The mixture contains 62 % one isomer and 38 % the enantiomer.
Let’s say that the mixture contains 62 % of the (<em>R</em>)-isomer.
Then % (<em>S</em>) = 100 % -62 % = 38 %
ee = % (<em>R</em>) - % (<em>S</em>) = 62 % -38 % = 24 %
Answer:
1040%
Explanation:
To solve this question we must convert the mass of Iron to moles in order to find limiting reactant. With limiting reactant we can find the theoretical moles of hydrogen and theoretical mass:
Percent yield = Actual yield (5.40g) / Theoretical yield * 100
<em>Moles Fe -Molar mass: 55.845g/mol-:</em>
10.3g * (1mol / 55.845g) = 0.184 moles of Fe will react.
For a complete reaction of these moles there are necessaries:
0.184 moles Fe* ( 3 mol H2SO4 / 2 mol Fe) = 0.277 moles H2SO4.
As there are 14.8 moles of the acid, <em>Fe is limiting reasctant.</em>
The moles of H2 produced are:
0.184 moles Fe* ( 3 mol H2 / 2 mol Fe) = 0.277 moles H2
The mass is:
0.277 moles H2 * (2.016g/mol) = 0.558g H2
Percent yield is:
5.40g / 0.558g * 100 = 1040%
It is possible the experiment wasn't performed correctly
In an alkene, cis and trans isomers are possible because the double band is rigid, cannot rotate, has groups attached to the carbons of the double bond that are fixed relative to each other, and only occurs with double bonds-possibility that molecule will have different geometries; two different molecules with slightly different properties.
-Trans-2 ends of chain across the double bond.
While naming Cis-Trans isomers the prefix cis or trans are placed in front of the alkene name when there are cis-trans isomers.
V ( NaOH ) = mL ?
M ( NaOH ) = 0.100 M
V ( HCl ) = 9.00 mL / 1000 => 0.009 L
M ( HCl ) = 0.0500 M
number of moles HCl:
n = M x V
n = 0.009 x 0.0500 => 0.00045 moles HCl
mole ratio:
<span>HCl + NaOH = NaCl + H2O
</span>
1 mole HCl ---------------- 1 mole NaOH
0.00045 moles HCl ----- ??
0.00045 x 1 / 1 => 0.00045 moles of NaOH
M = n / V
0.100 = 0.00045 / V
V = 0.00045 / 0.100
V = 0.0045 L
1 L ------------ 1000 mL
0.0045 L ----- ??
0.0045 x 1000 / 1 => 4.5 mL of NaOH
The density of a material is the mass of the material per unit volume. Here the weight of the same metal is 44.40g, 40.58g and 38.35g having volume 4.8 mL, 4.7 mL and 4.2 mL respectively. Thus the density of the metal as per the given data are, 
= 9.25g/mL, 
= 8.634g/mL and 
= 9.130g/mL respectively.
The equation of the standard deviation is √{∑(x - 
)÷N}
Now the mean of the density is {(9.25 + 8.634 + 9.130)/3} = 9.004 g/mL.
The difference of the density of the 1st metal sample (9.25-9.004) = 0.246 g/mL. Squaring the value = 0.060.
The difference of the density of the 2nd metal sample (9.004-8.634) =0.37 g/mL. Squaring the value = 0.136.
The difference of the density of the 3rd metal sample (9.130-9.004) = 0.126 g/mL. Squaring the value 0.015.
The total value of the squared digits = (0.060 + 0.136 + 0.015) = 0.211. By dividing the digit by 3 we get, 0.070. The standard deviation will be 
. Thus the standard deviation of the density value is 0.265g/mL.
