Answer:
Compound X has a molar mass of 316.25 g*mol^-1 and the following composition:
element & mass %
phosphorus & 39.18%
sulfur & 60.82%
Write the molecular formula of X.
Explanation:
The given molecule of phosphorus and sulfur has molar mass --- 316.25 g.
Empirical formula calculation:
element: phosphorus sulfur
co9mposition: 39.185% 60.82%
divide with
atomic mass: 39.185/31.0 g/mol 60.82/32.0g/mol
=1.26mol 1.90mol
smallest mole ratio: 1.26mol/1.26mol =1 1.90mol/1.26 mol =1.50
multiply with 2: 2 3
Hence, the empirical formula is:
P2S3.
Mass of empirical formula is:
158.0g/mol
Given, molecule has molar mass --- 316.25 g/mol
Hence, the ratio is:
316.25g/mol/158.0 =2
Hence, the molecular formula of the compound is :
2 x (P2S3)
=
Here we will use the general formula of Nernst equation:
Ecell = E°Cell - [(RT/nF)] *㏑Q
when E cell is cell potential at non - standard state conditions
E°Cell is standard state cell potential = - 0.87 V
and R is a constant = 8.314 J/mol K
and T is the temperature in Kelvin = 73 + 273 = 346 K
and F is Faraday's constant = 96485 C/mole
and n is the number of moles of electron transferred in the reaction=2
and Q is the reaction quotient for the reaction
SO42-2(aq) + 4H+(aq) +2Br-(aq) ↔ Br2(aq) + SO2(g) +2H2O(l)
so by substitution :
0 = -0.87 - [(8.314*346K)/(2* 96485)*㏑Q → solve for Q
∴ Q = 4.5 x 10^-26
Answer:
I think the teaspoon will cool down faster.
Explanation:
The teaspoon of water will cool down faster because it is a smaller amount of water so the entire [drop] will absorb heat energy faster.
Because they can't get trapped in.
Answer:
4.96 mol/dm³
Explanation:
From the question,
Mass of NaCl that dissolved in 0.5L of water = 500-346.8 = 153.2 g.
Therefore, 145.2(1/0.5)g of NaCl will dissolve in 1 L of water
mass of NaCl that will dissolve in 1 L of water = 290.4 g/dm³
Molar mass of NaCl = 58.5 g/mol.
Solubility is the amount of substance in mol that will dissolve in 1 L or 1 dm³ Solution.
solubility in (mol/dm³) = solubility in (g/dm³)/molar mass.
solubility in (mol/dm³) = 290.4/58.5
solubility in (mol/dm³) = 4.96 mol/dm³
