2.6H_2O) produces 1 mole of copper (II) ions and 2 moles of nitrate ions.

So, molarity of NO_3^- ions = (2 × 0.153) = 0.306 M

Hence, the molarity of NO_3^- ions in the solution is 0.306 M

3 0
3 years ago
Which equation correctly shows an endothermic reaction? *
Nady [450]
It’s A because +1500 kJ is before the arrow. it’s in the reactant side. so it is endothermic.
8 0
3 years ago
Question 26 Suppose a flask is filled with of and of . The following reaction becomes possible: The equilibrium constant for thi
Blizzard [7]

Answer:

0.36 M

Explanation:

There is some info missing. I think this is the complete question.

<em>Suppose a 250 mL flask is filled with 0.30 mol of N₂ and 0.70 mol of NO. The following reaction becomes possible: </em>

<em>N₂(g) +O₂(g) ⇄ 2 NO(g) </em>

<em>The equilibrium constant K for this reaction is 7.70 at the temperature of the flask.  Calculate the equilibrium molarity of O₂. Round your answer to two decimal places.</em>

<em />

Initially, there is no O₂, so the reaction can only proceed to the left to attain equilibrium. The initial concentrations of the other substances are:

[N₂] = 0.30 mol / 0.250 L = 1.2 M

[NO] = 0.70 mol / 0.250 L = 2.8 M

We can find the concentrations at equilibrium using an ICE Chart. We recognize 3 stages (Initial, Change, and Equilibrium) and complete each row with the concentration or change in the concentration.

    N₂(g) +O₂(g) ⇄ 2 NO(g)

I    1.2        0              2.8

C  +x         +x            -2x

E  1.2+x      x           2.8 - 2x

The equilibrium constant (K) is:

K=7.70=\frac{[NO]^{2}}{[N_{2}][O_{2}]} =\frac{(2.8-2x)^{2} }{(1.2+x).x}

Solving for x, the positive one is x = 0.3601 M

[O₂] = 0.3601 M ≈ 0.36 M

7 0
3 years ago
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