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3 years ago

What did ruthorford discover that thomson did not understand

Chemistry

1 answer:

Marianna [84]3 years ago

4 0

Answer: Atoms aren't entirely empty space

Explanation:

Rutherford did his scatter experiment to disprove Thomson's Plum Pudding Model. Rutherford shot helium nuclei at a sheet of gold and expected the particles to all go straight through to a sensor. The reason he expected this was because Thomson said that atoms are mostly made of empty space. Not all the particles went through the gold sheet, a lot of them actually reflected back to him. This meant that atoms aren't entirely empty space and something had to deflect the helium nuclei

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What is the pOH of<br> a<br> 2.6 x 10-6 M H+ solution?

scZoUnD [109]

Answer:

pH = -log 2.6 x 10-6 M

pH = 5.585

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8.4 as 2 sig figs

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3 years ago

What are some of the Differces between normal cells and cancer cells?

lilavasa [31]

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3 years ago

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Deffense [45]

Answer:

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Explanation:

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3 years ago

Using the molarity equation, calculate how many grams of salt you need in order to make 80ml of a 2M solution (MW = 58.44g/mole)

ollegr [7]

Answer:

9.35g

Explanation:

The molarity equation establishes that:

$\textrm{molarity}=\frac{\textrm{moles of solute}}{\textrm{liters of solution}}$

So, we have information about molarity (2M) and volume (80 ml=0.08 l), with that, we can find the moles of solute:

$\textrm{moles of solute}=\textrm{molarity}*\textrm{liters of solution}$

$\textrm{moles of solute}= 0.08 \textrm{ l} *2\textrm{ M} = 0.16 \textrm{ mol}$

The mathematical equation that establishes the relationship between molar weight, mass and moles is:

$\textrm{molar weight}= \frac{\textrm{mass}}{\textrm{moles}}$

$\textrm{MW}= \frac{\textrm{m}}{\textrm{n}}$

We have MW (58.44g/mole) and n (0.16 mol), and we need to find m (grams of salt needed) to solve the problem:

$\textrm{m} = \textrm{MW * n}= 58.44\frac{\textrm{g}}{\textrm{mol}} * 0.16 \textrm{ mol} = 9.35 \textrm{ g}$

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4 years ago

Draw a schematic diagram with a battery, connecting wires, a light bulb, and an ammeter in a series circuit.

Mila [183]

Answer:

here it is

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3 years ago