Answer:
A) 54.04%
B) 13-karat
Explanation:
A) From the problem we have
<em>1)</em> Mg + Ms = 9.40 g
<em>2)</em> Vg + Vs = 0.675 cm³
Where M stands for mass, V stands for volume, and g and s stand for gold and silver respectively.
We can rewrite the first equation using the density values:
<em>3)</em> Vg * 19.3 g/cm³ + Vs * 10.5 g/cm³ = 9.40
So now we have<em> a system of two equations</em> (2 and 3) <em>with two unknowns</em>:
We <u>express Vg in terms of Vs</u>:
We <u>replace the value of Vg in equation 3</u>:
- Vg * 19.3 + Vs * 10.5 = 9.40
- (0.675-Vs) * 19.3 + Vs * 10.5 = 9.40
- 13.0275 - 19.3Vs + 10.5Vs = 9.40
Now we <u>calculate Vg</u>:
- Vg + 0.412 cm³ = 0.675 cm³
We <u>calculate Mg from Vg</u>:
- 0.263 cm³ * 19.3 g/cm³ = 5.08 g
We calculate the mass percentage of gold:
- 5.08 / 9.40 * 100% = 54.04%
B)
We multiply 24 by the percentage fraction:
- 24 * 54.04/100 = 12.97-karat ≅ 13-karat
We need to increase the concentration of common ion first, in order to promote the common ion effect
<h3>What is the Common ion effect?</h3>
It is an effect that suppresses the dissociation of salt due to the addition of another salt having common ions.
For example, a saturated solution of silver chloride in equilibrium has Ag⁺ and Cl⁻ . Sodium Chloride is added to the solution and has a common ion Cl⁻. As a result, the equilibrium shifts to the left to form more silver chloride. Thus, solubility of AgCl decreases.
The Equilibrium law states that if a process is in equilibrium and is subjected to a change
- in temperature,
- pressure,
- the concentration of reactant or product,
then the equilibrium shifts in a particular direction, according to the condition.
Thus, an increase in the concentration of common ion promotes the common ion effect.
Learn more about common ion effect:
brainly.com/question/23684003
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It's just graphing. puck acceleration is the x axis and swing length is the y axis
Answer:
Concentration of the barium ions = ![[Ba^{2+}] = 0.4654 M](https://tex.z-dn.net/?f=%5BBa%5E%7B2%2B%7D%5D%20%3D%200.4654%20M)
Concentration of the chloride ions = ![[Cl^{-}]=0.9308 M](https://tex.z-dn.net/?f=%5BCl%5E%7B-%7D%5D%3D0.9308%20M%20)
Explanation:
Moles of hydrogen chloride = n
Volume of hydrogen chloride solution = 43.89 mL = 0.04389 L
Molarity of the hydrogen chloride = 0.1355 M
According to reaction, 2 moles of HCl reacts with 1 mole of barium hydroxide.
Then 0.05947 moles of HCl will react with:
barium hydroxide
Moles of barium hydroxide = 0.029735 mol
1 mole of barium hydroxide gives 1 mole of barium ion in an aqueous solution. Then 0.029735 moles of barium hydroxide will give:
Volume of solution after neutralization reaction :
= 20.0 mL + 43.89 mL = 63.89 mL = 0.06389 L
Concentration of the barium ions =![[Ba^{2+}]](https://tex.z-dn.net/?f=%5BBa%5E%7B2%2B%7D%5D)
![[Ba^{2+}]=\frac{0.029735 mol}{0.06389 L}=0.4654 M](https://tex.z-dn.net/?f=%5BBa%5E%7B2%2B%7D%5D%3D%5Cfrac%7B0.029735%20mol%7D%7B0.06389%20L%7D%3D0.4654%20M)
1 mole of barium chloride gives 1 mole of barium ions and 2 moles of chloride ions in an aqueous solution.
Then concentration of chloride ions will be:
![[Cl^-]=2\times [Ba^{2+}]=2\times 0.4654 M=0.9308 M](https://tex.z-dn.net/?f=%5BCl%5E-%5D%3D2%5Ctimes%20%5BBa%5E%7B2%2B%7D%5D%3D2%5Ctimes%200.4654%20M%3D0.9308%20M)
