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
The model of the atom has dramatically changed over many many years.We learned atoms make up different substances and are the smallest particles of matter, which have subatomic particles that are very small portions of matter. At first scientist only thought there were electrons which are negatively charged.
A. The molecules of solids are close together and compact, liquids are spread out but not too far apart, and gas molecules are really far apart.
B. Increase in temperature causes pressure to go up. Decrease in temperature cause pressure to go down
Unburned hydrocarbon on reacting with oxygen undergoes combustion reaction. However, the activation energy of this reaction is significantly high. When a catalyst like Pd is added to the reaction system, it provides active sites for the reaction to occur. It acts are a heterogeneous catalyst. It is pertinent of note that catalyst is refereed as heterogeneous, when it exist in different phase as compared to reactant and products. In present case, reactants and products are in gas phase, while catalyst is in solid phase. Due to availability of larger surface area at active site of Pd, activation energy of reaction decreases and decrease in activation energy favors higher reaction rates.
