Barium-131's radiation level won't reach 1/4 of its initial level for 24 hours.
ln[A] t = -kt + ln[A] 0 is the integrated rate rule for the first-order reaction A's products.
A straight line is produced when the natural log of [A] is plotted as a function of time since this equation has the form y = mx + b.
How is the length of a half-life determined?
The amount of time needed for the reactant concentration to drop to half its initial value is known as the half-life of a reaction. A first-order reaction's half-life is a constant that is correlated with its rate constant:
t 1/2 = 0.693/k.
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Length of 1 side 1.2*10^-5km =1.2*10^-5*10^5 =1.2cm
<span>volume of the cube (1.2)^3=1.728 cm^3 </span>
<span>density= mass/volume= 1.1/1.728=0.636 g/cm^3</span>
Answer:
hydronium concentration camps and the Base is it going there now and the Base
Answer:
a) 2.01 g
Explanation:
- Na₂CO₃ (s) + 2AgNO₃ (aq) → Ag₂CO₃ (s) + 2NaNO₃
First we <u>convert 0.0302 mol AgNO₃ to Na₂CO₃ moles</u>, in order to <em>calculate how many Na₂CO₃ moles reacted</em>:
- 0.0302 mol AgNO₃ *
= 0.0151 mol Na₂CO₃
So the remaining Na₂CO₃ moles are:
- 0.0340 - 0.0151 = 0.0189 moles Na₂CO₃
Finally we <u>convert Na₂CO₃ moles into grams</u>, using its <em>molar mass</em>:
- 0.0189 moles Na₂CO₃ * 106 g/mol = 2.003 g Na₂CO₃
The closest answer is option a).
Answer:
0.22 mol HClO, 0.11mol HBr.
0.25mol NH₄Cl, 0.12 mol HCl
Explanation:
A buffer is defined as a mixture in solution between weak acid and its conjugate base or vice versa.
Potassium hypochlorite (KClO) could be seen as conjugate base of HClO (Weak acid). That means the addition of <em>0.22 mol HClO </em>will convert the solution in a buffer. HBr reacts with KClO producing HClO, thus, <em>0.11mol HBr</em> will, also, convert the solution in a buffer. 0.23 mol HBr will react completely with KClO and in the solution you will have only HClO, no a buffering system.
Ammonia (NH₃) is a weak base and its conjugate base is NH₄⁺. That means the addition of <em>0.25mol NH₄Cl</em> will convert the solution in a buffer. Also, NH₃ reacts with HCl producing NH₄⁺. Thus, addition of<em> 0.12 mol HCl</em> will produce NH₄⁺. 0.25mol HCl consume all NH₃.
