Answer:
The initial volume of Ne gas is 261mL
Explanation:
This question can be answered using Ideal Gas Equation;
However, the following are the given parameters
Initial Pressure = 654mmHg
Finial Pressure = 345mmHg
Final Volume = 495mL
Required
Initial Volume?
The question says that Temperature is constant;
This implies that, we'll make use of Boyle's law ideal gas equation which states;
![P_1V_1 = P_2V_2](https://tex.z-dn.net/?f=P_1V_1%20%3D%20P_2V_2)
Where
represent the initial pressure
represent the final pressure
represent the initial temperature
represent the final temperature
![P_1 = 654mmHg\\P_2 = 345mmHg\\V_2 = 495mL](https://tex.z-dn.net/?f=P_1%20%3D%20654mmHg%5C%5CP_2%20%3D%20345mmHg%5C%5CV_2%20%3D%20495mL)
Substitute these values in the formula above;
![654 * V_1 = 345 * 495](https://tex.z-dn.net/?f=654%20%2A%20V_1%20%3D%20345%20%2A%20495)
![654V_1 = 170775](https://tex.z-dn.net/?f=654V_1%20%3D%20170775)
Divide both sides by 654
![\frac{654V_1}{654} = \frac{170775}{654}](https://tex.z-dn.net/?f=%5Cfrac%7B654V_1%7D%7B654%7D%20%3D%20%5Cfrac%7B170775%7D%7B654%7D)
![V_1 = \frac{170775}{654}](https://tex.z-dn.net/?f=V_1%20%3D%20%5Cfrac%7B170775%7D%7B654%7D)
![V_1 = 261.123853211](https://tex.z-dn.net/?f=V_1%20%3D%20261.123853211)
(Approximated)
<em>The initial volume of Ne gas is 261mL</em>
Answer:How many moles of NH3 are produced from 9 moles of hydrogen? a mot H, 12 mol NH3 = 16 mol NH₂). 3 motta. 3. How many moles of nitrogen are needed to make 6 moles of NH3? ... a. masses, in grams, of all reactants and products.
Explanation:
MORE POWER
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so given Molar mass= 32 g/mol
molar mass= (empirical formula) n
32 = (14x1 + 2x1) n
32 = 16 n , so n= 2
so, molecular formula= N2H4
Answer: D.Aluminium Oxide 0.10, Magnesium Oxide 0.50
Explanation:
Number of moles of NaOH= number of moles × volume
Number of moles= 100/1000 × 2 = 0.2 moles
Since;
2 moles of NaOH yield 1 mole of Al2O3
0.2 moles of NaOH will yield 0.2 × 1/2 = 0.1 moles of Al2O3.
Number of moles of HCl= 800/1000 × 2 = 1.6 moles
If 1 mole of Al2O3 requires 6 moles of HCl
0.1 moles of Al2O3 requires 0.1 × 6 = 0.6 moles of HCl.
Number of moles of HCl left after reaction with Al2O3 = 1.6- 0.6 = 1 mole
This leftover reacts with MgO
But;
1 mole of MgO reacts with 2 moles of HCl
x moles of MgO reacts with 1 mole of HCl
Thus; x= 0.5 moles of MgO