<span><span>N2</span><span>O3</span><span>(g)</span>→NO<span>(g)</span>+<span>NO2</span><span>(g)</span></span>
<span><span>[<span>N2</span><span>O3</span>]</span> Initial Rate</span>
<span>0.1 M r<span>(t)</span>=0.66</span> M/s
<span>0.2 M r<span>(t)</span>=1.32</span> M/s
<span>0.3 M r<span>(t)</span>=1.98</span> M/s
We can have the relationship:
<span>(<span><span>[<span>N2</span><span>O3</span>]/</span><span><span>[<span>N2</span><span>O3</span>]</span>0</span></span>)^m</span>=<span><span>r<span>(t)/</span></span><span><span>r0</span><span>(t)
However,
</span></span></span>([N2O3]/[N2O3]0) = 2
Also, we assume m=1 which is the order of the reaction.
Thus, the relationship is simplified to,
r(t)/r0(t) = 2
r<span>(t)</span>=k<span>[<span>N2</span><span>O3</span>]</span>
0.66 <span>M/s=k×0.1 M</span>
<span>k=6.6</span> <span>s<span>−<span>1</span></span></span>
Answer:
Mass = 114.26 g
Explanation:
Given data:
Number of gold atoms = 3.47×10²³ atoms
Mass in gram = ?
Solution:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance. The number 6.022 × 10²³ is called Avogadro number.
1 mole = 6.022 × 10²³ atoms
3.47×10²³ atoms × 1 mol /6.022 × 10²³ atoms
0.58 mol
Mass of gold:
Mass = number of moles × molar mass
Mass = 0.58 mol × 197 g/mol
Mass = 114.26 g
<span>C4H4
The compound in question has an equal ratio of hydrogen and carbon. The atomic weight of carbon is roughly 12 and the atomic weight of hydrogen is roughly 1. The mass of the compound in question is roughly 52.
52/13=4
C4H4</span>
Answer:
DUPLET RULE: The tendency of an atom to acquire an outer most shell of two electrons is called duplet rule. OCTET RULE: The tendency of an atom to acquire an outermost shell of eight electrons is called the octet rule.
