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Pachacha [2.7K]
2 years ago
13

If you have 400 grams of a substance that decays with a half-life of 14 days, then how much will you have after 56 days? To help

you answer the question, complete the table below. If appropriate, include units.
Chart:

Half Lives 0 1 ___ 3 ___

Total Time 0 ___ 28 days ___ ___

_______ ___ 200g ___ ___ ___
Chemistry
1 answer:
alisha [4.7K]2 years ago
3 0

Answer:

25 grams

Explanation:

You strat off with 400 grams of your substance. By day 14, half has dcayed and you only have 200 grams left. By day 28, there are 100 grams of the substance. On day 42, there are 50 grams left. Finally, on day 56, the substance has been through four half-lives and 25 grams remain.

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The cell potential of the following electrochemical cell depends on the gold concentration in the cathode half-cell: Pt(s)|H2(g,
Masja [62]

<u>Answer:</u> The concentration of Au^{3+} in the solution is 1.87\times 10^{-14}M

<u>Explanation:</u>

The given cell is:

Pt(s)|H_2(g.1atm)|H^+(aq.,1.0M)||Au^{3+}(aq,?M)|Au(s)

Half reactions for the given cell follows:

<u>Oxidation half reaction:</u> H_2(g)\rightarrow 2H^{+}(1.0M)+2e^-;E^o_{H^+/H_2}=0V ( × 3)

<u>Reduction half reaction:</u> Au^{3+}(?M)+3e^-\rightarrow Au(s);E^o_{Au^{3+}/Au}=1.50V ( × 2)

<u>Net reaction:</u> 3H_2(s)+2Au^{3+}(?M)\rightarrow 6H^{+}(1.0M)+2Au(s)

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=1.50-0=1.50V

To calculate the concentration of ion for given EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^6}{[Au^{3+}]^2}

where,

E_{cell} = electrode potential of the cell = 1.23 V

E^o_{cell} = standard electrode potential of the cell = +1.50 V

n = number of electrons exchanged = 6

[Au^{3+}]=?M

[H^{+}]=1.0M

Putting values in above equation, we get:

1.23=1.50-\frac{0.059}{6}\times \log(\frac{(1.0)^6}{[Au^{3+}]^2})

[Au^{3+}]=1.87\times 10^{-14}M

Hence, the concentration of Au^{3+} in the solution is 1.87\times 10^{-14}M

7 0
3 years ago
Gamma, beta and alpha radiation are all alike because:
timama [110]
Alpha, beta and gamma radiation are all types of radiation that can be emitted from the nucleus of an atom. Apart from this they are different. Alpha radiation consists of a helium nucleus, beta radiation consists of an electron (or positron) and gamma radiation consists of very high energy electromagnetic radiation. 
8 0
3 years ago
If a force of 1.4N is applied to a block with a mass of 7kg, what is the acceleration of the block in meters per second squared?
Anton [14]

Answer:

<h2>0.2 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

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a =  \frac{1.4}{7} =  \frac{1}{5}   \\

We have the final answer as

<h3>0.2 m/s²</h3>

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2 years ago
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Answer:

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Explanation:

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