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4 years ago

4.What did J. J. Thomson observe when he applied electric voltage to a cathode ray tube in his famous experiment?

Chemistry

2 answers:

Tpy6a [65]4 years ago

3 0

<h2>Answer 4:</h2>

Correct option is B

B.The glass opposite of the negative electrode started to glow.

<h2>Explanation:</h2>

In his experiment, tubes were two electric plates, producing a positive anode and a negative cathode, which he expected that it would deflect the rays. As he supposed, the rays were deflected by the electric charge, proving beyond doubt that the rays were made up of charged particles bearing a negative charge.

<h2>Answer 11:</h2>

Boron is a chemical element in the periodic table with symbol B and has atomic number 5.

Boron has two naturally occurring isotopes Boron-10 and Boron-11. The average atomic mass of these are calculated as

(10.01×0.198) + (11.01×0.802) = 10.81 amu

So, the average atomic mass of the Boron isotopes is 10.81 amu.

katrin [286]4 years ago

3 0

The answer to your question is <span>B.The glass opposite of the negative electrode started to glow.
</span>

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5÷966843966667777889578876

Phoenix [80]

Your answer my good friend should be:

9.668439666677778e+23

9.668439666677778e+23

(They are mean't to be fractions btw)

Exact Form: $\frac{5}{966843966667777889578876}$

Decimal Form: 5.17146527⋅ $10^{-24}$

4 0

3 years ago

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You observe the same spectral line in two stars that are identical in every way except that one rotates faster than the other. H

Tom [10]

Answer:

The line in the faster rotating star is broader.

Explanation:

7 0

3 years ago

The following physical constants are for water, H2O.

Delicious77 [7]

Answer:

$Q\approx6.4~kJ$

Explanation:

Quantity of heat required by 10 gram of ice initially warm it from -5°C to 0°C:

$Q_1=m.C_s.\Delta T$

here;

mass, m = 10 g

specific heat capacity of ice, $C_s=2.09~J.g^{-1}.^{\circ}C^{-1}$

change in temperature, $\Delta T=(5-0)=5^{o}C$

$Q_1=10\times2.09\times 5$

$Q_1=104.5~J$

Amount of heat required to melt the ice at 0°C:

$Q_2=m.\Delta H_{fus}$

where, $\Delta H_{fus}=6020~J/mol$

we know that no. of moles is = (wt. in gram) $\div$ (molecular mass)

$Q_2=\frac{10}{18} \times 6020$

$Q_2=3344.44~J$

Now, the heat required to bring the water to 70°C from 0°C:

$Q_3=m.C_L.\Delta T$

specific heat of water, $C_L=4.18~J/g/^oC$

change in temperature, $\Delta T=(70-0)=70^oC$

$Q_3=10\times 4.18\times 70$

$Q_3=2926~J$

Therefore the total heat required to warm 10.0 grams of ice at -5.0°C to a temperature of 70.0°C:

$Q=Q_1+Q_2+Q_3$

$Q=104.5+3344.44+2926$

$Q=6374.94~J$

$Q\approx6.4~kJ$

8 0

3 years ago

An object has a mass of 441 g and a volume

Ratling [72]

Answer:

<h3>The answer is 4.41 g/cm³</h3>

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

From the question we have

$density = \frac{441}{10} \\$

We have the final answer as

Hope this helps you

6 0

3 years ago

Given that sodium chloride is 39.0% sodium by mass, how many grams of sodium chloride are needed to have 400.mg of na present?

Galina-37 [17]

The mass percentage of Na in NaCl is 39.0%
This means that if there's 100 mg of NaCl , 39 mg of the compound is Na.
For compound to contain 39 mg Na - 100 mg of NaCl should be present.
Therefore for the compound to contain 400 mg - 100/39 x 400 - 1025.6 mg
The mass of NaCl present should be 1.026 g

4 0

3 years ago

Read 2 more answers