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Natali [406]
3 years ago
11

A box contains identical balls of which 12 are red, 18 white and 8 blue. Three balls are drawn from the box one after the other

without replacement. Find the probability that;
(a) three are red
(b)the first is blue and the other two are red.​
Chemistry
1 answer:
svet-max [94.6K]3 years ago
8 0

Answer:

a) 12/323

b) 8/233

Explanation:

a) The probability of a red ball being drawn is 12/38, or in a simplified fraction, 6/19. To find the probability that 3 are red you would multiply the probability of the fraction for each, except subtracting one from the total each time as the drawn is done without replacement. This is done as follows: 6/19 × 6/18 × 6/17= 12/323

b) The probability of drawing a blue ball is 8/38, or 4/19. To find that the first one is blue and the rest are red, the equation is done as follows: 4/19 × 6/18 × 6/17 = 8/233

(hopefully I did this right)

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oxygen is a common oxidizing agent in nature. what change in the oxidation number of oxygen must occur if it is to be an oxidizi
Minchanka [31]

Answer is: the oxidation number of oxygen must decrease.

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6 0
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What would be the empirical formula of a compound that is 25.5% carbon, 6.40% hydrogen, and 68.1% oxygen?
aleksandr82 [10.1K]
Empirical formula is the simplest ratio of whole numbers of components in a compound 
in 100 g of compound 
                                C                          H                          O
mass                   25.5 g                     6.40 g                 68.1 g
number of moles  25.5 g/12 g/mol      6.40 g/ 1 g/mol      68.1 g/ 16 g/mol 
                             = 2.13 mol              = 6.40 mol          = 4.26 mol
divide by least number of moles 
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8 0
3 years ago
Is the electrochemical cell spontaneous or not spontaneous as written at 25 ∘C ? spontaneous not spontaneous Calculate the poten
givi [52]

Answer:

The cell potential is 0.609 V. Given E > 0 the electrochemical cell is spontaneous as written.

Explanation:

Let's consider the oxidation and reduction half-reactions and the global reaction.

Anode (oxidation): Sn²⁺(0.0023 M) ⇒ Sn⁴⁺(0.13 M) + 2 e⁻

Cathode (reduction): 2 Fe³⁺(0.11 M) + 2 e⁻ ⇒ 2 Fe²⁺(0.0037 M)

Global reaction: Sn²⁺(0.0023 M) + 2 Fe³⁺(0.11 M) ⇒ Sn⁴⁺(0.13 M) + 2 Fe²⁺(0.0037 M)

The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E° = E°red,cat - E°red,an

E° = 0.771 V - 0.154 V = 0.617 V

The Nernst equation allows us to calculate the cell potential (E) under the given conditions.

E=E\° -\frac{0.05916}{n} logQ\\E=E\° -\frac{0.05916}{n} log\frac{[Sn^{+4}].[Fe^{2+}]}{[Sn^{2+}].[Fe^{3+} ]} \\E=0.617V-\frac{0.05916}{2} log\frac{(0.13).(0.0037)}{(0.0023).(0.11)} \\E=0.609V

The cell potential is 0.609 V. Given E > 0 the electrochemical cell is spontaneous as written.

4 0
3 years ago
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