Answer: the coefficient of volume expansion of glass = 0.86/(1000 * 52) = 0.00001654 per degree.
Explanation:
Original volume of mercury = 1000 cm3.
The final volume of mercury considering its volume expansion quotient = 1000 + 1000*(1.8*10^-4 *52) = 1000 + 9.36 = 1009.36 cm^3
Considering the glass as a non expanding substance, the complete excess volume of 9.36 cm3 of mercury should have overflown the container, but due to the expansion of glass, the capacity of mercury containment increases and so a lesser amount of mercury flows out.
The amount of mercury that actually flowed out = 8.50 cm3.
So, the expansion of the glass container = 9.36-8.50 = 0.86 cm3.
Using the formula for coefficient of expansion,
coefficient of volume expansion of glass = 0.86/(1000 * 52) = 0.00001654 per degree.
The molecular formula will be a multiple of the empirical CH2O. One unit of CH2O has a mass of 12+2*1+16 = 30 g. This means that if our compound has a molecular mass of 180 g/mol, we can divide 180 / 30 = 6 units, and our compound has 6 units of CH2O. This means that its molecular formula is C6H12O6.
The arrangement of particles that make up an ionic compound would be an ionic lattice type of crystal arrangement. An ionic lattice type of structure will be formed due to many of the ionic bonds formed between the oppositely charged ions of the metal and nonmetal.
Answer:
(1) addition of HBr to 2-methyl-2-pentene
Explanation:
In this case, we will have the formation of a <u>carbocation</u> for each molecule. For molecule 1 we will have a <u>tertiary carbocation</u> and for molecule 2 we will have a <u>secondary carbocation</u>.
Therefore the <u>most stable carbocation</u> is the one produced by the 2-methyl-2-pentene. So, this molecule would react faster than 4-methyl-1-pentene. (See figure)
NO is the limiting reagent and 4.34 g is the amount of the excess reagent that remains after the reaction is complete
<h3>What is a limiting reagent?</h3>
The reactant that is entirely used up in a reaction is called as limiting reagent.
The reaction:
→ 
Moles of nitrogen monoxide
Molecular weight:
=30g/mol



Moles of hydrogen
Molecular weight:
=30g/mol



Hydrogen gas is in excess.
NO is the limiting reagent.
The amount of the excess reagent remains after the reaction is complete.
(2.9 mol- 0.73 mol NO x
) x 
4.34 g
Learn more about limiting reagents here:
brainly.com/question/26905271
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