The chemical name of Hc2h3o2 is Acetic Acid.
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Answer:
HgO (empirical formula)
Explanation:
4.08 - 3.78 = 0.3g (oxygen)
0.02 : 0.02
0.02/0.02 : 0.02/0.02
1 : 1 (ratio)
HgO ( empirical formula)
2HgO ----> 2Hg + O2 ( your equation correct)
Answer:
0.120M is the concentration of the solution
Explanation:
<em>Assuming the mass of sodium nitrate dissolved was 2.552g</em>
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Molar concentration is an unit of concentration widely used in chemsitry defined as the moles of solute (In this case NaNO3) in 1L of solution.
To find this question we must find the moles of NaNO3 in 2.552g. With this mass and the volume (250mL = 0.250L) we can find molar concentration as follows:
<em>Moles NaNO3 -Molar mass: 84.99g/mol-</em>
2.552g * (1mol / 84.99g) = 0.0300 moles NaNO3
<em>Molar concentration:</em>
0.0300 moles NaNO3 / 0.250L =
<h3>0.120M is the concentration of the solution</h3>
Answers and Explanation:
a)- The chemical equation for the corresponden equilibrium of Ka1 is:
2. HNO2(aq)⇌H+(aq)+NO−2
Because Ka1 correspond to a dissociation equilibrium. Nitrous acid (HNO₂) losses a proton (H⁺) and gives the monovalent anion NO₂⁻.
b)- The relation between Ka and the free energy change (ΔG) is given by the following equation:
ΔG= ΔGº + RT ln Q
Where T is the temperature (T= 25ºc= 298 K) and R is the gases constant (8.314 J/K.mol)
At the equilibrium: ΔG=0 and Q= Ka. So, we can calculate ΔGº by introducing the value of Ka:
⇒ 0 = ΔGº + RT ln Ka
ΔGº= - RT ln Ka
ΔGº= -8.314 J/K.mol x 298 K x ln (4.5 10⁻⁴)
ΔGº= 19092.8 J/mol
c)- According to the previous demonstation, at equilibrium ΔG= 0.
d)- In a non-equilibrium condition, we have Q which is calculated with the concentrations of products and reactions in a non equilibrium state:
ΔG= ΔGº + RT ln Q
Q= ((H⁺) (NO₂⁻))/(HNO₂)
Q= ( (5.9 10⁻² M) x (6.7 10⁻⁴ M) ) / (0.21 M)
Q= 1.88 10⁻⁴
We know that ΔGº= 19092.8 J/mol, so:
ΔG= ΔGº + RT ln Q
ΔG= 19092.8 J/mol + (8.314 J/K.mol x 298 K x ln (1.88 10⁻⁴)
ΔG= -2162.4 J/mol
Notice that ΔG<0, so the process is spontaneous in that direction.
