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Alexus [3.1K]
2 years ago
13

What did Rutherford's gold foil experiment tell about the atom?​

Chemistry
1 answer:
VLD [36.1K]2 years ago
3 0

The alpha particles that were fired at the gold foil were positively charged. ... These experiments led Rutherford to describe the atom as containing mostly empty space, with a very small, dense, positively charged nucleus at the center, which contained most of the mass of the atom, with the electrons orbiting the nucleus.

hope this helped

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For a certain experiment, a student requires 100 milliliters of a solution that is 8% HCl (hydrochloric acid). The storeroom has
exis [7]

Answer:

70 mL of 5% HCl and 30 mL of 15% HCl

Explanation:

We will designate x to be the fraction of the final solution that is composed of 5% HCl, and y to be the fraction of the final solution that is composed of 15% HCl. Since the percentage of the final solution is 8%, we can write the following expression:

5x + 15y = 8

Since x and y are fractions of a total, they must equal one:

x + y = 1

This is a system of two equations with two unknowns. We will proceed to solve for x. First, an expression for y is found:

y = 1 - x

This expression is substituted into the first equation and we solve for x.

5x + 15(1 - x) = 8

5x+ 15 - 15x = 8

-10x = -7

x = 7/10 = 0.7

We then calculate the value of y:

y = 1 - x = 1 - 0.7 = 0.3

Thus 0.7 of the 100 mL will be the 5% HCl solution, so the volume of 5% HCl we need is:

(100 mL)(0.7) = 70 mL

Similarly, the volume of 15% HCl we need is:

(100 mL)(0.3) = 30 mL

3 0
3 years ago
Explain the process of fossil fuel formation.
topjm [15]

Explanation:

Fossil fuel is an overall term for covered ignitable geologic stores of natural materials, framed from rotted plants and creatures that have been changed over to unrefined petroleum, coal, flammable gas, or weighty oils by introduction to warmth and weight in the world's outside more than a huge number of years.

The consuming of petroleum products by people is the biggest wellspring of emanations of carbon dioxide, which is one of the ozone depleting substances that permits radiative compelling and adds to an unnatural weather change.

A little bit of hydrocarbon-based powers are biofuels gotten from climatic carbon dioxide, and consequently don't build the net measure of carbon dioxide in the environment.

6 0
3 years ago
How does the presence of a catalyst affect the enthalpy of a reaction? Group of answer choices 1. It depends on whether you are
Otrada [13]

Answer:

Option 3. The catalyst does not affect the enthalpy change (\Delta H_\text{rxn}) of a reaction.

Explanation:

As its name suggests, the enthalpy change of a reaction (\Delta H_\text{rxn}) is the difference between the enthalpy of the products and the reactants.

On the other hand, a catalyst speeds up a reaction because it provides an alternative reaction pathway from the reactants to the products.

In effect, a catalyst reduces the activation energy of the reaction in both directions. The reactants and products of the reaction won't change. As a result, the difference in their enthalpies won't change, either. That's the same as saying that the enthalpy change \Delta H_\text{rxn} of the reaction would stay the same.

Refer to an energy profile diagram. Enthalpy change of the reaction \Delta H_\text{rxn} measures the difference between the two horizontal sections. Indeed, the catalyst lowered the height of the peak. However, that did not change the height of each horizontal section or the difference between them. Hence, the enthalpy change of the reaction stayed the same.

6 0
3 years ago
How many grams of CO2 are produced from 51 g of CH4
Elina [12.6K]

Answer:

use the equation Mass= RFM*Moles

Explanation:

use your periodic table

and create a little table

3 0
3 years ago
Suppose a 1.0 L solution contains 0.020 M in Cu(NO3)2, then 0.40 moles of NH3 are added. Assuming no change in volume, what is t
Rasek [7]

Answer:

4.6*10^{-14} M

Explanation:

Concentration of Cu^{2+} = [Cu(NO_3)_2] = 0.020 M

Constructing an ICE table;we have:

                                 Cu^{2+}+4NH_3_{aq} \rightleftharpoons [Cu(NH_3)_4]^{2+}_{(aq)}

Initial (M)             0.020          0.40                        0

Change (M)         -  x                - 4 x                       x

Equilibrium (M)  0.020 -x        0.40 - 4 x              x

Given that: K_f =1.7*10^{13}

K_f } = \frac{[Cu(NH_3)_4]^{2+}}{[Cu^{2+}][NH_3]^4}

1.7*10^{13} = \frac{x}{(0.020-x)(0.40-4x)^4}

Since x is so small; 0.40 -4x = 0.40

Then:

1.7*10^{13} = \frac{x}{(0.020-x)(0.0256)}

1.7*10^{13} = \frac{x}{(5.12*10^{-4}-0.0256x)}

1.7*10^{13}(5.12*10^{-4} - 0.0256x) = x

8.704*10^9-4.352*10^{11}x =x

8.704*10^9 = 4.352*10^{11}x

x = \frac{8.704*10^9}{4.352*10^{11}}

x = 0.0199999999999540

Cu^{2+}= 0.020 - 0.019999999999954

Cu^{2+} = 4.6*10^{-14} M

8 0
2 years ago
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