Answer:
4.43L is final volume of the ballon
Explanation:
Avogadro's law of ideal gases states that <em>equal volumes of gases, at the same temperature and pressure, have the same number of molecules</em>.
The formula is:

Where V and n are volume and moles of the gas in initial and final conditions.
If the initial conditions are 0.0145 moles and 2.54L and final amount of moles is 0.0253moles, final volume is:

V₂ = <em>4.43L is final volume of the ballon</em>
Answer:
Ka = 4.76108
Explanation:
- CO(g) + 2H2(g) ↔ CH3OH(g)
∴ Keq = [CH3OH(g)] / [H2(g)]²[CO(g)]
[ ]initial change [ ]eq
CO(g) 0.27 M 0.27 - x 0.27 - x
H2(g) 0.49 M 0.49 - x 0.49 - x
CH3OH(g) 0 0 + x x = 0.11 M
replacing in Ka:
⇒ Ka = ( x ) / (0.49 - x)²(0.27 - x)
⇒ Ka = (0.11) / (0.49 - 0.11)² (0.27 - 0.11)
⇒ Ka = (0.11) / (0.38)²(0.16)
⇒ Ka = 4.76108
The structure of tofacitinib with explicit carbon and hydrogen is the first attached picture.
The number of carbon atoms in tofacitinib are 16.
The molecular formula of tofacitinib is
.
The structure of tofacitinib with all the filled hydrogen atoms is the second attached picture.
Explanation:
Mg(s) + Cr(C2H3O2)3 (aq)
Overall, balanced molecular equation
Mg(s) + Cr(C2H3O2)3(aq) --> Mg(C2H3O2)3(aq) + Cr(s)
To identify if an element has been reduced or oxidized, the oxidation number is observed in both the reactant and product phase.
An increase in oxidation number denotes that the element has been oxidized.
A decrease in oxidation number denotes that the element has been reduced.
Oxidation number of Mg:
Reactant - 0
Product - +3
Oxidation number of Cr:
Reactant - +3
Product - 0
Note: C2H3O2 is actually acetate ion; CH3COO- The oxidatioon number of C, H and O do not change.
Oxidized : Mg
Reduced : Cr
Answer : The volume of oxygen occupy at 173° would be, 703.2 mL
Explanation :
Charles' Law : It states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.
Mathematically,

where,
are the initial volume and temperature of the gas.
are the final volume and temperature of the gas.
We are given:

Now put all the given values in above equation, we get:

Therefore, the volume of oxygen occupy at 173° would be, 703.2 mL