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Ratling [72]
3 years ago
10

Help me with 4,5 plz

Chemistry
1 answer:
Helen [10]3 years ago
7 0
The answer to 4 is A.                                              
The answer to 5 is C.
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The end product will depend upon

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b) the final treatment of the reaction

If we have just taken methylmagnesium iodide and p-hydroxyacetophenone, then we will get methane and hydroxyl group substituted with MgI in place of hydrogen

Figure 1

However if we have taken excess of methylmagnesium iodide which is Grignard's reagent followed by hydrolysis we will get different product

Figure 2

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Choose the correct coefficients to balance the following equation:
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You placed 6.35 g of a mixture containing unknown amounts of BaO(s) and MgO(s) in a 3.50-L flask containing CO₂(g) at 30.0°C and
adelina 88 [10]

Mass of BaO in  initial mixture = 3.50g

Explanation:

Let mass of BaO in mixture be x g

mass of MgO in mixture be (6.35 - x) g

Initially CO_2

Volume = 3.50 L

Temp = 303 K

Pressure = 750 torr = 750 / 760 atm

Applying ideal gas equation

PV = nRT

n = PV / RT

(n)_CO_2 = ((750/760)* 3.50) / 0.0821 * 303

(n)_CO_2 = 0.139 mole

Finally; mole of CO_2

n= PV /RT

((245/760) *3.5) / 303* 0.0821

(n)_CO_2 = 0.045 mole

Mole of CO_2 reacted = 0.139 - 0.045

=0.044 mole

BaO + CO_2  BaCO_3

Mgo + CO_2  MgCO_3

moles of CO_2 reacted = ( moles of BaO + moles of MgO)

moles of BaO in mixture = x / 153 mole

moles of MgO in mixture = 6.35 - x mole / 40

Equating,

x/ 153 +6.35/40 = 0.094

= x/153 + 6.35 / 40 - x/40 =0.094

= x (1/40 - 1153) = (6.35/40 - 0.094)

= x * 10.018464

= 0.06475

mass of BaO in mixture = 3.50g

5 0
3 years ago
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