Answers:
<span>Answer 1: 10.03 g of siver metal can be formed.</span>
Answer 2: 3.11 g of Co are left over.
Work:
1) Unbalanced chemical equation (given):
<span>Co + AgNO3 → Co(NO3)2 + Ag
2) Balanced chemical equation
</span>
<span>Co + 2AgNO3 → Co(NO3)2 + 2Ag
3) mole ratios
1 mol Co : 2 mole AgNO3 : 1 mol Co(NO3)2 : 2 mol Ag
4) Convert the masses in grams of the reactants into number of moles
4.1) 5.85 grams of Co
# moles = mass in grams / atomic mass
atomic mass of Co = 58.933 g/mol
# moles Co = 5.85 g / 58.933 g/mol = 0.0993 mol
4.2) 15.8 grams of Ag(NO3)
# moles Ag(NO3) = mass in grams / molar mass
molar mass AgNO3 = 169.87 g/mol
# moles Ag(NO3) = 15.8 g / 169.87 g/mol = 0.0930 mol
5) Limiting reactant
Given the mole ratio 1 mol Co : 2 mol Ag(NO3) you can conclude that there is not enough Ag(NO3) to make all the Co react.
That means that Ag(NO3) is the limiting reactant, which means that it will be consumed completely, whilce Co is the excess reactant.
6) Product formed.
Use this proportion:
2 mol Ag(NO3) 0.0930mol Ag(NO3)
--------------------- = ---------------------------
2 mol Ag x
=> x = 0.0930 mol
Convert 0.0930 mol Ag to grams:
mass Ag = # moles * atomic mass = 0.0930 mol * 107.868 g/mol = 10.03 g
Answer 1: 10.03 g of siver metal can be formed.
6) Excess reactant left over
1 mol Co x
----------------------- = ----------------------------
2 mole Ag(NO3) 0.0930 mol Ag(NO3)
=> x = 0.0930 / 2 mol Co = 0.0465 mol Co reacted
Excess = 0.0993 mol - 0.0465 mol = 0.0528 mol
Convert to grams:
0.0528 mol * 58.933 g/mol = 3.11 g
Answer 2: 3.11 g of Co are left over.
</span>
Answer:
The maximum amount of work that can be done by this system is -2.71 kJ/mol
Explanation:
Maximum amount of work denoted change in gibbs free energy 
during the reaction.
Equilibrium concentration of B = 0.357 M
So equilibrium concentration of A = (1-0.357) M = 0.643 M
So equilibrium constant at 253 K, ![K_{eq}= \frac{[B]}{[A]}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%20%5Cfrac%7B%5BB%5D%7D%7B%5BA%5D%7D)
[A] and [B] represent equilibrium concentrations
When concentration of A = 0.867 M then B = (1-0.867) M = 0.133 M
So reaction quotient at this situation, 
We know, 
where R is gas constant and T is temperature in kelvin
Here R is 8.314 J/(mol.K), T is 253 K, Q is 0.153 and 
is 0.555
So, 
= -2710 J/mol
= -2.71 kJ/mol
Answer:
Robert Boyle
Explanation:
En muchos textos se suele considerar a Robert Boyle como el científico que introdujo en la Química un concepto de elemento diferente al empleado por los aristotélicos o por los alquimistas y que sirvió de antecedente al que formulara Lavoisier en 1789.
Answer:
The pH of the solution is 11.48.
Explanation:
The reaction between NaOH and HCl is:
NaOH + HCl → H₂O + NaCl
From the reaction of 3.60x10⁻³ moles of NaOH and 5.95x10⁻⁴ moles of HCl we have that all the HCl will react and some of NaOH will be leftover:
Now, we need to find the concentration of the OH⁻ ions.
![[OH^{-}] = \frac{n_{NaOH}}{V}](https://tex.z-dn.net/?f=%20%5BOH%5E%7B-%7D%5D%20%3D%20%5Cfrac%7Bn_%7BNaOH%7D%7D%7BV%7D%20)
Where V is the volume of the solution = 1.00 L
![[OH^{-}] = \frac{n_{NaOH}}{V} = \frac{3.01 \cdot 10^{-3} moles}{1.00 L} = 3.01 \cdot 10^{-3} mol/L](https://tex.z-dn.net/?f=%20%5BOH%5E%7B-%7D%5D%20%3D%20%5Cfrac%7Bn_%7BNaOH%7D%7D%7BV%7D%20%3D%20%5Cfrac%7B3.01%20%5Ccdot%2010%5E%7B-3%7D%20moles%7D%7B1.00%20L%7D%20%3D%203.01%20%5Ccdot%2010%5E%7B-3%7D%20mol%2FL%20)
Finally, we can calculate the pH of the solution as follows:
![pOH = -log([OH^{-}]) = -log(3.01 \cdot 10^{-3}) = 2.52](https://tex.z-dn.net/?f=%20pOH%20%3D%20-log%28%5BOH%5E%7B-%7D%5D%29%20%3D%20-log%283.01%20%5Ccdot%2010%5E%7B-3%7D%29%20%3D%202.52%20)
Therefore, the pH of the solution is 11.48.
I hope it helps you!
