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3 years ago

How did rutherford's experimental evidence lead to the development of a new atomic number

1 answer:

8 0

Rutherford's experiment was the gold foil experiment.

The gold foil experiment was him shooting alpha particles (you could think of this as a Helium atom without its electrons) into a gold foil. The whole experiment was surrounded with something called Zinc Sulfide which sparked when the alpha particles hit it.

Most of the alpha particles went through, approximately 1 in 8000 alpha particles deflected at a large angle (almost right back to where it was shot).

This constant ratio caused him to conclude that:-the atom was mostly empty space (since most alpha particles went through)-there was something very positive in the atom (the proton)-the proton was very dense (since it made something going light speed deflect back at a large angle)-The proton was also very small (since only 1 in 8000 hit it)

Prior to the discovery of the proton, John Dalton's periodic table was used. Having "elements" such as soda and potash. Now that we have discovered the proton and found out that each atom's number of protons is unique, we used that to classify each element's identity.

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I bet no one can solve this

Nikitich [7]

The answer is 89.4 g/mol cuz the two units carry positives so that means 8.50 flows for 2 more hours than usual.

3 0

2 years ago

What must be the molarity of an aqueous solution of trimethylamine, (ch3)3n, if it has a ph = 11.20? (ch3)3n+h2o⇌(ch3)3nh++oh−kb

Stolb23 [73]

0.040 mol / dm³. (2 sig. fig.)

<h3>Explanation</h3>

$(\text{CH}_3)_3\text{N}$ in this question acts as a weak base. As seen in the equation in the question, $(\text{CH}_3)_3\text{N}$ produces $\text{OH}^{-}$ rather than $\text{H}^{+}$ when it dissolves in water. The concentration of $\text{OH}^{-}$ will likely be more useful than that of $\text{H}^{+}$ for the calculations here.

Finding the value of $[\text{OH}^{-}]$ from pH:

Assume that $\text{pK}_w = 14$ ,

$\begin{array}{ll}\text{pOH} = \text{pK}_w - \text{pH} \\ \phantom{\text{pOH}} = 14 - 11.20 &\text{True only under room temperature where }\text{pK}_w = 14 \\\phantom{\text{pOH}}= 2.80\end{array}$ .

$[\text{OH}^{-}] =10^{-\text{pOH}} =10^{-2.80} = 1.59\;\text{mol}\cdot\text{dm}^{-3}$ .

Solve for $[(\text{CH}_3)_3\text{N}]_\text{initial}$ :

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}$

Note that water isn't part of this expression.

The value of Kb is quite small. The change in $(\text{CH}_3)_3\text{N}$ is nearly negligible once it dissolves. In other words,

$[(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}$ .

Also, for each mole of $\text{OH}^{-}$ produced, one mole of $(\text{CH}_3)_3\text{NH}^{+}$ was also produced. The solution started with a small amount of either species. As a result,

$[(\text{CH}_3)_3\text{NH}^{+}] = [\text{OH}^{-}] = 10^{-2.80} = 1.58\times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}$ .

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\textbf{initial}} = \text{K}_b = 1.58\times 10^{-3}$ ,

$[(\text{CH}_3)_3\text{N}]_\textbf{initial} =\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{\text{K}_b}$ ,

$[(\text{CH}_3)_3\text{N}]_\text{initial} =\dfrac{(1.58\times10^{-3})^{2}}{6.3\times10^{-5}} = 0.040\;\text{mol}\cdot\text{dm}^{-3}$ .

8 0

3 years ago

What is the mass in grams of BaCl2 that is needed to prepare 200 mL of a 0.500 M solution

DENIUS [597]

Answer:

= 20.82 g of BaCl2

Explanation:

Given,

Volume = 200 mL

Molarity = 0.500 M

Therefore;

Moles = molarity × volume

= 0.2 L × 0.5 M

= 0.1 mole

But; molar mass of BaCl2 is 208.236 g/mole

Therefore; 0.1 mole of BaCl2 will be equivalent to;

= 208.236 g/mol x 0.1 mol

= 20.82 g

Therefore, the mass of BaCl2 in grams required is 20.82 g

6 0

2 years ago

What is the mass of six of these marbles? What is the volume? What is the<br> density?

chubhunter [2.5K]

Answer:

All right. So let's calculate the density of a glass marble. Remember that the formula for density is mass over volume. So if I know that the masses 18.5 g. And I know that the um volume is 6.45 cubic centimeters. I can go ahead and answer this to three significant figures. So it's going to be 2.87 grams per cubic centimeter. Okay, that's our density. Now, density is an intensive process. Okay. We're an intensive property. I really should say. It doesn't depend on how much you have. Mhm. If I have one marble, its density is going to be 2.87 g per cubic centimeter. If I have two marbles, the density will be the same because I'll double the mass and I'll also double the volume. So when I divide them I'll get the same number. Okay, that's what makes it an intensive property. No matter how many marbles I have, they'll have the same density. Mass though is not an intensive property. So if I have six marbles and I want to know what the massive six marbles is. Well, I know the mass of each marble is 18.5 g. So the mass of six marbles Is going to be 100 11 g. Because mass is an extensive property. It depends on how much you have. If I change the number of marbles, I'm going to change the mass. That's an extensive property. All right. So we've calculated the density. We've calculated the mass and then what happens to the density of one marble compared to six marbles as we mentioned before. Since densities and intensive property, the densities will be the same, no matter how may.

Explanation:

5 0

2 years ago

During an experiment, Juan rolled a six-sided number cube 18 times. The number two occurred four times. Juan claimed the experim

Karolina [17]

Answer:

Because u would have to find the undercorse of 010-1 witch makes the out of part by 6

Explanation:

Given :

Juan rolled a six-sided number cube 18 times.

The number two occurred four times.

To Find: Juan claimed the experimental probability of rolling a two was approximately 1/9. Why is Juan’s experimental probability incorrect?

Solution:

Total events = number of times cube rolled = 18

Favorable events = The number two occurred four times. = 4

So, Experimental probability of rolling a two was approximately 1/9

6 0

2 years ago