<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:
the average answer in maybe 60
<span>Answer:
For this problem, you would need to know the specific heat of water, that is, the amount of energy required to raise the temperature of 1 g of water by 1 degree C. The formula is q = c X m X delta T, where q is the specific heat of water, m is the mass and delta T is the change in temperature. If we look up the specific heat of water, we find it is 4.184 J/(g X degree C). The temperature of the water went up 20 degrees.
4.184 x 713 x 20.0 = 59700 J to 3 significant digits, or 59.7 kJ.
Now, that is the energy to form B2O3 from 1 gram of boron. If we want kJ/mole, we need to do a little more work.
To find the number of moles of Boron contained in 1 gram, we need to know the gram atomic mass of Boron, which is 10.811. Dividing 1 gram of boron by 10.811 gives us .0925 moles of boron. Since it takes 2 moles of boron to make 1 mole B2O3, we would divide the number of moles of boron by two to get the number of moles of B2O3.
.0925/2 = .0462 moles...so you would divide the energy in KJ by the number of moles to get KJ/mole. 59.7/.0462 = 1290 KJ/mole.</span>
Answer:
sample B contains the larger density
Explanation:
Given;
volume of sample A, V = 300 mL = 0.3 L
Molarity of sample A, C = 1 M
volume of sample B, V = 145 mL = 0.145 L
Molarity of sample B, C = 1.5 M
molecular mass of sodium chloride, Nacl = 23 + 35.5 = 58.5 g/mol
Molarity is given as;
The reacting mass for sample A = 0.3mol x 58.5 g/mol = 17.55 g
The reacting mass for sample B = 0.2175 mol x 58.5 g/mol = 12.72 g
The density of sample A 
The density of sample B 
Therefore, sample B contains the larger density
