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slavikrds [6]
3 years ago
14

Which of the following statements is true? A. Democritus discovered the proton. B. Dalton discovered the proton. C. Thomson disc

overed the electron. D. Rutherford discovered the neutron.
Chemistry
1 answer:
iren2701 [21]3 years ago
8 0

It’s C.thomson discovered the electron

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pashok25 [27]

Answer:

1 no. herbivore ....... 2 no. exception......3 no. diameter........4 no. heart........ 5 no. catastrophe...... 6 no. develop...... mark me brainliest plzzzz

8 0
2 years ago
Read 2 more answers
It all adds up worksheet
Komok [63]
Atomic number and the number of protons are the same...
Neutrons = Mass number - number of protons
Electrons are same # unless there is a charge 
The whole number you see on the periodic table is the atomic number of the element which is also same as the number of protons

1) carbon - 14  ;  Mass number = 14 , Protons = 6 , Neutrons = 14 - 6 = 8 
                                Electrons = 6 

2) Lead - 208  ; Mass # = 208 , Protons = 82 , Neutrons = 208 - 82 = 126
                          Electrons = 82

3) Uranium - 239 ; Mass # = 239  , Protons  =  92,Neutrons = 239 - 92 = 147
                                 Electrons =  92

4) Uranium - 238  ; Mass # = 238 , Protons = 92 , Neutrons = 238 - 92 = 146
                                   
Electrons = 92 

5) Tin - 118  ; Mass # = 118 , Protons = 50 , Neutrons = 118 - 50 = 68
                         Electrons = 50
8 0
3 years ago
We might think of a porous material as being a composite wherein one of the phases is a pore phase. Estimate upper and lower lim
eduard

Answer:

The upper and lower limits for the room-temperature thermal conductivity of a magnesium oxide material having a volume fraction of 0.10 of pores that are filled with still air are

Ku = 38.252 W/mK

K lower = 0.199 W/mK

Explanation:

As we know  

Ku = Vp * Kair + Vmagnesium * K metal  

Ku = 0.10 *0.02 + (1-0.25) * 51

Ku = 38.252 W/mK

The lower limit  

K lower = Kmetal* Kair/( Vp * Kmetal + Vmetal * K air)

K lower = (0.02*51)/(0.10*51 + 0.90 * 0.02)

K lower = 0.199 W/mK

8 0
2 years ago
Please answer this question ASAP
Nezavi [6.7K]

Answer:

put a test tube over the opening, remove it and quickly put a lit splint near the mout or in the tube. if you hear a squeaky pop it is hydrogen.

Explanation:

hydrogen ignites in air.

4 0
3 years ago
An oxide of aluminum contains 0.545 g of Al and 0.485 g of O. Find the empirical formula
uysha [10]

Answer:

The answer is: <u>Al2O3</u>

Explanation:

The data they give us is:

  • 0.545 gr Al
  • 0.485 gr O.

To find the empirical formula without knowing the grams of the compound, we find it per mole:

  • 0.545 g Al * 1 mol Al / 27 g Al = 0.02 mol Al
  • 0.485 g O * 1 mol O / 16 g O = 0.03 mol O

Then we must divide the results obtained by the lowest result, which in this case is 0.02:

  • 0.02 mol Al / 0.02 = 1  Al
  • 0.03 mol O / 0.02 = 1.5  O

Since both numbers have to give an integer, multiply by 2 until both remain integers:

  • 1Al * 2 = 2Al
  • 1.5O * 2 = 3O

Now the answer is given correctly:

  • Al2O3

8 0
3 years ago
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