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
Answer:
190.4g
Explanation:
1.6mol of KBr (119.002g KBr/1 mol) = 190.4g
since you want to find grams, take the molar mass of KBr (119.002) per 1 mol and use it as your conversion factor (119.002g KBr/1 mol) which will then cancel out mols and leave you with grams.
Answer : The volume of 
required to neutralize is, 340 mL
Explanation :
To calculate the volume of base (NaOH), we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is 
are the n-factor, molarity and volume of base which is NaOH.
We are given:
Putting values in above equation, we get:
Hence, the volume of required to neutralize is, 340 mL
Magnesium is classified as a metal. Magnesium has two valence electrons
Answer:
2.64 × 10⁶ g
Explanation:
We can find the mass of air using the ideal gas equation.
where,
P is the pressure (P = 1.00 atm)
V is the volume (V = 2.95 × 10⁶ L)
n is the number of moles
R is the ideal gas constant (0.08206atm.L/mol.K)
T is the absolute temperature (121°C + 273 = 394 K)
m is the mass
M is the molar mass (28.09 g/mol)
