Answer:
The only thing that will not affect the potential is the adition of solid Sn.
Explanation:
The potencial of a cell is linked to the concentration of the substances involved in the reactions by the equation of Nernst. So a change of one of them would change the cell potential.
The Keq for this reaction is:
![K_{eq}=\frac{[Sn^{2+}]*[H_2]}{[H^+]}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5BSn%5E%7B2%2B%7D%5D%2A%5BH_2%5D%7D%7B%5BH%5E%2B%5D%7D)
<em>Sn is not included because it's in solid state. </em>
As can be seen, changing the concentrations of H2 (increasing the pressure), H+ (lowering the pH) or Sn2+ will affect the potential.
The only thing that will not affect it is the adition of solid Sn.
<span>Ionic compounds form crystal lattices, not molecules. The term formula unit is used to indicate the simplest whole-number ratio of ions in the compound.</span>
Answer:
![\large \boxed{\text{495.5 mg }}](https://tex.z-dn.net/?f=%5Clarge%20%5Cboxed%7B%5Ctext%7B495.5%20mg%20%7D%7D)
Explanation:
This is an example of back titration.
You add a known excess of HCl to the CaCO₃. Some of it reacts with the CaCO₃.
You titrate the left-over HCl.
From the difference you can calculate the amount of CaCO₃.
100.09
CaCO₃ + 2HCl ⟶ CaCl₂ + H₂CO₃
m/g: 1.2450
V/mL: 65.00
c/mol·L⁻¹: 0.4984
NaOH + HCl ⟶ NaCl + H₂O
V/mL: 37.15
c/mol·L⁻¹: 0.2065
1. Total moles of HCl
![\text{Moles of HCl}= \text{65.00 mL HCl} \times \dfrac{\text{0.4984 mmol HCl}}{\text{1 mL HCl}} = \text{32.40 mmol HCl}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20HCl%7D%3D%20%5Ctext%7B65.00%20mL%20HCl%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B0.4984%20mmol%20HCl%7D%7D%7B%5Ctext%7B1%20mL%20HCl%7D%7D%20%3D%20%5Ctext%7B32.40%20mmol%20HCl%7D)
2. Excess moles of HCl
![\text{Moles of HCl} = \text{37.15 mL NaOH} \times \dfrac {\text{0.2065 mmol NaOH}}{\text{1 mL NaOH}} \times \dfrac{\text{1 mmol HCl}}{\text{1 mmol NaOH}}\\\\= \text{7.671 mmol HCl}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20HCl%7D%20%3D%20%5Ctext%7B37.15%20mL%20NaOH%7D%20%5Ctimes%20%5Cdfrac%20%7B%5Ctext%7B0.2065%20mmol%20NaOH%7D%7D%7B%5Ctext%7B1%20mL%20NaOH%7D%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1%20mmol%20HCl%7D%7D%7B%5Ctext%7B1%20mmol%20NaOH%7D%7D%5C%5C%5C%5C%3D%20%5Ctext%7B7.671%20mmol%20HCl%7D)
3. Moles of HCl that reacted
n = (32.40 - 7.671) mmol HCl = 24.72 mmol HCl
4. Moles of CaCO₃
![\text{Moles of CaCO}_{3} = \text{24.72 mmol HCl} \times \dfrac{\text{1 mmol CaCO}_{3}}{\text{2 mmol HCl}} = \text{12.36 mmol CaCO}_{3}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20CaCO%7D_%7B3%7D%20%3D%20%5Ctext%7B24.72%20mmol%20HCl%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1%20mmol%20CaCO%7D_%7B3%7D%7D%7B%5Ctext%7B2%20mmol%20HCl%7D%7D%20%3D%20%5Ctext%7B12.36%20mmol%20CaCO%7D_%7B3%7D)
5. Mass of Ca
![\text{Mass of Ca} = \text{12.36 mmol CaCO}_{3} \times \dfrac{\text{1 mmol Ca}}{\text{ 1 mmol CaCO}_{3}} \times \dfrac{\text{40.08 mg Ca}}{\text{1 mmol Ca}}\\\\= \textbf{495.5 mg Ca}\\\\\text{The calcium content of the tablet is $\large \boxed{\textbf{495.5 mg }}$}](https://tex.z-dn.net/?f=%5Ctext%7BMass%20of%20Ca%7D%20%3D%20%5Ctext%7B12.36%20mmol%20CaCO%7D_%7B3%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1%20mmol%20Ca%7D%7D%7B%5Ctext%7B%201%20mmol%20CaCO%7D_%7B3%7D%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B40.08%20mg%20Ca%7D%7D%7B%5Ctext%7B1%20mmol%20Ca%7D%7D%5C%5C%5C%5C%3D%20%5Ctextbf%7B495.5%20mg%20Ca%7D%5C%5C%5C%5C%5Ctext%7BThe%20calcium%20content%20of%20the%20tablet%20is%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B495.5%20mg%20%7D%7D%24%7D)
Answer:The mole ratio is A:B:A+B
Explanation:when substance reacts according to John Dalton's theory,their combining ratio is often a replica of the combining moles of the reactants and that of the products. So when Amoles of X combines withB moles of Y ,they produce A+B moles of XY to get a balance reaction e.g if 1moles of X2 reacts with 1moles of Y2 to form XY
The balance equation is seen as
X2+Y2_____2XY
Transform boundaries are places where plates slide sideways past each other. At transform boundaries lithosphere is neither created nor destroyed. Many transform boundaries are found on the sea floor, where they connect segments of diverging mid-ocean ridges. California's San Andreas fault is a transform boundary.