The 1996 NHTSA study, Crash Outcome Data Evaluation System
(CODES), related traffic flow and medicinal accounts in seven states to measure
entire costs of wound from motor vehicle crashes. The research originate that the
mean inpatient costs for crash fatalities who were not using safety belts
were 55 percent <span>higher than for those who were belted.</span>
You did not include the options but I can tell you the product ratio.
The product ratio is the mole ratio of the products of the reaction.
From the balanced chemical equation you have all the mole ratios:
The given equation is: 2 C6H5COOH + 15O2 --> 14 CO2 + 6H2O
The mole ratios are: 2 C6H5COOH: 15 O2: 14 CO2 : 6 H2O
The products are CO2 and H2O
Their mole ratio = 14 CO2 : 6 H2O
That can be expressed as:
14 mol CO2 7 mol CO2
----------------- = -----------------
6 mol H2O 3 mol H2O
It is also the same that:
6 mol H2O : 14 mol CO2
6 mol H2O 3 mol H2O
------------------ = -------------------
14 mol CO2 7 mol CO2
So, compare your options to the ratios show above and pick the proper ratio.
Answer:
(1) 0.10 (2) 17.8 g
Explanation:
Since the reaction ratio is 1:1 what we need is to convert the given masses to moles and you will have the answer:
MW anthracene = 178.23 g/mol
MW maleic anhydride = 98.06 g/mol
a) mass anthracene = 178 mg x 1 g/ 1000 mg = 0.178 g anthracene
Moles anthracene = 0.178 g anthracene/ 178.23 g/mol
= 0.001 mol anthracene
0.001 mol anthracene x 1 mol maleic acid/mol anthracene
= 0.001 mol maleic anhydride
mass maleic anhydride = 0.001 mol x 98.06 g/mol = 0.10 g
b) moles maleic anhydride = 9.8 g/ 98.06 g/mol = 0.099 moles
0.099 moles maleic anhydride x 1 mol anthracene/mol maleic anhydride =
0.099 mol anthracene
g anthracene = 0.10mol x 178 g/mol = 17.8 g
Answer:
Akash
Explanation:
it could be a magnet with the same poles facing eachoher
<span>In order to do this, you have change the alkene into an
alkyne. That is the aim of Br2/CH2Cl2 trailed by NaNH2. The Br2 with form a vic
dihalide (3,4-dibromo octane). Adding of NaNH2 will execute two E2 reactions.
-NH2 will eliminate an H from carbons 3 and 4. This double elimination will make
the alkyne. Then handling the alkyne with H2/Lindlar will form the cis alkene. The
final product will be CIS-3-octene.</span>
