All categories

2 years ago

What was thompson working with when he discovered the cathode rays

2 answers:

6 0

Thomas discovered cathode rays when he was working with electrons. He was performing experiments with electrons when one of them became electrically charged and a beam of light came from it, this creating a cathode ray.

vlabodo [156]2 years ago

3 0

In 1897 British physicist J.J. Thomson showed the rays were composed of a previously unknown negatively charged particle, which was later named the electron.

You might be interested in

Setting up a one-step unit conversion A student sets up the following equation to convert a measurement. (The ? stands for a num

Aneli [31]

Answer:

0.0008 m

Explanation:

We are given that 0.080 cm

We have to convert 0.080 cm into meter

To find the value of 0.080 cm in meter we are using unitary method

We know that

100 cm =1 m

1 cm = $\frac{1}{100} m$

Therefore, 0.080 cm = $\frac{1}{100}\times 0.080 m$

0.080 cm = $\frac{1}{100}\times \frac{80}{1000}m$

0.080 cm = $\frac{8}{10000}m$

0.08cm=0.0008 m

Hence, the value of 0.080 cm is equal to 0.0008 meter .

7 0

2 years ago

What class of compounds do fats belong to

NARA [144]

Fats are referred to as lipids, or cellular compounds that are insoluble in water.

7 0

3 years ago

Read 2 more answers

Which of these careers would a student studying biotechnology

matrenka [14]

I think it’s c I could be wrong

3 0

2 years ago

If the products of a reaction are Co2 and 2H20, what is known about the<br>reactants?

Svetradugi [14.3K]

Answer:

Reactant : A combustion of hydrocarbon.

Explanation:

It is known that when hydrocarbons are involved in combustion, they produce carbon dioxide and water.

CxHy + (x+y/4)O2 ===> xCO2 + y/2H2O

4 0

2 years ago

If 20.0 g of NaOH is added to 0.750 L of 1.00 M Cd(NO₃)₂, how many grams of Cd(OH)₂ will be formed in the following precipitatio

bulgar [2K]

Answer:

$m_{Cd(OH)_2}=36.6 gCd(OH)_2$

Explanation:

Hello.

In this case, for the given chemical reaction, in order to compute the grams of cadmium hydroxide that would be yielded, we must first identify the limiting reactant by computing the yielded moles of that same product, by 20.0 grams of NaOH (molar mass = 40 g/mol) and by 0.750 L of the 1.00-M solution of cadmium nitrate as shown below considering the 1:2:1 mole ratios respectively:

$n_{Cd(OH)_2}^{by\ NaOH}=20.0gNaOH*\frac{1molNaOH}{40gNaOH} *\frac{1molCd(OH)_2}{2molNaOH} =0.25molCd(OH)_2\\\\n_{Cd(OH)_2}^{by\ Cd(NO_3)_2}=0.750L*1.00\frac{molCd(NO_3)_2}{L}*\frac{1molCd(OH)_2}{1molCd(NO_3)_2} =0.75molCd(OH)_2$

Thus, since 20.0 grams of NaOH yielded less of moles of cadmium hydroxide, NaOH is the limiting reactant, therefore the mass of cadmium hydroxide (molar mass = 146.4 g/mol) is:

$m_{Cd(OH)_2}=0.25molCd(OH)_2*\frac{146.4gCd(OH)_2}{1molCd(OH)_2} \\\\m_{Cd(OH)_2}=36.6 gCd(OH)_2$

Best regards.

4 0

2 years ago