Hey there!
Molarity = 4.8/850 x 1000 =5.647mol/lit or 5.65M
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<span>Charles' law says "at a constant pressure, the volume of a fixed amount of gas is directly proportional to its absolute temperature".
V </span>α T
Where V is the volume and T is the temperature in Kelvin of the gas. We can use this for two situations as,
V₁/T₁ = V₂/T₂
V₁ = 2.00 L
T₁ = 40.0 ⁰C = 313 K
V₂ = ?
T₂ = 30.0 ⁰C = 303 K
By applying the formula,
2.00 L / 313 K = V₂ / 303 K
V₂ = (2.00 L / 313 K) x 303 K
V₂ = 1.94 L
Hence, the volume of the balloon at 30.0 ⁰C is 1.94 L
Answer:
4.823 x 10^-19 J
Explanation:
Energy is calculated by E = hv where h - Planck's constant in joule.s
v - frequency.
in this particular question the wave length is 4.12 x 10^-7 m. to exhaustively use this we need a relation between wave length & frequency. c=wv where C is approximately 3 x 10^8m/s
-v = c/w = 3x10^8m/s / 4.12 x 10^-7m = 7.28 x 10^14 Hz or 1/sec
now we can simply use Planck's constant in E=hv =
(6.626 x 10^-34) x (7.28 x 10^14Hz) = 4.823 x 10^-19 J.
Answer:
2KCl + F₂ → 2KF + Cl₂
Explanation:
Law of conservation of mass:
According to the law of conservation mass, mass can neither be created nor destroyed in a chemical equation.
This law was given by French chemist Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.
2KCl + F₂ → 2KF + Cl₂
In this equation mass of reactant and product is equal. There are 2 potassium 2 chlorine and fluorine atoms on both side of equation it means mass remain conserved.
All other options are incorrect because mass is not conserved.
Mg₂ + LiBr ---> LiMg + Br
In this equation mass of magnesium is more on reactant side.
Na +O₂ ---> Na₂O
In this equation there is more oxygen and less sodium on reactant side while there is more sodium and less oxygen on product side.
H₂O ---> H₂ + O₂
In this equation there is less oxygen on reactant side while more oxygen on product side.
Half reaction :
Oxidation
Cu ⇒ Cu²⁺ + 2e⁻
Reduction
2Ag⁺+ 2e⁻ ⇒2Ag
<h3>Further explanation</h3>
Given
Reaction
Cu+2AgNO₃⇒Cu(NO₃)₂+2Ag
Required
Oxidation and reduction half-reactions
Solution
Oxidation is an increase in oxidation number, while reduction is a decrease in oxidation number.
Cu ⇒ Cu²⁺ + 2e⁻
0 to +2
2Ag⁺+ 2e⁻ ⇒2Ag
+1 to 0