Answer:
i. n = 5
ii. ΔE = 7.61 ×
KJ/mole
Explanation:
1. ΔE = (1/λ) = -2.178 ×
(
-
)
(1/434 ×
) = -2.178 ×
(
)
⇒ 434 ×
= (1/-2.178 ×
)
But,
= 2
434 ×
= (1/2.178 ×
)
434 ×
× 2.178 ×
= 
⇒
= 5
Therefore, the initial energy level where transition occurred is from 5.
2. ΔE = hf
= (hc) ÷ λ
= (6.626 × 10−34 × 3.0 ×
) ÷ (434 ×
)
= (1.9878 ×
) ÷ (434 ×
)
= 4.58 ×
J
= 4.58 ×
KJ
But 1 mole = 6.02×
, then;
energy in KJ/mole = (4.58 ×
KJ) ÷ (6.02×
)
= 7.61 ×
KJ/mole
The answer is 3.39 mol.
<span>Avogadro's number is the number of molecules in 1 mol of substance.
</span><span>6.02 × 10²³ molecules per 1 mol.
</span>2.04 × 10²⁴<span> molecules per x.
</span>6.02 × 10²³ molecules : 1 mol = 2.04 × 10²⁴ molecules : x
x = 2.04 × 10²⁴ molecules * 1 mol : 6.02 × 10²³ molecules
x = 2.04/ 6.02 × 10²⁴⁻²³ mol
x = 0.339 × 10 mol
<span>x = 3.39 mol
</span>
Answer:
The answer is 465.6 mg of MgI₂ to be added.
Explanation:
We find the mole of ion I⁻ in the final solution
C = n/V -> n = C x V = 0.2577 (L) x 0.1 (mol/L) = 0.02577 mol
But in the initial solution, there was 0.087 M KI, which can be converted into mole same as above calculation, equal to 0.02242 mol.
So we need to add an addition amount of 0.02577 - 0.02242 = 0.00335 mol of I⁻. But each molecule of MgI₂ yields two ions of I⁻, so we need to divide 0.00335 by 2 to find the mole of MgI₂, which then is 0.001675 mol.
Hence, the weight of MgI₂ must be added is
Weight of MgI₂ = 0.001675 mol x 278 g/mol = 0.4656 g = 465.6 mg
<span>ideal gas law is: PV = nRT
P = pressure (torr) = 889 torr
V = volume (Liters) = 11.8 L
n = moles of gas = 0.444 mol
R = gas constant = 62.4 (L * torr / mol * k)
solve for T (in kelvin)
T = PV/nR
T = (889*11.8)/(.444*62.4)
T = 378.6 K
convert to C (subtract 273)
T = 105.6 deg C</span>
Answer:
0.66g of water
Explanation:
Molar heat of vaporization of any substance is defined as the heat necessary to vaporize 1 mole of the substance.
If heat of vaporization of water is 40.79kJ/mol and you add 1.50kJ, the moles you vaporize are:
1.50kJ × (1mol / 40.79kJ) = 0.0368 moles of water.
As molar mass of water is 18.01g/mol, mass of water that can be vaporized are:
0.0368 moles × (18.01g / mol) = <em>0.66g of water</em>