Due to pyro-electric properties and molarity Li2SO4 cannot be written as lithium sulfur oxide.
Explanation:
Lithium sulfate is a white inorganic salt with the formula Li2SO4. It is the lithium salt of sulfuric acid.
Lithium sulfate has water solubility, though it does not follow the usual trend of solubility versus temperature — its solubility in water decreases with increasing temperature, as its dissolution is an exothermic process. This property is shared with few inorganic compounds, such as the lanthanoid sulfates.
Lithium sulfate has pyro-electric properties. When aqueous lithium sulfate is heated, the electrical conductivity also increases. The molarity of lithium sulfate also plays a role in the electrical conductivity optimal conductivity is achieved at 2M and then decreases.
Lithium sulfate has a rapid gastrointestinal absorption rate and complete following oral administration of tablets or the liquid form.
1) we calculate the molar mass of He (helium) and Kr (Krypton).
atomic mass (He)=4 u
atomic mas (Kr)=83.8 u
Therefore the molar mass will be:
molar mass(He)=4 g/mol
molar mass(Kr)=83.8 g/mol.
1) We can find the next equation:
mass=molar mass x number of moles.
x=number of moles of helium
y=number of moles of helium.
(4 g/mol) x +(83.8 g/mol)y=103.75 g
Therefore, we have the next equation:
(1)
4x+83.8y=103.75
2) We can find other equation:
We have 30% helium atoms and 70% Kryptum atoms, therefore we have 30% Helium moles and 70% of Krypton moles.
1 mol is always 6.022 * 10²³ atoms or molecules, (in this case atoms).
Then:
x=number of moles of helium
y=number of moles of helium.
(x+y)=number of moles of our sample.
x=30% of (x+y)
Therefore, we have this other equation:
(2)
x=0.3(x+y)
With the equations(1) and (2), we have the next system of equations:
4x+83.8y=103.75
x=0.3(x+y) ⇒ x=0.3x+0.3y ⇒ x-0.3x=0.3y ⇒ 0.7 x=0.3y ⇒ x=0.3y/0.7
⇒x=3y/7
We solve this system of equations by substitution method.
x=3y/7
4(3y/7)+83.8y=103.75
lower common multiple)7
12y+586.6y=726.25
598.6y=726.25
y=1.21
x=3y/7=3(1.21)/7=0.52
We have 0.52 moles of helium and 1.21 moles of Krypton.
1 mol=6.022 * 10²³ atoms
Total number of particles=(6.022 *10²³ atoms /1 mol) (number of moles of He+ number of moles of Kr).
Total number of particles=6.022 * 10²³ (0.52+1.21)=6.022 * 10²³ (1.73)=
=1.044 * 10²⁴ atoms.
Answer: The sample have 1.044 * 10²⁴ atoms.
For this problem we can use half-life formula and radioactive decay formula.
Half-life formula,
t1/2 = ln 2 / λ
where, t1/2 is half-life and λ is radioactive decay constant.
t1/2 = 8.04 days
Hence,
8.04 days = ln 2 / λ
λ = ln 2 / 8.04 days
Radioactive decay law,
Nt = No e∧(-λt)
where, Nt is amount of compound at t time, No is amount of compound at t = 0 time, t is time taken to decay and λ is radioactive decay constant.
Nt = ?
No = 1.53 mg
λ = ln 2 / 8.04 days = 0.693 / 8.04 days
t = 13.0 days
By substituting,
Nt = 1.53 mg e∧((-0.693/8.04 days) x 13.0 days))
Nt = 0.4989 mg = 0.0.499 mg
Hence, mass of remaining sample after 13.0 days = 0.499 mg
The answer is "e"
Answer:
3.3 mol
Explanation:
5.2 mol * 14.9 L / 23.5 L
Answer: 0.0164 molar concentration of hydrochloric acid in the resulting solution.
Explanation:
1) Molarity of 0.250 L HCl solution : 0.0328 M
Moles of HCl in 0.250 L solution = 0.0082 moles
2) Molarity of 0.100 L NaOH solution : 0.0245 M
Moles of NaOH in 0.100 L solution = 0.00245 moles
3) Concentration of hydrochloric acid in the resulting solution.
0.00245 moles of NaOH will neutralize 0.00245 moles of HCl out of 0.0082 moles of HCl.
Now the new volume of the solution = 0.100 L +0.250 L = 0.350 L
Moles of HCl left un-neutralized = 0.0082 moles - 0.00245 moles = 0.00575 moles
Molarity of HCl left un-neutralized :
0.0164 molar concentration of hydrochloric acid in the resulting solution.
