First find the no. of moles of NaOH :
<span>30/1000 = 0.3 dm3 so no. of moles = 0.3*0.5 = 0.15 moles </span>
<span>as NaOH reacts with HNO3 in a ratio of one to one, there must have been 0.15 moles of HNO3 too </span>
<span>moles/volume = concentration </span>
<span>volume= 15/1000 = 0.15 dm3 </span>
<span>concentration = 1.15/0.15 = 1 mol.dm-3 </span>
<span>The quicker way would be to realize that you used twice as much NaOH so the HNO3 had to be twice as strong</span>
Answer:
0.0468 g.
Explanation:
- The decay of radioactive elements obeys first-order kinetics.
- For a first-order reaction: k = ln2/(t1/2) = 0.693/(t1/2).
Where, k is the rate constant of the reaction.
t1/2 is the half-life time of the reaction (t1/2 = 1620 years).
∴ k = ln2/(t1/2) = 0.693/(1620 years) = 4.28 x 10⁻⁴ year⁻¹.
- For first-order reaction: <em>kt = lna/(a-x).</em>
where, k is the rate constant of the reaction (k = 4.28 x 10⁻⁴ year⁻¹).
t is the time of the reaction (t = t1/2 x 8 = 1620 years x 8 = 12960 year).
a is the initial concentration (a = 12.0 g).
(a-x) is the remaining concentration.
∴ kt = lna/(a-x)
(4.28 x 10⁻⁴ year⁻¹)(12960 year) = ln(12)/(a-x).
5.54688 = ln(12)/(a-x).
Taking e for the both sides:
256.34 = (12)/(a-x).
<em>∴ (a-x) = 12/256.34 = 0.0468 g.</em>
Answer: ------------------------------------------------------------------------------------------------------------------------
Explanation: True, Ammonia is a base
The types of intermolecular forces that occur in a substance will affect its physical properties, such as its phase, melting point and boiling point.
Answer:
B - The atoms form a bond with a bond length of 75 pm
