Answer:
MEANS:
1 = 97.7
2 = 74.3
3 = 50
4 = 30
5 = 13
UNCERTAINTY:
gimme a sec, ill put it in the comments under this
Answer:
ΔH = -20kJ
Explanation:
The enthalpy of formation of a compound is defined as the change of enthalpy during the formation of 1 mole of the substance from its constituent elements. For H₂S(g) the reaction that describes this process is:
H₂(g) + S(g) → H₂S(g)
Using Hess's law, it is possible to sum the enthalpies of several reactions to obtain the change in enthalpy of a particular reaction thus:
<em>(1) </em>H₂S(g) + ³/₂O₂(g) → SO₂(g) + H₂O(g) ΔH = -519 kJ
<em>(2) </em>H₂(g) + ¹/₂O₂(g) → H₂O(g) ΔH = -242 kJ
<em>(3) </em>S(g) + O₂(g) → SO₂(g) ΔH = -297 kJ
The sum of -(1) + (2) + (3) gives:
<em>-(1) </em>SO₂(g) + H₂O(g) → H₂S(g) + ³/₂O₂(g) ΔH = +519 kJ
<em>(2) </em>H₂(g) + ¹/₂O₂(g) → H₂O(g) ΔH = -242 kJ
<em>(3) </em>S(g) + O₂(g) → SO₂(g) ΔH = -297 kJ
<em>-(1) + (2) + (3): </em><em>H₂(g) + S(g) → H₂S(g) </em>
<em>ΔH =</em> +519kJ - 242kJ - 297kJ = <em>-20 kJ</em>
<em />
I hope it helps!
Answer:
a
. eight tenths of her cookies
Explanation:
Let the total number of Lakesha's cookies be represented by x.
So that;
She gave three tenths to Bailey = 
of x
= 
She gave five tenths to Helen = 
of x
= 
Fraction of Lakesha's cookies given away = +
= 
= 
Thus, the fraction of cookies given away by Lakesha is 
.
Answer:
Initial concentration of HI is 5 mol/L.
The concentration of HI after 
is 0.00345 mol/L.
Explanation:
Rate Law: ![k[HI]^2 ](https://tex.z-dn.net/?f=k%5BHI%5D%5E2%0A)
Rate constant of the reaction = k = 
Order of the reaction = 2
Initial rate of reaction = 
Initial concentration of HI =![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
![1.6\times 10^{-7} mol/L s=(6.4\times 10^{-9} L/mol s)[HI]^2](https://tex.z-dn.net/?f=1.6%5Ctimes%2010%5E%7B-7%7D%20mol%2FL%20s%3D%286.4%5Ctimes%2010%5E%7B-9%7D%20L%2Fmol%20s%29%5BHI%5D%5E2)
![[A_o]=5 mol/L](https://tex.z-dn.net/?f=%5BA_o%5D%3D5%20mol%2FL)
Final concentration of HI after t = [A]
t =
Integrated rate law for second order kinetics is given by:
![\frac{1}{[A]}=kt+\frac{1}{[A_o]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%3Dkt%2B%5Cfrac%7B1%7D%7B%5BA_o%5D%7D)
![\frac{1}{[A]}=6.4\times 10^{-9} L/mol s\times 4.53\times 10^{10} s+\frac{1}{[5 mol/L]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%3D6.4%5Ctimes%2010%5E%7B-9%7D%20L%2Fmol%20s%5Ctimes%204.53%5Ctimes%2010%5E%7B10%7D%20s%2B%5Cfrac%7B1%7D%7B%5B5%20mol%2FL%5D%7D)
![[A]=0.00345 mol/L](https://tex.z-dn.net/?f=%5BA%5D%3D0.00345%20mol%2FL)
The concentration of HI after is 0.00345 mol/L.
Answer is: <span>concentration of NOCl is 3.52 M.
</span>
Balanced chemical reaction: 2NOCl(g) ⇄ 2NO(g) + Cl₂<span>(g).
Kc = 8.0.
</span>[NOCl] = 1.00 M; equilibrium concentration.
[NO] = x.
[Cl₂] = x/2; equilibrium concentration of chlorine.<span>
Kc = </span>[Cl₂] ·[NO]² / [NOCl].
8.00 = x/2 · x² / 1.
x³/2 = 8.
x = ∛16.
x = 2.52 M.
co(NOCl) = [NOCl] + x.
co(NOCl) = 1.00 M + 2.52 M.
co(NOCl) = 3.52 M; the initial concentration of NOCl.
