Correct answer is
D. 4.032 g of Hydrogen atoms contain same number of atoms as that contained by 2 moles of hydrogen molecules. Solution is below,
Answer:
The final volume is 990.8 L
Explanation:
Let calculate the moles of gas in the first situation:
P . V = n . R . T
1.20 atm . 2.90 L = n . 0.082 . 293K
(1.20 atm . 2.90 L) / (0.082 . 293K) = 0.145 moles
This are the same moles in the second situation:
P . V = n . R . T
0.003atm . V = 0.145 moles . 0.082 . 250K
V = (0.145 moles . 0.082 . 250K) / 0.003atm
V = 990.8 L
Answer:
- [HOCl] = 0.00909 mol/liter
- [H₂O] = 0.03901 mol/liter
- [Cl₂O] = 0.02351 mol/liter
Explanation:
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<u>1. Chemical reaction:</u>

<u>2. Initial concentrations:</u>
i) 1.3 g H₂O
- Number of moles = 1.3g / (18.015g/mol) = 0.07216 mol
- Molarity, M = 0.07216 mol / 1.5 liter = 0.0481 mol/liter
ii) 2.2 g Cl₂O
- Number of moles = 2.2 g/ (67.45 g/mol) = 0.0326 mol
- Molarity = 0.0326mol / 1.5 liter = 0.0217 mol/liter
<u>3. ICE (Initial, Change, Equilibrium) table</u>

I 0.0481 0.0326 0
C -x -x +x
E 0.0481-x 0.0326-x x
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<u>4. Equilibrium expression</u>
![K_c=\dfrac{[HOCl]^2}{[H_2O].[Cl_2O]}](https://tex.z-dn.net/?f=K_c%3D%5Cdfrac%7B%5BHOCl%5D%5E2%7D%7B%5BH_2O%5D.%5BCl_2O%5D%7D)

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<u>5. Solve:</u>

Use the quadatic formula:

The positive result is x = 0.00909
Thus the concentrations are:
- [HOCl] = 0.00909 mol/liter
- [H₂O] = 0.0481 - 0.00909 = 0.03901 mol/liter
- [Cl₂O] = 0.0326 - 0.00909 = 0.02351 mol/liter
Answer:
Protons, neutrons, and electrons
Explanation:
Both protons and neutrons have a mass of 1 amu and are found in the nucleus. However, protons have a charge of +1, and neutrons are uncharged. Electrons have a mass of approximately 0 amu, orbit the nucleus, and have a charge of -1.
So first find the moles of the H₂SO₄: Mass = Moles x RFM
so mass = 5.25 x 98 = 514.5g of <span>H₂SO₄</span>
so to find how many Liters of solution use:
Volume = Density x Grams of solute (per kg +1000)
density = 1.266 x 514.5 +1000 = 1917.357kg/l
now use equation: Conc. = Moles / Volume of solution to find the conc.
Conc. = 5.25 x 1917.357 = 4.39Mol⁻¹
Hope that helps