To solve for the empirical formula, we write first all the data.
Given:
Compound 1: 76 wt% Ru and 24wt% O
Compound 2: 61.2 wt% Ru and 38.8 wt% O
Required: Empirical Formula of Compound 1
Solution:
Assume total mass of the compound is 100 g
Solving for Compound 1,
76 g Ru x <u>1 mol Ru </u> = 0.75195 mol Ru
101.07 g Ru
24 g O x <u>1 mol O </u> = 1.5 mol O
16 g O
Then, divide each mole with the smallest number of moles calculated
Ru = 0.75195 mol/0.75195 mol = 1
O = 1.5 mol/0.75195 mol = 2
Therefore, the empirical formula for Compound 1 is RuO2.
<em>ANSWER: RuO2</em>
Only C is correct.
A will produce either an methylbromide + alcohol
B will produce alcohol and alkyl bromide as well
C cyclic ether when reacted with HBr will only produce 1 product which has alcohol group (-OH group) on one end and Bromide group on the other end
Mass of the water is 2.63 g.
<u>Explanation:</u>
Mass of the water, m = ? g
Temperature, ΔT = 15 °C
Heat absorbed, q = 165 J
Specific heat capacity, c = 4.18 J / g °C
q = m × c × ΔT
Now, we have to find the mass of the water by rewriting the above equation as,
m = 
Now Plugin the above values in the equation as,
m = 
= 2.63 g
So the mass of the water is found as 2.63 g.
Answer:
1.32 moles.
Explanation:
From the question given above, the following data were obtained:
Density of Al = 2.70 g/cm³
Volume of Al = 13.2 cm³
Number of mole of Al =.?
Next, we shall determine the mass of Al.
This can be obtained as follow:
Density of Al = 2.70 g/cm³
Volume of Al = 13.2 cm³
Mass of Al =?
Density = mass / volume
2.7 = mass of Al / 13.2
Cross multiply
Mass of Al = 2.7 × 13.2
Mass of Al = 35.64 g
Finally, we shall determine the number of mole of Al. This can be obtained as follow:
Mass of Al = 35.64 g
Molar mass of Al = 27 g/mol
Number of mole of Al =?
Mole = mass / molar mass
Number of mole of Al = 35.64 / 27
Number of mole of Al = 1.32 moles
Thus, 1.32 moles of aluminum are present in the block of the metal.
is this a multiple choice question?
