Answer:
if you are working with hazardous materials.
Explanation:
A properly operating and correctly used fume hood can reduce or eliminate exposure to volatile liquids, dusts, and mists. It is advisable to use a laboratory hood when working with all hazardous substances.
Answer:
1 mol
Explanation:
Using the general gas law equation as follows:
PV = nRT
Where;
P = pressure (atm)
V = volume (L)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/molK)
T = temperature (K)
According to the provided information in the question;
V = 22.4L
T = 273K
P = 1 atm
R = 0.0821 Latm/molK
n = ?
Using PV = nRT
n = PV/RT
n = (1 × 22.4) ÷ (0.0821 × 273)
n = 22.4 ÷ 22.4
n = 1mol
Answer:
I think the answer is option B
CH3 is the empirical formula for the compound.
A sample of a compound is determined to have 1.17g of Carbon and 0.287 g of hydrogen.
The number of atom or moles in the compound is
1.17 g C X 1 mol of C / 12.011 g C = 0.097411 mol of C.
0.287 g H x 1 mol of H / 1 g H = 0.28474 mol H.
This compound contains 0.097411 mol of carbon and 0.28474 mol of Hydrogen.
So we can represent the compound with the formula C0.974H0.284.
Subscripts in formulas can be made into whole numbers by multiplying the smaller subscript by the larger subscript.
we can divide 0.284 by 0.0974.
0.284 / 0.0974 = 3.
So here, Carbon is one and hydrogen is 3.
We can write the above formula as a CH3.
Hence the empirical formula for the sample compound is CH3.
For a detailed study of the empirical formula refer given link brainly.com/question/13058832.
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