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Brut [27]
3 years ago
15

How did Ernest Rutherford's experiment relate to J.J. Thomson's work?

Chemistry
1 answer:
olasank [31]3 years ago
4 0

Answer:

D

Explanation:

I am sorry if wrong

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1 point Berny has a 0.15-kilogram candle, a 1.00-kilogram jar, and a 0.25-kilogram candle holder. He places the candle and candl
evablogger [386]

Answer:

1.40

Explanation:

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3 years ago
Help with question 01.4 pleaseeee
Alina [70]

Answer:

Is there answer choices?

Explanation:

5 0
3 years ago
What is the mass of 3.022 moles of cobalt?
mote1985 [20]

Answer:

178.1g

Explanation:

m= n × MM

where

m is mass

n is moles &

MM is molecular mass

The molecular mass of cobalt is 58.933195. With this information and the number of moles given in the question, we can perform the calculation

m= 3.022 × 58.933195

m=178.09611529

m= 178.1g

7 0
3 years ago
Explain how to label a compound as a Bronstead-Lowry acid or base .
-BARSIC- [3]

Explanation:

According to the Bronsted-Lowry conjugate acid-base theory:

An acid is defined as a substance which looses donates protons and thus forming conjugate base.

A base is defined as a substance which accepts protons and thus forming conjugate acid.

For example:

HCl(aq)+NH_3(aq)\rightarrow Cl^-(aq)+NH_4^{+}(aq)

Here, HCl is loosing a proton, thus it is considered as an acid and after losing a proton, it forms Cl^- which is a conjugate base.

And, NH_3 is gaining a proton, thus it is considered as a base and after gaining a proton, it forms NH_4^{+} which is a conjugate acid.

7 0
3 years ago
What is the electric force on a proton 2.5 fmfm from the surface of the nucleus? Hint: Treat the spherical nucleus as a point ch
sammy [17]

Explanation:

It is known that charge on xenon nucleus is q_{1} equal to +54e. And, charge on the proton is q_{2} equal to +e. So, radius of the nucleus is as follows.

            r = \frac{6.0}{2}

              = 3.0 fm

Let us assume that nucleus is a point charge. Hence, the distance between proton and nucleus will be as follows.

              d = r + 2.5

                 = (3.0 + 2.5) fm

                 = 5.5 fm

                 = 5.5 \times 10^{-15} m     (as 1 fm = 10^{-15})

Therefore, electrostatic repulsive force on proton is calculated as follows.

              F = \frac{1}{4 \pi \epsilon_{o}} \frac{q_{1}q_{2}}{d^{2}}

Putting the given values into the above formula as follows.

           F = \frac{1}{4 \pi \epsilon_{o}} \frac{q_{1}q_{2}}{d^{2}}

              = (9 \times 10^{9}) \frac{54e \times e}{(5.5 \times 10^{-15})^{2}}

              = (9 \times 10^{9}) \frac{54 \times (1.6 \times 10^{-19})^{2}}{(5.5 \times 10^{-15})^{2}}

              = 411.2 N

or,           = 4.1 \times 10^{2} N

Thus, we ca conclude that 4.1 \times 10^{2} N is the electric force on a proton 2.5 fm from the surface of the nucleus.

8 0
4 years ago
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