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LekaFEV [45]
3 years ago
9

If I toss a coin, how does the fact that I got tails affect the probility that I will get tails again?

Chemistry
2 answers:
artcher [175]3 years ago
6 0

Answer:

  The probability would be 50%.                                                    

alexgriva [62]3 years ago
4 0

Answer:

Your first flip has a 1/2 or 50% chance of getting tails. In order to get 2 tails in a row you have to get tails first and then again. The first flip is 1/2 and the second one is another 1/2. Combined you have a 1/4 chance of getting 2 tails in a row.

Each time you flip you have 1/2 a chance, but when you look at all the flips at once, each consecutive flip of the same side has a smaller probability.

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Explanation:

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3 years ago
"What quantitative measurement and qualitative observations can be made about the interactions of matter and energy?" can someon
kirza4 [7]

Quantitative measurements are numerical values, they involve amounts and units like measuring things. Qualitative observations appeal to the five senses, like what does the interaction look and sound like

5 0
4 years ago
What is the pH of 0.10 M NaF(aq). The Ka of HF is 6.8 x 10-4
dsp73

<u>Given information:</u>

Concentration of NaF = 0.10 M

Ka of HF = 6.8*10⁻⁴

<u>To determine:</u>

pH of 0.1 M NaF

<u>Explanation:</u>

NaF (aq) ↔ Na+ (aq) + F-(aq)

[Na+] = [F-] = 0.10 M

F- will then react with water in the solution as follows:

F- + H2O ↔ HF + OH-

Kb = [OH-][HF]/[F-]

Kw/Ka = [OH-][HF]/[F-]

At equilibrium: [OH-]=[HF] = x and [F-] = 0.1 - x

10⁻¹⁴/6.8*10⁻⁴ = x²/0.1-x

x = [OH-] = 1.21*10⁻⁶ M

pOH = -log[OH-] = -log[1.21*10⁻⁶] = 5.92

pH = 14 - pOH = 14-5.92 = 8.08

Ans: (b)

pH of 0.10 M NaF is 8.08

6 0
3 years ago
The half-life of a radioactive isotope is the amount of time it takes for a quantity of that isotope to decay to one half of its
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Radio active decay reactions follow first order rate kinetics.

a) The half life and decay constant for radio active decay reactions are related by the equation:

t_{\frac{1}{2}} =\frac{ln 2}{k}

t_{\frac{1}{2}} = \frac{0.693}{k}

Where k is the decay constant

b) Finding out the decay constant for the decay of C-14 isotope:

Decay constant (k) = \frac{0.693}{t_{\frac{1}{2}}}

k = \frac{0.693}{5230 years}

k = 1.325 * 10^{-4} yr^{-1}

c) Finding the age of the sample :

35 % of the radiocarbon is present currently.

The first order rate equation is,

[A] = [A_{0}]e^{-kt}

\frac{[A]}{[A_{0}]} = e^{-kt}

\frac{35}{100} = e^{-(1.325 *10^{-4})t}

ln(0.35) = -(1.325 *10^{-4})(t)

t = 7923 years

Therefore, age of the sample is 7923 years.

3 0
3 years ago
A recent study has revealed that chlorinated hydrocarbons, gasoline and other volatile organic compounds (VOC's) have become sig
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Answer:

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