Chemistry, 10.07.2019 05:30 elijahjacksonrp6z2o7

Group a used 5.00 moles of styrene and 0.0010 moles of benzoyl peroxide while group b used the same amount of benzoyl peroxide as group a but used only 1.50 moles styrene. what would the average molecular weight of each group's polymer sample be

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Answer from: kira96

Answer : Average molecular weight of group A = 104.0276 g

Average molecular weight of group B = 104.0920 g

Solution : Given, Moles of group A styrene = 5 moles

Moles of group A benzoyl peroxide = 0.0010 moles

Moles of group B styrene = 1.5 moles

Moles of group B benzoyl peroxide = 0.0010 moles

Molecular mass of styrene = 104 g/mole

Molecular mass of benzoyl peroxide = 242.23 g/mol

Formula used : Average molar mass = Total mass / Total moles

For group A,

Average molar mass of group A = $\frac{\text{Total mass of group A}}{\text{Total moles of group A}}$

= $\frac{\left ( 5\times 104 \right )+\left ( 0.0010\times 242.23 \right )}{5+0.0010}$

= $\frac{520+0.24223}{5.0010}$

= $\frac{520.24223}{5.0010}$

= 104.0276 g

Average molar mass of group A = 104.0276 g

For group B,

Average molar mass of group B = $\frac{\text{Total mass of group B}}{\text{Total moles of group B}}$

= $\frac{\left ( 1.5\times 104 \right )+\left ( 0.0010\times 242.23 \right )}{1.5+0.0010}$

= $\frac{156+0.24223}{1.5010}$

= $\frac{156.24223}{1.5010}$

= 104.0920 g

Average molar mass of group B = 104.0920 g

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

Answer from: Quest

air

in a tire you fill a tire with air

then a couple days later you check the air pressure in the tire

hope this mark me brainliest

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Group a used 5.00 moles of styrene and 0.0010 moles of benzoyl peroxide while group b used the same...