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olga2289 [7]
2 years ago
9

Which scientist determined the charge of the electron?.

Chemistry
1 answer:
Alexus [3.1K]2 years ago
3 0

Answer:

J.J. Thomson

Explanation:

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Adding acid and and catching the solution that drains through.
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Mercury’s natural state is where the atoms are close to each other but are still free to pass by each other. In which state(s) c
kvv77 [185]

Mercury naturally exists in Liquid state.

On Condensing it can exist in Solid state as well.

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Please help me complete the following word equations:) also write the balanced equation, full ionic equation, and net ionic equa
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Explanation:

The states may differ depending on the reactions

6 0
3 years ago
in a mixture of 1.90 mol of gas, 0.85 mol are nitrogen (n2) molecules. what is the mole fraction of n2 in this mixture?
Nostrana [21]

0.447 is the mole fraction of Nitrogen in this mixture.

mole fraction of nitrogen= moles of nitrogen/total moles

mole fraction of nitrogen=0.85/1.90

mole fraction of nitrogen=0.447

The product of the moles of a component and the total moles of the solution yields a mole fraction, which is a unit of concentration measurement. Because it is a ratio, mole fraction is a unitless statement. The sum of the components of the mole fraction of a solution is one. In a mixture of 1 mol benzene, 2 mol carbon tetrachloride, and 7 mol acetone, the mole fraction of the acetone is 0.7. This is computed by dividing the sum of the moles of acetone in the solution by the total number of moles of the solution's constituents:

To know more about mole fraction visit : brainly.com/question/8076655

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4 0
1 year ago
A coffee-cup calorimeter initially contains 125 g water at 24.28C. Potassium bromide (10.5 g), also at 24.28C, is added to the w
iren2701 [21]

Answer:

The solution is given below

Explanation:

Heat, q= mc∆T

q= 125g x 4.18 J/g∙°C x (21.18x- 24.28) °C

q=  -1619.75J

NEGATIVE SIGN INDICATES THAT HEAT IS ABSORBED.

Enthalpy Change, ∆H = 1619.75 7/ 10.5 g

                                     = 154.26 J/g

No. of moles of KBr = Mass of KBr/ Molecular Weight of KBr

                                =10.5g/119gmol-1

                                =0.088 mol

∆H= 1619.75 J/ 0.088 mol

      = 18.41 kJ/mol  

6 0
3 years ago
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