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 combustion reaction of propane would be expressed as:
C3H8 + 5O2 = 3CO2 + 4H2O
To determine the mass of water that is produced from the given amount of propane, we use the mass of propane and the relation of the substances from the balanced reaction. We do as follows:
moles propane = 22 g C3H8 ( 1 mol / 44.1 g ) = 0.50 mol C3H8
moles H2O = 0.50 mol C3H8 ( 4 mol H2O / 1 mol C3H8) = 2 mol H2O
mass H2O = 2 mol H2O ( 18.02 g / 1 mol ) = 36.04 g H2O
Therefore, the mass of water that is produced from 22 grams of propane would be 36.04 g.
i believe it would be B '' tetrahedral compound ''
Explanation:
hexaaqua iron (III) trinitrate