The element chlorine has an atomic weight of 35.5 and consists of two stable isotopes chlorine-35 and chlorine-37. The isotope chlorine-35 has a mass of 35.0 amu and a percent natural abundance of 75.5 %. The isotope chlorine-37 has a percent natural abundance of 24.5 %. What is the mass of chlorine-37? amu

The answer is 37amu

Explanation:

We are given:

Chlorine-35: atomic mass  =35 amu  and percent abundance  =75.5% ≡ 0.755

Chlorine-37: atomic mass  = x amu  and percent abundance  =24.5% ≡ 0.25

From table, relative atomic mass of Chlorine is 35.5

⇒(0.755 × 35) + (24.5 × x) = 35.5

26.402 + 24.5x = 35.5

∴ x = 9.075 ÷ 24.5 = 0.37 ≡ 37.

∴ mass of Chlorine 37 = 37amu

The mass of Chlorine-37 is 37.04amu.

Explanation:

To calculate the atomic mass of a element with two isotopes we use:

(%abundance * mass)of chlorine-35 + (%abundance * mass) of chlorine-37 = atomic mass

Since the atomic mass has been given, we can equate the values as shown below and calculate:

(75.5/100 * 35) + (24.5/100 * Y) = 35.5

(0.755 * 35) + (24.5Y / 100) =35.5

26.425 + 24.5Y / 100 = 5.5

24.5Y/ 100 = 35.5 - 26.425

24.5Y/100 = 9.075

24.5Y = 9.075 * 100

24.5Y = 907.5

Y = 907.5 / 24.5

Y = 37.04amu

The mass of Chlorine-37 is 37.04amu.

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

if you're asking what comes out, it's steam.