The answer is 34.1 mL.
Solution:
Assuming ideal behavior of gases, we can use the universal gas law equation
P1V1/T1 = P2V2/T2
The terms with subscripts of one represent the given initial values while for terms with subscripts of two represent the standard states which is the final condition.
At STP, P2 is 760.0torr and T2 is 0°C or 273.15K. Substituting the values to the ideal gas expression, we can now calculate for the volume V2 of the gas at STP:
(800.0torr * 34.2mL) / 288.15K = (760.0torr * V2) / 273.15K
V2 = (800.0torr * 34.2mL * 273.15K) / (288.15K * 760.0torr)
V2 = 34.1 mL
The name of the compound by using the <u>IUPAC nomenclature of organic compounds</u> is 1 -octene. The correct option is the last option - 1-octene.
<h3>Nomenclature of Organic compounds</h3>
From the question, we are to determine the name of the given molecule.
To name the compound, we will follow the IUPAC rules.
Some of IUPAC rules are
- Find the longest continuous carbon chain. Determine the root name for this parent chain.
- For Alkenes (organic compounds with double bond), number the chain of carbons that includes the C=C so that the C=C has the lower position number. Change “ane” to “ene” and assign a position number to the first carbon of the C=C.
The given compound has 8 carbons and a double bond. The root name of the compound is octane.
By <u>IUPAC rules</u>, the compound is an <u>Octene</u>.
Since the double bond is between carbon-1 and carbon-2. The compound becomes 1-octene.
Hence, the name of the compound by using the <u>IUPAC nomenclature of organic compounds</u> is 1 -octene. The correct option is the last option - 1-octene.
Learn more on Nomenclature of Organic compounds here: brainly.com/question/26754333
The diagram for the compound is attached below.
Crystallization metamorphism sedimentation and erosion
0.040 mol / dm³. (2 sig. fig.)
<h3>Explanation</h3>
in this question acts as a weak base. As seen in the equation in the question,
produces
rather than
when it dissolves in water. The concentration of
will likely be more useful than that of
for the calculations here.
Finding the value of
from pH:
Assume that
,
.
.
Solve for
:
![\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D_%5Ctext%7Bequilibrium%7D%5Ccdot%5B%28%5Ctext%7BCH%7D_3%29_3%5Ctext%7BNH%7D%5E%7B%2B%7D%5D_%5Ctext%7Bequilibrium%7D%7D%7B%5B%28%5Ctext%7BCH%7D_3%29_3%5Ctext%7BN%7D%5D_%5Ctext%7Bequilibrium%7D%7D%20%3D%20%5Ctext%7BK%7D_b%20%3D%201.58%5Ctimes%2010%5E%7B-3%7D)
Note that water isn't part of this expression.
The value of Kb is quite small. The change in
is nearly negligible once it dissolves. In other words,
.
Also, for each mole of
produced, one mole of
was also produced. The solution started with a small amount of either species. As a result,
.
,
,
.