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grin007 [14]
3 years ago
14

What did Ernest Rutherford s gold foil experiment demonstrate about atoms?

Chemistry
1 answer:
Ghella [55]3 years ago
5 0
-Positively charged nucleus 
-Empty spaced
-Dense core
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Which substance will most likely dissociate when it is dissolved in water?
Katyanochek1 [597]

Answer: CaC12

Explanation: Calcium chloride not 100% sure tho

6 0
2 years ago
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How many seconds would it take to deposit 17.3 g of ag (atomic mass = 107.87) from a solution of agno3 using a current of 10.00
Brums [2.3K]
Data Given:

Time = t = ?

Current = I = 10 A

Faradays Constant = F = 96500

Chemical equivalent = e = 107.86/1 = 107.86 g

Amount Deposited = W = 17.3 g

Solution:
             According to Faraday's Law,

                                          W  =  I t e / F

Solving for t,

                                          t  =  W F / I e
Putting values,
                                          t  =  (17.3 g × 96500) ÷ (10 A × 107.86 g)

                                          t  =  1547.79 s

                                          t  = 1.54 × 10³ s
5 0
2 years ago
The colour change of methyl orange with neutral substance​
Arisa [49]

Answer:

Pink

Explanation:

Because at first its orange then neutral its pink

3 0
3 years ago
The amount of dissolved salt in water is know as
Aleksandr [31]

Answer:

The water is solvent and the salt is solute

5 0
3 years ago
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A sample of sodium-24 with an activity of 14 mCi is used to study the rate of blood flow in the circulatory system. If sodium-24
SOVA2 [1]

Answer:

See explanation

Explanation:

From;

0.693/t1/2 = 2.303/t log (Ao/A)

Where;

t1/2 = half life of the sodium-24

t = time taken

Ao = initial activity of sodium-24

A= activity of sodium-24 at time t

a)

0.693/15 = 2.303/15 log (14/A)

0.0462 = 0.1535 log (14/A)

0.0462/0.1535 =  log (14/A)

log (14/A) = 0.0462/0.1535

14/A = Antilog(0.3)

14/A= 1.995

A = 14/1.995

A = 7.0 mCi

b)

0.693/15 = 2.303/30 log (14/A)

0.0462 =0.0768 log(14/A)

0.0462/0.0768 =log (14/A)

(14/A) =Antilog (0.6)

A = 14/Antilog (0.6)

A = 3.5 mCi

c)

0.693/15 = 2.303/45 log (14/A)

0.0462= 0.0512 log (14/A)

log (14/A) = 0.0462/0.0512

log (14/A) = 0.9

(14/A) = Antilog (0.9)

A= 14/Antilog (0.9)

A = 14/7.9

A = 1.77  mCi

d)

2.5 days = 2.5 * 24 hours = 60 hours

0.693/15 = 2.303/60 log (14/A)

0.0462 = 0.03838 log (14/A)

log (14/A) = 0.0462/0.03838

(14/A) = Antilog(1.2)

A= 14/Antilog(1.2)

A = 14/15.8

A = 0.886 mCi

Note that activity (A) decreases as time increases.

5 0
3 years ago
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