1 carbon and 2 oxygen atoms CO2
Answer:
Answer: a) 20g of H2O (18.02 g/mol) molecules=6.68x10^23
Explanation:
In order to find the amount of molecules of each of the options, we need to follow the following equation.

So, let´s get the number of molecules for each of the options.





the smalest number is in option a)
Best of luck.
For chapter 4 is the strongest because he is 100 and bonding strongest is which liquid is hydrogen bonding strong it so that means it
Answer:
The answer to your question is letter B, 2-methylhexane.
Explanation:
Remember that for naming organic compounds first, we need to look for the largest chain of carbons.
In your example, the largest chain is horizontal and has 6 carbons.
Later, we need to circle all the branches, in your example there is only one branch located close to the left side
After that, we number the carbons of the main chain, starting in the corner with more branches, in your example we start from the first carbon on the left.
Finally, start naming the number of the carbon branch, later hte name of the branch and finally the name of the main chain.
Answer:
Explanation:
Use one of your experimentally determined values of k, the activation energy you determined, and the Arrhenius equation to calculate the value of the rate constant at 25 °C. Alternatively, you can simply extrapolate the straight line plot of ln(k) vs. 1/T in your notebook to 1/298 , read off the value of ln(k), and determine the value of k. Please put your answer in scientific notation. slope=-12070, Ea=100kJ/mol, k= 0.000717(45C), 0.00284(55C), 0.00492(65C), 0.0165(75C), 0.0396(85C)
Explanation;
According to Arrhenius equation:
i.e. ln(k2/k1) = -Ea/R (1/T2 - 1/T1)
Where, k1 = 0.000717, T1 = 45 oC = (45+273) K = 318 K
T2 = 25 oC = (25 + 273) K = 298 K
i.e. ln(k2/0.000717) = -12070 (1/298 - 1/318)
i.e. ln(k2/0.000717) = -2.54738
i.e. k2/0.000717 = 
= 0.078286
Therefore, the required constant (k2) = 0.078286 * 0.000717 = 