<span>
If mass remains the same while the volume of a substance increases, the
density of the substance will decrease. The increase in the volume is called expansion, while the decrease in the volume is called contraction.
</span><span>The density measures how much matter (mass) there is in a given amount of space (volume). So, if the mass stays the same and the volume increases then there is the same amount of mass ad before but in bigger volume so the density is decreased. </span>
Answer:
Half life = 1 / k[Ao]
Explanation:
From:
1/ [A] = kt + 1/ [Ao]
Isolating t on its own, we have:
kt = 1 / [A] - 1 / [Ao]
t = 1 / [Ao] / k
Re-arranging we have:
t = 1 / k [Ao]
The t represents the t=half life of the second order reaction and the formula can be re-written as:
t1/2 = 1 / k [Ao]
This is so because second order reaction decreases at a much faster rate than zero and first order reactions and there slopes decreases to zero at a much faster rate.
Answer:
false
Explanation:
There are a equal number of protons and electrons in neutral atoms.
Answer:
The enthalpy of the nitrogen-nitrogen bond in N2H4 is 162.6 kJ
Explanation:
For the reaction: N2H4(g)+H2(g)→2NH3(g), the enthalpy change of reaction is
ΔH rxn = 2 ΔHºf NH3 - ΔHºf N2H4
but we also know that the ΔH rxn is calculated by accounting the sum of number of bonds formed and bonds broken as follows:
ΔH rxn = 6H (N-H) + 4 (N-H) + 2H (H-H)
where H is the bond enthalpy .When bonds are broken H is positive, and negative when formed, in the product there are 6 N-H bonds , and in the reactants 4 N-H and 1 H-H bonds).
Consulting an appropiate reference handbook or table the following values are used:
ΔHºf (NH3) = -46 kJ/mol
ΔHºf (N2H4) = 95.94 kJ/mol
(The enthalpy of fomation of hydrogen in its standard state is zero)
H (N-H) = 391 kJ
H (H-H) = 432 kJ
H (N-N) = ?
So plugging our values:
ΔH rxn = 2mol ( -46.0 kJ/mol) - 1mol(95.4 kJ/mol) = -187.40 kJ
-187.40 kJ = 6(-391 kJ) + 4 (391 kJ) + 432 + H(N-N)
-187.40 kJ = -350 kJ + H(N-N)
H(N-N) = 162.6 kJ
To be able to answer this equations, we must set given information. Suppose the reaction to yield NO is:
N₂ + O₂ → 2 NO
Next, suppose you have 1 g of each of the reactants. Determine first which is the limiting reactant.
1 g N₂ (1 mol N₂/ 28 g)(2 mol NO/1 mol N₂)= 0.07154 mol NO present
Number of molecules = 0.07154 mol NO(6.022×10²³ molecules/mol)
<em>Number of molecules = 4.3×10²² molecules NO present</em>
