Answer : The enthalpy of the reaction = -1839.6 KJ
Solution : Given,
= -520.0 KJ/mole
= -1699.8 KJ/mole
The balanced chemical reaction is,
Formula used :
We know that the standard enthalpy of formation of the element is equal to Zero.
Therefore, the enthalpy of formation of (Mn) and (Al) is equal to zero.
Now, put all the values in above formula, we get
= (-3399.6) + (1560)
= -1839.6 KJ
Answer:
No.of moles of C is , n = mass/molar mass = 75.46 g / 12 (g/mol) = 6.3 moles No.of moles of H is , n' = mass/molar mass = 4.43 g / 1.0(g/mol) = 4.43 moles No.of moles of O is , n'' = mass/molar mass = 20.10 g / 16(g/mol) =1.25 moles Ratio to the no.of moles of C,H& O is 6.3 : 4.43 : 1.25 In the simple integer ratio is ( 6.3/1.25) : ( 4.43/1.25) : (1.25/1.25) 5.04 :3.5 : 1
Explanation:
Answer: 1:4.69
Explanation:
The ratio can be expressed as:
Ua/Ub= √(Mb/Ma)
Where Ua/Ub is the ratio of velocity of hydrogen to carbon dioxide and Ma is the molecular mass of hydrogen gas= 2
Mb is the molecular mass of CO2 = 44
Therefore
Ua/Ub= √(44/2)
Ua/Ub = 4.69
Therefore the ratio of velocity of hydrogen gas to carbon dioxide = 1:4.69
which implies hydogen is about 4.69 times faster than carbon dioxide.
Answer:
The air molecules that are surrounding the metal will speed up, and the molecules in the metal will slow down.
Explanation:
As molecules heat they begin to move faster. The heat from the metal plate will make the molecules at room temperature move faster. And the room temperature makes the hot place cool, making the hot plate molecules slow down.
To visualize this, you can use this link
https://phet.colorado.edu/en/simulations/gas-properties
click the play button to activate the activity
Answer:
0.56 liters
Explanation:
First we <u>convert 0.80 grams of O₂ into moles</u>, using its molar mass:
- 0.80 g ÷ 32 g/mol = 0.025 mol
At STP, 1 mol of any given mass occupies 22.4 L. With that information in mind we <u>calculate the volume that 0.025 moles of O₂ gas would occupy</u>:
- 0.025 mol * 22.4 L/mol = 0.56 L
Thus the answer is 0.56 liters.