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Cloud [144]
3 years ago
10

A 5.00-g sample of aluminum pellets (specific heat capacity = 0.89 j/°c · g) and a 10.00-g sample of iron pellets (specific heat

capacity = 0.45 j/°c · g) are heated to 100.0°c. the mixture of hot iron and aluminum is then dropped into 93.1 g of water at 20.3°c. calculate the final temperature of the metal and water mixture, assuming no heat loss to the surroundings.
Chemistry
1 answer:
kow [346]3 years ago
7 0
when (M*C*ΔT)Al + (M*C*ΔT)Fe = - (M*C*ΔT)w

when water is gaining heat and Al& Fe losing heat

when M(Al) = 5 g

C(Al) = 0.89 

ΔT = 100 - Tf

and when M(Fe) = 10 g

C(Fe) = 0.45 

ΔT= 100 - Tf

and Mw = 93.1 g

Cw = 4.181

ΔT = Tf - 20.3

by substitution:

∴ 5 * 0.89 * ( Tf-100) + 10 * 0.45 * ( Tf - 100) = 93.1 * 4.181 * (Tf-20.3)

∴ Tf = 18.4 °C
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Kindly refer the attachment for complete reaction and products.

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Sodium tert-butoxide would approach from the less hindered side that is through the primary centre and hence would lead to the formation of 1-butene .The major product formed in this reaction would be 1-butene .

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The major product formed does not follow the zaitsev rule of forming a more substituted alkene as sodium tert-butoxide cannot approach to abstract the secondary proton due to steric hindrance.

5 0
4 years ago
Henry finds an element that is light blue, breaks easily in his hand, and does not reflect light. How should Henry classify the
pshichka [43]

Answer:

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b. a metalloid

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