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The temperatures of the gases will not be equal.
- The ideal gas law specifies the mechanical behavior of ideal gases. It is capable of calculating the volume of gases produced or consumed.
- In chemical equations, this equation is widely used to convert between volumes and molar quantities.
- The ideal gas law states that the pressure, temperature, and volume of gas are related to each other
∵V is the same for both
∴T= same for both.
When n increases, T decreases, so, n is 1 for hydrogen gas and 0.5 for oxygen gas ,implying that oxygen gas will have more temperature than hydrogen gas since its number of moles are lower than hydrogen gas.
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Answer:- 29.6 moles of carbon.
Solution:- We have been given with 3.7 moles of and asked to calculate the moles of C.
Looking at the formula of the compound, there are 8 carbons in it means 1 mol of he compound has 8 moles of C. So, if we multiply the given moles of the compound by 8 then we get the moles of C.
= 29.6 mol C
Hence. there are 29.6 moles of C in 3.7 moles of .
First a balanced reaction equation must be established:
→ 
Now if mass of aluminum = 145 g
the moles of aluminum = (MASS) ÷ (MOLAR MASS) = 145 g ÷ 30 g/mol
= 4.83 mols
Now the mole ratio of Al : O₂ based on the equation is 4 : 3
[4Al + 3 O₂ → 2 Al₂O₃]
∴ if moles of Al = 4.83 moles
then moles of O₂ = (4.83 mol ÷ 4) × 3
= 3.63 mol (to 2 sig. fig.)
Thus it can be concluded that 3.63 moles of oxygen is needed to react completely with 145 g of aluminum.
Now if mass of aluminum = 145 g
the moles of aluminum = (MASS) ÷ (MOLAR MASS) = 145 g ÷ 30 g/mol
= 4.83 mols
Now the mole ratio of Al : O₂ based on the equation is 4 : 3
[4Al + 3 O₂ → 2 Al₂O₃]
∴ if moles of Al = 4.83 moles
then moles of O₂ = (4.83 mol ÷ 4) × 3
= 3.63 mol (to 2 sig. fig.)
Thus it can be concluded that 3.63 moles of oxygen is needed to react completely with 145 g of aluminum.