Answer:
The correct answer is "-268.667°C".
Explanation:
Given:
Temperature,
= 4.483 K (below)
Now,
The formula of temperature conversion will be:
⇒ 
By putting the values, we get
⇒ 
⇒ 
Thus the above is the correct answer.
Answer:
Explanation:
Volume of silver cube = 2.42³ = 14.17 cm³
mass of silver cube = volume x density
= 14.17 x 10.49 = 148.64 gm
Volume of gold cube = 2.75³ = 20.8 cm³
mass of gold cube = 20.8 x 19.3 = 401.44 gm
specific heat of silver and gold are .24 and .129 J /g°C
mass of 112 mL water = 112 g
Heat absorbed = heat lost = mass x specific heat x temperature fall or rise
Heat lost by metals
= 148.64 x .24 x ( 85.4 -T) + 401.44 x .129 x ( 85.4 - T )
= (35.67 + 51.78 ) x ( 85.4 - T )
87.45 x ( 85.4 - T )
= 7468.23 - 87.45 T
Heat gained by water
= 112 x 1 x ( T - 20.5 )
= 112 T - 2296
Heat lost = heat gained
7468.23 - 87.45 T = 112 T - 2296
199.45 T = 9764.23
T = 48.95° C
Water goes into the sky by condensation I think
Answer:
3.0 moles Al₂O₃
Explanation:
We do not know which of the reactants is the limiting reactant. Therefore, you need to convert both of the given mole values into the product. This can be done using the mole-to-mole ratio made up of the balanced equation coefficients.
4 Al + 3 O₂ -----> 2 Al₂O₃
6.0 moles Al 2 moles Al₂O₃
---------------------- x ------------------------- = 3.0 moles Al₂O₃
4 moles Al
4.0 moles O₂ 2 moles Al₂O₃
---------------------- x ------------------------- = 2.7 moles Al₂O₃
3 moles O₂
As you can see, O₂ produces the smaller amount of product. This means O₂ is the limiting reactant. Remember, the limiting reactant is the reactant which runs out before the other reactant(s) are completely reacted. As such, the actual amount of Al₂O₃ produced is 2.7 moles.
However, since this problem is directly addressing how much Al₂O₃ is produced from Al, the answer you most likely are looking for is 3.0 moles Al₂O₃.
