We are given with the mass of Arsine (
The mass of arsine is 15g
there is a relation between moles, mass and molar mass of any compound which is
The molar mass of Arsine = atomic mass of As + 3X atomic mass of H
the molar mass of Arsine = 74.92 + 3X 1 = 77.92 g/mol
Let us calculate the moles as
The reaction between K₂SO₄(aq) and SrI₂(aq) produces KI(aq) and SrSO₄(s) as products.
The reaction is
K₂SO₄(aq) + SrI₂(aq) → KI(aq)+ SrSO₄(s)
To balance the equation both side of the reaction should have same number of atoms in each element.
Right hand side of the reaction has 1 K, 1 I, 1 Sr, 1 S and 4 O atoms while 2 K, 2 I, 1 Sr,1 S and 4 O present in left hand side of the reaction.
Hence, number of I atoms and number of K atoms are not balanced.
To balance the K atoms we should add 2 before KI. Then I atoms will be 2 at the right hand side.
Hence, the balanced reaction equation is
K₂SO₄(aq) + SrI₂(aq) → 2KI(aq)+ SrSO₄(s)
Acceleration is the rate of change and is constantly positive
The answers for Acceleration:
m/s^2
Change in velocity over change in time
Velocity is the speed of the object but also involves direction can be both negative and positive
The answers for Velocity:
m/s
Can be positive or negative
Change in displacement over change in time
Both Acceleration and Velocity:
A rate of change
Answer:
A and D are true , while B and F statements are false.
Explanation:
A) True. Since the standard gibbs free energy is
ΔG = ΔG⁰ + RT*ln Q
where Q= [P1]ᵃ.../([R1]ᵇ...) , representing the ratio of the product of concentration of chemical reaction products P and the product of concentration of chemical reaction reactants R
when the system reaches equilibrium ΔG=0 and Q=Keq
0 = ΔG⁰ + RT*ln Q → ΔG⁰ = (-RT*ln Keq)
therefore the first equation also can be expressed as
ΔG = RT*ln (Q/Keq)
thus the standard gibbs free energy can be determined using Keq
B) False. ΔG⁰ represents the change of free energy under standard conditions . Nevertheless , it will give us a clue about the ΔG around the standard conditions .For example if ΔG⁰>>0 then is likely that ΔG>0 ( from the first equation) if the temperature or concentration changes are not very distant from the standard conditions
C) False. From the equation presented
ΔG⁰ = (-RT*ln Keq)
ΔG⁰>0 if Keq<1 and ΔG⁰<0 if Keq>1
for example, for a reversible reaction ΔG⁰ will be <0 for forward or reverse reaction and the ΔG⁰ will be >0 for the other one ( reverse or forward reaction)
D) True. Standard conditions refer to
T= 298 K
pH= 7
P= 1 atm
C= 1 M for all reactants
Water = 55.6 M
Answer:
293.1 mL.
Explanation:
- Boyle's law states that: at a constant temperature the pressure of a given mass of an ideal gas is inversely proportional to its volume.
- It can be expressed as: <em>P₁V₁ = P₂V₂,</em>
P₁ = 546.0 mm Hg, V₁ = 350.0 mL.
P₂ = 652.0 mm Hg, V₂ = ??? mL.
<em>∴ V₂ = (P₁V₁)/(P₂)</em> = (546.0 mm Hg)(350.0 mL) / (652.0 mm Hg) = <em>293.1 mL.</em>
