Answer: The factor that lead to cyclopropane being less stable than the other cycloalkanes is the presence of a RING STRAIN.
Explanation:
In organic chemistry, the end carbon atoms of an open aliphatic chain can join together to form a closed system or ring to form cycloalkanes. Such compounds are known as cyclic compounds. Examples include cyclopropane, cyclobutane, cyclopentane and many among others.
Cyclopropane is less stable than other cycloalkanes mentioned above because of the presence of ring strain in its structural arrangement. The ring strain is the spatial orientation of atoms of the cycloalkane compounds which tend to give off a very high and non favourable energy. The release of heat energy which is stored in the bonds and molecules cause the ring to be UNSTABLE and REACTIVE.
The presence of the ring strain affects mainly the structures and the conformational function of the smaller cycloalkanes. cyclopropane, which is the smallest cycloalkane than the rest mentioned above, contains only 3 carbons with a small ring.
Explanation:
i found this the question is different but I think the situation is same
Answer: 27.09 ppm and 0.003 %.
First, <u>for air pollutants, ppm refers to parts of steam or gas per million parts of contaminated air, which can be expressed as cm³ / m³. </u>Therefore, we must find the volume of CO that represents 35 mg of this gas at a temperature of -30 ° C and a pressure of 0.92 atm.
Note: we consider 35 mg since this is the acceptable hourly average concentration of CO per cubic meter m³ of contaminated air established in the "National Ambient Air Quality Objectives". The volume of these 35 mg of gas will change according to the atmospheric conditions in which they are.
So, according to the <em>law of ideal gases,</em>
PV = nRT
where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the constant gas (0.082057 atm L / mol K)
The moles of CO will be,
n = 35 mg x
x
→ n = 0.00125 mol
We clear V from the equation and substitute P = 0.92 atm and
T = -30 ° C + 273.15 K = 243.15 K
V = 
→ V = 0.0271 L
As 1000 cm³ = 1 L then,
V = 0.0271 L x
= 27.09 cm³
<u>Then the acceptable concentration </u><u>c</u><u> of CO in ppm is,</u>
c = 27 cm³ / m³ = 27 ppm
<u>To express this concentration in percent by volume </u>we must consider that 1 000 000 cm³ = 1 m³ to convert 27.09 cm³ in m³ and multiply the result by 100%:
c = 27.09
x
x 100%
c = 0.003 %
So, <u>the acceptable concentration of CO if the temperature is -30 °C and pressure is 0.92 atm in ppm and as a percent by volume is </u>27.09 ppm and 0.003 %.
Answer:
13.53 kJ
Explanation:
The energy of a gas can be calculated by the equation:
E = (3/2)*n*R*T
Where n is the number of moles, R is the gas constant (8.314 J/mol.K), and T is the temperature.
E = (3/2)*3.5*8.314*310
E = 13,531.035 J
E = 13.53 kJ
Answer:
0.00268 M
Explanation:
To find the new molarity, you need to (1) find the moles of CuSO₄ (via the molarity equation using the beginning molarity and volume) and then (2) find the new molarity (using the moles and combined volume). Your final answer should have 3 sig figs to match the given values.
<u>Step 1:</u>
3.00 mL / 1,000 = 0.00300 L
Molarity = moles / volume (L)
0.0250 M = moles / 0.00300 L
(0.0250 M) x (0.00300 L) = moles
7.50 x 10⁻⁵ = moles
<u>Step 2:</u>
25.0 mL / 1,000 = 0.0250 L
0.0250 L + 0.00300 L = 0.0280 L
Molarity = moles / volume (L)
Molarity = (7.50 x 10⁻⁵ moles) / (0.0280 L)
Molarity = 0.00268 M