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3 years ago

5

In Rutherford's golf foil experiment, no a particles stuck to the atom. If a particles had stuck to the atom, what might you hav

e
deduced about the atom's structure?

1 answer:

borishaifa [10]3 years ago

4 0

Answer

:yes or no

Explanation:

you needd tol try it yolurself

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What is Lewis symbol for K+

Alexxx [7]

Lewis symbol for K+ is given by a dot and K.

<h3>Meaning of Lewis symbol</h3>

Lewis symbols can be defined as a symbol the is used in describing the valence electron configurations of an atom ions.

A Lewis symbol consists of the symbol of the element and then surrounded by one dot for each of its valence electrons.

In conclusion, Lewis symbol for K+ is given by a dot and K.

Learn more about Lewis symbol: brainly.com/question/6605265

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7 0

2 years ago

What is the proper name for Zn(CIO)2?

Sergio [31]

Answer:

ZINC CHLORATE

Explanation:

Zinc chloride is the name of chemical compounds with the formula ZnCl2 and its hydrates.

3 0

3 years ago

Read 2 more answers

He rate constant of a reaction is 4.55 × 10−5 l/mol·s at 195°c and 8.75 × 10−3 l/mol·s at 258°c. what is the activation energy o

Xelga [282]

Answer : The activation energy of the reaction is, $17.285\times 10^4kJ/mole$

Solution :

The relation between the rate constant the activation energy is,

$\log \frac{K_2}{K_1}=\frac{Ea}{2.303\times R}\times [\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_1$ = initial rate constant = $4.55\times 10^{-5}L/mole\text{ s}$

$K_2$ = final rate constant = $8.75\times 10^{-3}L/mole\text{ s}$

$T_1$ = initial temperature = $195^oC=273+195=468K$

$T_2$ = final temperature = $258^oC=273+258=531K$

R = gas constant = 8.314 kJ/moleK

Ea = activation energy

Now put all the given values in the above formula, we get the activation energy.

$\log \frac{8.75\times 10^{-3}L/mole\text{ s}}{4.55\times 10^{-5}L/mole\text{ s}}=\frac{Ea}{2.303\times (8.314kJ/moleK)}\times [\frac{1}{468K}-\frac{1}{531K}]$

$Ea=17.285\times 10^4kJ/mole$

Therefore, the activation energy of the reaction is, $17.285\times 10^4kJ/mole$

8 0

3 years ago

Read 2 more answers

An element has five isotopes. Calculate the atomic mass of this element using the information below. Show all your work. Using t

Katena32 [7]

Answer: Sol:-

Data provided in the question is :-

Atomic mass of isotope -1 = 64 amu

Atomic mass of isotope -2 = 66 amu

Atomic mass of isotope -3 = 67 amu

Atomic mass of isotope -4 = 68 amu

Atomic mass of isotope - 5 = 70 amu

Percentage abundace of isotope - 1 = 48.89 %

Percentage abundance of isotope -2 = 27.81 %

Percentage abundance of isotope - 3 = 4.11%

Percentage abundance of isotope-4 = 18.57%

Percentage abundance of isotope - 5 = 0.62 %

Formula used :-

Average atomic mass of an element =[ {(atomic mass of isotope-1 * percentage abundance of isotope-1) + ( atomic mass of isotope-2 * percentage abundance of isotope -2) + ( atomic mass of isotope -3 * percantege abundance of isotope-3 ) + ( atomic mass of isotope-4 * percentage abundance of isotope-4) + (atomic mass of isotope-5 * percentage abundance of isotope-5)} / 100]

Calculation :-

Put all the value in the formula :-

Average atomic mass of an element = [{(64 * 48.89) + (66 * 27.81) + (67 * 4.11) + (68 * 18.57) + (70 * 0.62)} / 100] amu

= [{(3128.96) + (1835.46) +(257.37) + (1262.76) + (43.4)} / 100] amu

= {(6528.04) / 100} amu

= 65.2804 amu

Average atomic mass of an element is = 65.2804 amu

Then this mass is approximatly equal to atomic mass of zinc so this element would be zinc

atomic mass of zinc = 65.38 \approx 65.2804 amu

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3 years ago

PLEASE HELP!!!!!!! CHEMISTRY!!!!

AnnZ [28]

Answer:

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Explanation:

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