Answer:
A and C
Explanation:
The <u>monochlorination</u> is a reaction in which we have to add only one Cl to the molecule. In this case we only have <u>two types of carbons</u>. <u>Primary</u> (the methyl groups) and the <u>tertiary</u> (the carbons in the middle).
Therefore we only have 2 type of <u>isomers</u>. The first one in which the Cl would bond to the primary carbons, the <u>primary alkyl halide</u> (1-chloro-2,3-dimethylbutane) and the second one in which the Cl would bond to the tertiary carbons, the <u>tertiary alkyl halide</u> (2-chloro-2,3 -dimethylbutane).
See the figure
Bromothymol blue is acid base indicator which has pH intervals 6 - 7.6 and its two colors are yellow in acid and blue in basic.
The pH of CaO is about 11.5 to 12.5 this will make the indicator has blue color at the beginning.
By bubbling CO₂ to CaO this leads to formation of CaCO₃ which also basic (pH about 9.5 - 10) so the color of indicator remains blue
No color change of the indicator will observed (remains blue)
Finding the volume of regular geometric objects uses standard formulas. The volume of a box equals length times width times height, for example. Not every object, however, fits a formula. Water displacement is best only used for irregular shaped objects.
The mass in grams of 8.94×10²² formula units of CuF₂ is 15.07 g
<h3>Avogadro's hypothesis </h3>
6.02×10²³ formula units = 1 mole of CuF₂
But
1 mole of CuF₂ = 101.5 g
Thus,
6.02×10²³ formula units = 101.5 g of CuF₂
<h3>How to determine the mass of 8.94×10²² formula units of CuF₂</h3>
6.02×10²³ formula units = 101.5 g of CuF₂
Therefore,
8.94×10²² formula units = (8.94×10²² × 101.5 ) / 6.02×10²³
8.94×10²² formula units = 15.07 g of CuF₂
Thus, 8.94×10²² formula units are present in 15.07 g of CuF₂
Learn more about Avogadro's number:
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We can use the ideal gas law equation to find the number of moles in the gas
PV = nRTwhere P - pressure - 1.2 atm x 101 325 Pa/atm = 121 590 Pa
V - volume - 3.94 x 10⁻³ m³
n - number of moles
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature - 15 °C + 273 = 288 K
substituting the values in the equation
121 590 Pa x 3.94 x 10⁻³ m³ = n x 8.314 Jmol⁻¹K⁻¹ x 288 K
n = 0.200 mol
molar mass of gas is = mass / number of moles
molar mass = 12.8 g / 0.200 mol = 64 g/mol
molar mass of gas is 64 g/mol
