The given mass of cobalt chloride hydrate = 2.055 g
A sample of cobalt chloride hydrate was heated to drive off waters of hydration and the anhydrate was weighed.
The mass of anhydrous cobalt chloride = 1.121 g anhydrate.
The mass of water lost during heating = 2.055 g - 1.121 g = 0.934 g
Converting mass of water of hydration present in the hydrate to moles using molar mass:
Mass of water = 0.934 g
Molar mass of water = 18.0 g/mol
Moles of water = 
Answer:
SO₃(g) + H₂O(l) → H₂SO₄(aq)
Explanation:
The<em> molecular formula for the involved species</em> are:
- Sulfur trioxide = SO₃. ("trioxide" indicates the presence of 3 oxygen atoms).
With the above information in mind we can proceed to write the reaction equation:
- SO₃(g) + H₂O(l) → H₂SO₄(aq)
Answer:
0.482 ×10²³ molecules
Explanation:
Given data:
Volume of gas = 2.5 L
Temperature of gas = 50°C (50+273 = 323 k)
Pressure of gas = 650 mmHg (650/760 =0.86 atm)
Molecules of N₂= ?
Solution:
PV= nRT
n = PV/RT
n = 0.86 atm × 2.5 L /0.0821 atm. mol⁻¹. k⁻¹. L × 323 k
n = 2.15 atm. L /26.52 atm. mol⁻¹.L
n = 0.08 mol
Number of moles of N₂ are 0.08 mol.
Number of molecules:
one mole = 6.022 ×10²³ molecules
0.08×6.022 ×10²³ = 0.482 ×10²³ molecules
<span>The equation you used is KE=hv-hv0, where h=6.63*10^-34 (constant). You multiply h by 1.5*10^15. Multiply h by the threshold freq of cesium (from part A). Subtract the second answer from the first answer, and you get the kinetic energy. Hope this helps.</span>
Answer:
Vapour pressure of cyclohexane at 50°C is 490torr
Vapour pressure of benzene at 50°C is 90torr
Explanation:
Using Raoult's law, pressure of a solution is defined by the sum of the product sbetween mole fraction of both solvents and pressure of pure solvents.
In the first solution:
<em>(1)</em>
For the second equation:
<em>(2)</em>
Replacing (2) in (1):
-122.5torr = -0.250P°A
<em>Vapour pressure of cyclohexane at 50°C is 490torr</em>
And for benzene:
<em>Vapour pressure of benzene at 50°C is 90torr</em>
