Answer:
CH₂
Explanation:
From the question given above, the following data were obtained:
Mass of compound = 1 g
Mass of CO₂ = 3.14 g
Mass of H₂O = 1.29 g
Empirical formula =?
Next, we shall determine the mass of Carbon and hydrogen present in the compound. This can be obtained as follow:
For Carbon, C:
Mass of CO₂ = 3.14 g
Molar mass of CO₂ = 12 + (2×16)
= 12 + 32
= 44 g/mol
Molar mass of C = 12 g/mol
Mass of C =?
Mass of C = molar mass of C/ Molar mass of CO₂ × Mass of CO₂
Mass of C = 12/44 × 3.14
Mass of C = 0.86 g
For hydrogen, H:
Mass of C = 0.86 g
Mass of compound = 1 g
Mass of H =?
Mass of H = (Mass of compound) – (mass of C)
Mass of H = 1 – 0.86
Mass of H = 0.14 g
Finally, we shall determine the empirical formula of the cyclopropane. This can be obtained as follow:
Mass of C = 0.86 g
Mass of H = 0.14 g
Divide by their molar mass
C = 0.86 / 12 = 0.07
H = 0.14 / 1 = 0.14
Divide by the smallest
C = 0.07 / 0.07 = 1
H = 0.14 / 0.07 = 2
Thus, the empirical formula of cyclopropane is CH₂
Answer:
Explanation:
Pv = n RT
P =7.65, T= 310K, for glucose n = moles= weight/molar mass= weight/180, R is gas constant
7.65* V =W/180 *0.0821 * 310
w/v= 7.65*180/0.0821*310 = 1377/25.45 = 54.10 =5.4%
The answer would be, D. because ive taken the test.
Answer:
groups
Explanation:
periods are left and right groups are up and down
Explanation:
<h3>Atomic Number and Mass</h3>
Each element has its own unique properties. Each contains a different number of protons and neutrons, giving it its own atomic number and mass number. The atomic number of an element is equal to the number of protons that element contains.