According to the law of conservation of mass, the amount of BARIUM present of the reactants is the same as the amount present in the products (the precipitate).

(11.21 g BaSO4) / (233.4 g/mol BaSO4) = 0.0480 mol BaSO4 and original barium salt

(10.0 g) / (0.0480 mol) = 208.3 g/mol

So it must have been BaCl2, because the molar mass of Barium is 137 which leave 71 grams left. Since Barium is a +2 charge, it means the atom next to it must be twice. Chlorine mass is 35, which twice is 71

7 0

3 years ago

What is fiber and fabric

Sidana [21]

Answer:

Fibre , it is a long thin strand or something like thread of material.

Fabric, it is a cloth material made by weaving or knitting threads together.

6 0

2 years ago

Read 2 more answers

A 4.36 g sample of an unknown alkali metal hydroxide is dissolved in 100.0 ml of water. an acid-base indicator is added and the

Fofino [41]

Answer:

(a) $102.6g/mol$

(b) Rubidium

Explanation:

Hello,

This titration is carried out by assuming that the volume of base doesn't have a significant change when the mass is added, thus, we state the following data a apply the down below formula to compute the molarity of the base solution:

$V_{base}=0.1L; M_{acid}=2.5M, V_{acid}=0.017L\\V_{base}M_{base}=V_{acid}M_{acid}$

Solving for the molarity of base we've got:

$M_{base}=\frac{M_{acid}*V_{acid}}{V_{base}}=\frac{2.50M*0.017L}{0.1L} =0.425M=0.425mol/L$

Now, we can compute the moles of the base as:

$n_{base}=0.425mol/L*0.1L=0.0425mol$

(a) Now, one divides the provided mass over the previously computed moles to get the molecular mass of the unknown base:

$\frac{4.36g}{0.0425mol} =102.6g/mol$

(b) Subtracting the atomic mass of oxygen and hydrogen, the metal's atomic mass turns out into:

$102.6g/mol-16g/mol-1g/mol=85.6g/mol$

So, that atomic mass dovetails to the Rubidium's atomic mass.

Best regards.

8 0

3 years ago