The ratio is 1.000 mol of iron to 1.500 mol of oxygen (Fe1O1.5). Finally, multiply the ratio by two to get the smallest possible whole number subscripts while still maintaining the correct iron-to-oxygen ratio: The empirical formula is Fe2O3.

7 0

2 years ago

Read 2 more answers

Calculate the volume of 4.50 mol of SO2 (g) measured at STP. At STP: P = 101.3 kPa, T = 273.15 K A. 87.3 L B. 101 L C. 127 L D.

miss Akunina [59]

Answer:

101L

Explanation:

We use the ideal gas relation

PV = nRT

P = pressure = 101.3KPa

V = volume = ?

n = number of moles = 4.5moles

T = Temperature = 273.15K

R = molar gas constant = 8.314J/mol.k

Rearranging the equation to make V the subject of the formula yields :

V = nRT/P

= ( 4.5 × 8.314 × 273.15) ÷ 101.3

= 10,219.361 ÷ 101.3 = 100.88L which is apprx 101L

8 0

2 years ago

When water was added to a 4.00 gram mixture of potassium oxalate hydrate (molar mass 184.24 g/mol) and calcium hydrate shown bel

Sliva [168]

Answer:

% (COOK)2H2O = 37.826 %

Explanation:

mix: (COOK)2H2O + Ca(OH)2 → CaC2O4 + H2O

∴ mass mix = 4.00 g

∴ mass (CaC2O4)H2O = 1.20 g

∴ Mw (COOK)2H2O = 184.24 g/mol

∴ Mw (CaC2O4)H2O = 146.12 g/mol

∴ r = mol (COOK)2H2O / mol (CaC2O4)H2O = 1

% (COOK)2H2O = (mass (COOK)2H2O / mass Mix) × 100

⇒ mass (COOK)2H2O = (1.20 g (CaC2O4)H2O)×(mol (CaC2O4)H2O / 146.12 g (CaC2O4)H2O)×(mol (COOK)2H2O/mol (CaC2O4)H2O)×(184.24 g (COOK)2H2O/mol (COOK)2H2O)

⇒ mass (COOK)2H2O = 1.513 g

⇒ % (COOK)2H2O = ( 1.513 g / 4 g )×100

⇒ % (COOK)2H2O = 37.826 %

7 0

3 years ago

In order for a circuit work there can be no _______ within the circuit?

Luden [163]

I’m not very good at this subject, however I have a idea of what it could be but if wrong My apologies “ Placed wires “.

Have a blessed day and God bless.

8 0

2 years ago