Answer:
Explanation:
The question will be easier to solve if we interpret it as, " How long will it take until one-fourth of a sample of the element remains,?"
The half-life of the element is the time it takes for half of it to decay.
After one half-life, half (50 %) of the original amount will remain.
After a second half-life, half of that amount (25 %) will remain, and so on.
We can construct a table as follows:
Answer:
density=6.74g/ml
:320g÷47.5ml
d=6.74g/ml
thank you
<em><u>I </u></em><em><u>hope</u></em><em><u> </u></em><em><u>this </u></em><em><u>is </u></em><em><u>helpful</u></em>
Answer:
i think so
Ba(OH)2 + H2SO4 ------> BaSO4 + 2H2O
1) Moles of Ba(OH)2 = moles of H2SO4 = 0.025L x 2)0.02M = 5.0 x 10^-4M
Concn of Ba(OH)2 in g/L = 5.0 x 10^-4M x 171.33g/mol = 0.086g/mol
Answer:
Cathode: Ag
Anode: Br₂
Explanation:
In the cathode must occur a reduction, so it's more likely to a metal atom be in the cathode. For the metals given the reduction reactions and the potential of reduction are:
Ag⁺ + e⁻ ⇒ Ag⁰ E° = + 0.80 V
Fe⁺² + 2e⁻ ⇒ Fe⁰ E° = - 0.44 V
Al⁺³ + 3e⁻ ⇒ Al⁰ E° = -1.66 V
As the potential for Ag is the higher, the reduction will occur for it first, so in the cathode will produce Ag.
For the anode an oxidation must occurs, so the reactions for the nonmetals are:
F₂ + 2e⁻ ⇒ 2F⁻ E° = +2.87 V
Cl₂ + 2e⁻ ⇒ 2Cl⁻ E° = +1.36 V
Br₂ + 2e⁻ ⇒ 2Br⁻ E° = +1.07 V
For oxidation, the less the E°, the faster the reaction will occur, so Br₂ will be formed in the anode.
Answer:
you should study.
Explanation:
it depends on when the exam is, but remember to make sure you study well for it and dont rely on studying it all the night of the exam.
