Answer to Question #265245 in Chemistry for garey

Question

If there are 2.50 moles of an unknown atomic element and the mass of this amount is 60.78 g, the unknown element must be:

A - Sodium

B - Magnesium

C - Aluminium

D - Silicon

The mass of a sample of 0.124 mol of tin(II) fluoride is _____g. Remember to provide your answer using significant digits.

In 50g of ammonia, there are _____ mol. Remember to use significant digits for your answer

The molar mass of iron (III) sulfate is ______ g/mol.

The expected products after combining sodium iodide with lead(II) nitrate would likely be:

A - NaNO3(aq) + PbI(s)

B - Na(NO3)2(aq) + PbI2(s)

C - Na(NO3)2(aq) + PbI(s)

D - NaNO3(aq) + PbI2(s)

Convert 540000000 to scientific notation in the form of a.b X 10c. Enter the digits a, b, and c in this order for your 3-digit numerical answer.

When performing the calculation below, the final answer would be rounded to 10, according to the rounding and significant digit rules.

   (4.36 X 2.2) + 0.8580

True or false?

Answer 1

Solution:

(1):

The molar mass of he unknown element = (60.78 g) / (2.50 mol) = 24.312 g/mol

It is magnesium (Mg)

Answer (1): B - Magnesium

(2):

tin(II) fluoride = SnF₂

The molar mass of SnF₂ is 156.69 g/mol

Hence,

(0.124 mol) × (156.69 g mol^-1) = 19.42956 g SnF₂ = 19.4 g SnF₂

Answer (2): 19.4 g

(3):

ammonia = NH₃

The molar mass of NH₃ is 17.031 g/mol

Hence,

(50 g NH₃) × (1 mol NH₃ / 17.031 g NH₃) = 2.9356 mol NH₃ = 3 mol NH₃

Answer (3): 3 mol

(4):

iron(III) sulfate = Fe₂(SO₄)₃

The molar mass of iron(III) sulfate is 399.88 g/mol

Answer (4): 399.88 g/mol

(5):

sodium iodide = NaI

lead(II) nitrate = Pb(NO₃)₂

Balanced chemical equation: 2NaI(aq) + Pb(NO₃)₂(aq) → 2NaNO₃(aq) + PbI₂(s)

Answer (5): D - NaNO₃(aq) + PbI₂(s)

(6):

540000000 = 5.4×10⁸

a.b×10^c

(a = 5; b = 4; c = 8) = 548

Answer (6): 548

(7):

(4.36 × 2.2) + 0.8580 = 10.45 = 10.5 - True

Answer (7): True