Answer:
a mixture of two these
Explanation:
The number of isomeric monochlorides depends on the structure and number of equivalent hydrogen atoms in each isomer of pentane.
n-pentane has three different kinds of equivalent hydrogen atoms leading to three isomeric monochlorides formed.
Isopentane has four different types of equivalent hydrogen atoms hence four isomeric monochlorides are formed.
Lastly, neopentane has only one type of equivalent hydrogen atoms that yields one mono chlorination product.
Hence the cylinder must contain a mixture of isopentane and neopentane which yields four and one isomeric monochlorides giving a total of five identifiable monochloride products as stated in the question.
First gap is BY USING MAGNETS
2nd is ELECTROLYSIS
3rd is EVAPORATION
<em>Answer :</em> 72.05 g/mol
<span>
<em>Explanation : </em>
Let's </span>assume that the given gas is an ideal gas. Then we can use ideal gas equation,<span>
PV = nRT<span>
</span>
Where,
P = Pressure of the gas (Pa)
V = volume of the gas (m³)
n = number of moles (mol)
R = Universal gas constant (8.314 J mol</span>⁻¹ K⁻¹)<span>
T = temperature in Kelvin (K)
<span>
The given data for the gas </span></span>is,<span>
P = 777 torr = 103591 Pa
V = </span>125 mL = 125 x 10⁻⁶ m³<span>
T = (</span>126 + 273<span>) = 399 K
R = 8.314 J mol</span>⁻¹ K⁻¹<span>
n = ?
By applying the formula,
103591 Pa x </span>125 x 10⁻⁶ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 399 K<span>
n = 3.90 x 10</span>⁻³<span> mol
</span>Moles (mol) = mass (g) /
molar mass (g/mol)<span>
Mass of the gas = </span><span>0.281 g
</span>Moles of the gas = 3.90 x 10⁻³ mol
<span>Hence,
molar mass of the gas = mass / moles
= 0.281 g / </span>3.90 x 10⁻³ mol
<span> = 72.05 g/mol
</span>
Answer:Rubidium and Strontium
Explanation:
We found cesium, strontium, aluminum, sulfur, chlorine, and fluorine on the periodic table. Cesium is the farthest left and the lowest, while fluorine is the farthest right and the highest, so we know they have the highest metallic character and the lowest metallic character, respectively.