PLEASE ITS FOR MY FINAL How would you go about calculating the molarity of the following solution: 19.5 grams of Be(OH)2 is dissolved in enough water to make a 1.5 L solution.

0.302M

Explanation:

Molarity, which is the molar concentration of a solution can be calculated as follows:

Molarity = number of moles (n) ÷ volume (V)

According to this question, 19.5 grams of Be(OH)2 is dissolved in enough water to make a 1.5 L solution.

Using mole = mass/molar mass to calculate the number of moles in 19.5g of Be(OH)2

Molar mass of Be(OH)2 = 9 + (16 + 1)2

= 9 + (17)2

= 9 + 34

= 43g/mol

mole = 19.5/43

mole = 0.453moles

Molarity = n/V

Molarity = 0.453/1.5

Molarity = 0.302M

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.