Answer:
<h2>1.89 atm</h2>
Explanation:
The new volume can be found by using the formula for Boyle's law which is
Since we are finding the new volume
From the question we have
We have the final answer as
<h3>1.89 atm</h3>
Hope this helps you
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.
#SPJ1.
<u>Explanation:</u>
The number of moles that are present in a liter of a solution is called Molarity. Mole is the unit of Molarity.The concentration of the solutes that are present in a solution refers to Osmolarity . The unit is osmol. This helps in observing the movement of water from one side to another side of a semipermeable membrane.
The main property of salts that helps to understand the differentiation between the molarity and osmolarity is the salts Ionization. One mole Na+ and one mole of Cl- is produced by the dissociation of a mole of sodium chloride occurs. The diffusion of water is the osmosis. Here, the one molar sodium chloride solution produces a osmotic pressure which is high than one molar glucose solution and this will not undergo dissociation.
The balanced equation
for the reaction is
CO(g) + 2H₂(g) ⇄ CH₃OH(g)
The given
concentrations are at equilibrium state. Hence we can use them directly in
calculation with the expression for the equilibrium constant, k.
expression for k can be written as
k = [CH₃OH(g)] / [CO(g)] [H₂<span>(g) ]²
</span>[H₂<span>]=0.072 M
[CO]= 0.020M
[CH</span>₃OH]= 0.030 M
From substitution,
k = 0.030
M / 0.020 M x (0.072 M)²
k =
289.35 M⁻²
<span>
Hence, equilibrium constant for the given reaction at 700 K is 289.35 M</span>⁻².
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