Answer: You multiply and divide when rounding division significant figures
Explanation: Both multiplying and dividing significant figures have the same rule. That rule is, the FINAL ANSWER of a multiplication and division problem should be rounded to the number of significant figures that is the least amount of any figures used in the multiplication or division. Let us demonstrate below.
Answer:
(a) The equilibrium partial pressure of BrCl (g) will be greater than 2.00 atm.
Explanation:
Q is the coefficient of the reaction and is calculated the same of the way of the equilibrium constant, but using the concentrations or partial pressures in any moment of the reaction, so, for the reaction given:
Q = (pBrCl)²/(pBr₂*pCl₂)
Q = 2²/(1x1)
Q = 4
As Q < Kp, the reaction didn't reach the equilibrium, and the value must increase. As we can notice by the equation, Q is directly proportional to the partial pressure of BrCl, so it must increase, and be greater than 2.00 atm in the equilibrium.
The partial pressures of Br₂ and Cl₂ must decrease, so they will be smaller than 1.00 atm. And the total pressure must not change because of the stoichiometry of the reaction: there are 2 moles of the gas reactants for 2 moles of the gas products.
Because is a reversible reaction, it will not go to completion, it will reach an equilibrium, and as discussed above, the partial pressures will change.
Answer:
They can be dated using radioactive carbon. But for rocks older than 50 thousand years they look for layers of igneous rock or volcanic ash above and Bellow the layer of rock/fossil.
Answer:
(i)The mole fractions are :
(ii)
(iii)ΔG = 1.974kJ
Explanation:
The given equation is :
⇄
Let
be the number of moles dissociated per mole of 
Thus ,
<em>The initial number of moles of :</em>
+
⇄
+ 
And finally the number of moles of ![C[tex] is 0.9Thus ,[tex]3\alpha=0.9\\\alpha=0.3[tex]The final number of moles of:[tex]A = 1-2\alpha=1-2*0.3=0.4mol[tex] [tex]B=2(1-\alpha)=2(1-0.3)=1.4mol[tex][tex]D=1+2\alpha=1+2*0.3=1.6mol[tex]Thus , total number of moles are : 0.4+1.4+0.9+1.6=4.3(i)The mole fractions are : [tex]A=\frac{0.4}{4.3} \\=0.0930](https://tex.z-dn.net/?f=C%5Btex%5D%20is%200.9%3C%2Fp%3E%3Cp%3EThus%20%2C%3C%2Fp%3E%3Cp%3E%5Btex%5D3%5Calpha%3D0.9%5C%5C%5Calpha%3D0.3%5Btex%5D%3C%2Fp%3E%3Cp%3E%3Cem%3E%3Cstrong%3EThe%20final%20number%20of%20moles%20of%3A%3C%2Fstrong%3E%3C%2Fem%3E%3C%2Fp%3E%3Cul%3E%3Cli%3E%3Cem%3E%3Cstrong%3E%5Btex%5DA%20%3D%201-2%5Calpha%3D1-2%2A0.3%3D0.4mol%5Btex%5D%20%3C%2Fstrong%3E%3C%2Fem%3E%3C%2Fli%3E%3C%2Ful%3E%3Cul%3E%3Cli%3E%3Cem%3E%3Cstrong%3E%5Btex%5DB%3D2%281-%5Calpha%29%3D2%281-0.3%29%3D1.4mol%5Btex%5D%3C%2Fstrong%3E%3C%2Fem%3E%3C%2Fli%3E%3C%2Ful%3E%3Cul%3E%3Cli%3E%3Cem%3E%3Cstrong%3E%5Btex%5DD%3D1%2B2%5Calpha%3D1%2B2%2A0.3%3D1.6mol%5Btex%5D%3C%2Fstrong%3E%3C%2Fem%3E%3C%2Fli%3E%3C%2Ful%3E%3Cp%3EThus%20%2C%20total%20number%20of%20moles%20are%20%3A%200.4%2B1.4%2B0.9%2B1.6%3D4.3%3C%2Fp%3E%3Cp%3E%3Cstrong%3E%28i%29The%20mole%20fractions%20are%20%3A%20%3C%2Fstrong%3E%3C%2Fp%3E%3Cul%3E%3Cli%3E%3Cstrong%3E%5Btex%5DA%3D%5Cfrac%7B0.4%7D%7B4.3%7D%20%5C%5C%3D0.0930)
(ii)

Where ,
are the partial pressures of A,B,C,D respectively.
Total pressure = 1 bar .
∴
<em>
</em>
<em>
</em>
<em>
</em>
<em>
</em>

(iii)
Δ
ΔG = 
Answer:
0.9 mole of Fe(OH)3.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
Fe(NO3)3 + 3NaOH —> Fe(OH)3 + 3NaNO3
Now, we can determine the moles of iron (III) hydroxide formed from the reaction as follow:
From the balanced equation above,
3 moles of NaOH reacted to produce 1 mole of Fe(OH)3.
Therefore, 2.7 moles of NaOH will react to produce = 2.7/3 = 0.9 mole of Fe(OH)3.
Therefore, 0.9 mole of Fe(OH)3 is produced from the reaction.