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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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How do I solve for V1 in this equation

arlik [135]

Answer:

Explanation:

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$

<u>First of all cross multiply</u>

That's

$P_{1}V_{1}T_{2} = P_{2}V_{2}T_{1}$

<u>Next divide both sides by </u> $P_{1}T_{2}$ <u> in order to isolate </u> $V_{1}$

We have

$\frac{P_{1}V_{1}T_{2}}{P_{1}T_{2}} = \frac{P_{2}V_{2}T_{1}}{P_{1}T_{2}}$

We have the final answer as

$V_{1} = \frac{P_{2}V_{2}T_{1}}{P_{1}T_{2}}$

Hope this helps you

3 0

4 years ago

2) A track star ran 1500 meters. How many feet did they run?

horrorfan [7]

The number of feet the track star ran is 4921.26 feet

<h3>Conversion scale </h3>

We can convert meter to feet by using the following conversion scale:

0.3048 meter = 1 feet

<h3>How to cconvert 1500 meters to feet</h3>

The following data were obtained from the question:

Distance (in meter) = 1500 meter
Distance (in feet) =?

We can convert 1500 meter to feet as illustrated below:

0.3048 meter = 1 feet

Therefore,

1500 meter = (1500 meter × 1 feet) / 0.3048 meter

1500 meter = 4921.26 feet

Thus, 1500 meter is equivalent to 4921.26 feet

Learn more about conversion:

brainly.com/question/1560145

#SPJ1

5 0

2 years ago

What is a low pressure system?

KonstantinChe [14]

A low pressure system has lower pressure at its center than the areas around it. Winds blow towards the low pressure, and the air rises in the atmosphere where they meet. As the air rises, the water vapor within it condenses, forming clouds and often precipitation.

<u>Explanation</u>:

Wind flow towards the low pressure and the air rises in the atmosphere. As the air increases, the water vapor within it solidifies, forming clouds and undergo precipitation. Low pressure formed in the center areas.

The atmospheric circulations of air up and down in a low-pressure area remove a small amount of atmosphere. This usually happens between warm and cold air masses by flowing air which tries to reduce the contrast of temperature.

5 0

3 years ago

What is the most likely product of hydration of CH3CH=CH2 ?

nikdorinn [45]

Answer is: C. CH3CHOHCH3.

Balanced chemical reaction of hydration of propene:

CH₃CH=CH₂ + H₂O → CH₃CHOHCH₃.

A hydroxyl group (OH⁻) attaches to one carbon of the double bond and a proton (H⁺) adds to the other carbon of the double bond according to Markovnikov's rule (hydrogen gets attached to the carbon with more hydrogen substituents).

Product of this reaction is propan-2-ol (isopropyl alcohol).

7 0

3 years ago

Two linear hydrocarbons, Hexane (C6H14) and Heptane (C7H16), form pretty much an ideal solution at any composition. A solution i

Anettt [7]

Answer:

$y_{C_6H_{14}}=0.92$

Explanation:

Hello,

At first, we compute liquid-phase molar fractions:

$n_{C_6H_{14}}=463.96 g*\frac{1mol}{86g} =5.3949molC_6H_{14}\\n_{C_7H_{16}}=667.71 g*\frac{1mol}{100g} =6.6771molC_7H_{16}\\x_{C_6H_{14}}=\frac{5.3949}{5.3949+6.6771} =0.447\\x_{C_7H_{16}}=1-x_{C_6H_{14}}=0.553$

Now, by means of the fugacity concept, for hexane, for instance, we have:

$f_{C_6H_{14}}^V=f_{C_6H_{14}}^L\\y_{C_6H_{14}}p_T=x_{C_6H_{14}}p_{C_6H_{14}}$

In this manner, at 25 °C the vapor pressure of hexane and heptane are 0.198946 atm and 0.013912 atm repectively, thus, the total pressure is:

$p_T=x_{C_6H_{14}}p_{C_6H_{14}}+x_{C_7H_{16}}p_{C_7H_{16}}\\p_T=0.447*0.198946 atm +0.553*0.013912 atm=0.096622atm$

Finally, from the hexane's fugacity equation, we find its mole fraction in the vapour as:

$y_{C_6H_{14}}=\frac{x_{C_6H_{14}}p_{C_6H_{14}}}{p_T}=\frac{0.447*0.198946 atm}{0.096622atm} \\y_{C_6H_{14}}=0.92$

Best regards.

3 0

3 years ago

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