Answer:
26.2g = Mass of water in the calorimeter
Explanation:
The heat absorbed for the water is equal to the heat released for the metal. Based on the equation:
Q = m*C*ΔT
<em>Where Q is heat, m is the mass of the sample, C is specific heat of the material and ΔT is change in temperature</em>
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Replacing we can write:
13.9g * 0.449J/g°C * (54.2°C-15.6°C) = m(H₂O) * 4.184J/g°C * (15.6°C-13.4°C)
240.9J = m(H₂O) * 9.2J/g
<h3>26.2g = Mass of water in the calorimeter</h3>
You start by finding the mol of each
59.9g C x (mol C / 12.01 g C) = 4.98 mol C
8.06g H x (mol H / 1.00 g H) = 8.06 mol H
32.0 g O x (mol O / 16.0 g O) = 2 mol O
So when you set it up you have
C4.98 H8 O2
You divide each by the smallest mol. The smallest mol is 2
C2 H4 O2.5
However you can’t have half a mol in the empirical formula. If the value ends in 0.5, you multiply everything by 2
You’re left with
C2H8O5
The EMPIRICAL formula for lucite is C2H8O5
Note empirical is not the same as chemical formula.
Potassium carbonate<span> (K</span>2CO3<span>) is a white salt, </span>soluble in water<span> (</span>insoluble<span> in ethanol) which forms a strongly alkaline solution. </span>
Answer:
<h3>B. CH3CH2CH2CH2CH2C1</h3>
Explanation:
I HOPE IT HELPS :)
The plant makes 3.50 × 10⁷ bearings per day
<em>Pounds of bearings</em> = 1 da × (4355 lb bearings/5 da) = 871.0 lb bearings
<em>Grams of bearings</em> = 871.0 lb bearings × (453.6 g bearings/1 lb bearings)
= 3.951 × 10⁵ g bearings
<em>No. of bearings</em> = 3.951 × 10⁵ g bearings × (1 bearing/0.0113 g bearing)
= 3.50 × 10⁷ bearings
