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Artyom0805 [142]
3 years ago
15

A 23.5g aluminum block is warmed to 65.9°C and plunged into an insulated beaker containing 55.0g water initially at 22.3°C. The

aluminum and the water are allowed to come to thermal equilibrium.
Assuming that no heat is lost, what is the final temperature of the water and aluminum?
Chemistry
1 answer:
atroni [7]3 years ago
4 0

Answer:

25.97oC

Explanation:

Heat lost by aluminum = heat gained by water

M(Al) x C(Al) x [ Temp(Al) – Temp(Al+H2O) ] = M(H2O) x C(H2O) x [ Temp(Al+H2O) – Temp(H2O) ]

Where M(Al) = 23.5g, C(Al) = specific heat capacity of aluminum = 0.900J/goC, Temp(Al) = 65.9oC, Temp(Al+H2O)= temperature of water and aluminum at equilibrium = ?, M(H2O) = 55.0g, C(H2O)= specific heat capacity of liquid water = 4.186J/goC

Let Temp(Al+H2O) = X

23.5 x 0.900 x (65.9-X) = 55.0 x 4.186 x (X-22.3)

21.15(65.9-X) = 230.23(X-22.3)

1393.785 - 21.15X = 230.23X – 5134.129

230.23X + 21.15X = 1393.785 + 5134.129

251.38X = 6527.909

X = 6527.909/251.38

X = 25.97oC

So, the final temperature of the water and aluminum is = 25.97oC

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Magnesium sulfate heptahydrate is heated until all the water is driven off. The sample loses 11.80 grams upon heating. What was
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Hello!

The initial mass of Magnesium Sulfate Heptahydrate (MgSO₄·7H₂O) is 23,08 g

The chemical reaction for the dehydrating of Magnesium Sulfate Heptahydrate (MgSO₄·7H₂O) is the following:

MgSO₄·7H₂O(s) + Δ → MgSO₄(s) + 7H₂O(g)

We know that the sample loses 11,80 g upon heating. That mass is the mass of Water that is released as vapor. Knowing that piece of information, we can apply the following conversion factor to go from the mass of water to the moles of water and back to the mass of the original compound (mi).

mi=11,80gH_2O* \frac{1 mol H_2O}{18gH_2O}* \frac{1molMgSO_4.7H_2O}{7molH_2O}* \frac{246,47 g MgSO_4.7H_2O}{1 mol MgSO_4.7H_2O}  \\  \\ mi=23,08gMgSO_4.7H_2O

Have a nice day!

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3 years ago
What would be the mass of a 33.5dm3 sample of O2 at STP?
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Answer:

• One mole of oxygen is equivalent to 16 grams.

→ But at STP, 22.4 dm³ are occupied by 1 mole.

{ \tt{22.4 \:  {dm}^{3}   \: \dashrightarrow \: 16 \: grams}} \\  { \tt{33.5 \:  {dm}^{3}  \:  \dashrightarrow \: ( \frac{33.5 \times 16}{22.4} ) \: grams}} \\  \\  \dashrightarrow \: { \boxed{ \tt{23.94 \: g \approx \: 24 \: grams}}}

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