Missing in your question:
Picture (1)
when its an open- tube manometer and the h = 52 cm.
when the pressure of the atmosphere is equal the pressure of the gas plus the pressure from the mercury column 52 Cm so, we can get the pressure of the gas from this formula:
P(atm) = P(gas) + height (Hg)
∴P(gas) = P(atm) - height (Hg)
= 0.975 - (520/760)
= 0.29 atm
Note: I have divided 520 mm Hg by 760 to convert it to atm
Picture (2)
The pressure of the gas is the pressure experts by the column of mercury and when we have the Height (Hg)= 67mm
So the pressure of the gas =P(atm) + Height (Hg)
= 0.975 + (67/ 760) = 1.06 atm
Picture (3)
As the tube is closed SO here the pressure of the gas is equal the height of the mercury column, and when we have the height (Hg) = 103 mm. so, we can get the P(gas) from this formula:
P(gas) = Height(Hg)
= (103/760) = 0.136 atm
Answer: D
Explanation:
Chlorine is in group 7 or (VII) in Roman numerals, which means it has 7 balance electrons. It only needs one electron to become stable, hence it is next to the noble gases
The answer is 0.975 L
Volume = mol/Molarity
We have molarity (0.788 M) and we need mol and volume. Let's first calculate number of moles of CaCl2 in 85.3 g:
Molar mass of CaCl2 is sum of atomic masses of Ca and Cl:
Mr(CaCl2) = Ar(Ca) + 2Ar(Cl) = 40 + 2 * 35.45 = 40 + 70.9 = 110.9 g/mol
So, if 110.9 g are in 1 mol, 85.3 g will be in x mol:
110.9 g : 1 mole = 85.3 g : x
x = 85.3 g * 1 mole / 110.9
x = 0.769 moles
Now, calculate the volume:
V = 0.769/0.788
V = 0.975 L
Answer:
150g
Explanation:
Assuming they are ideal gases at the same temperature and pressure, equal moles of gasses have equal volume. IN this case, if we have 10g of hydrogen gas, that is 5 moles of H2 gas. That means 5 moles og NO2 will occupy the same volume which is 5*(14.0 + 16.0*2) = 230 g
The original concentration : 800%
<h3>Further explanation</h3>
Given
Diluted 1/40
Final concentration 20%
Required
The original concentration
Solution
Dilution is the process of adding solvent to get a more dilute solution.
The moles(n) before and after dilution are the same.
Can be formulated :
n₁ = n₂
M₁.V₁ = M₂.V₂
diluted 1/40(dilution factor)⇒ V₁/V₂=1/40⇒V₂/V₁=40
M₂ = 0.2(20%)
Input the value for M₁ :
M₁=M₂ x (V₂/V₁)
M₁ = 0.2 x 40
M₂ = 8(800%)
=
