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2 years ago

How many equivalents are in 0.60 mole of Mg2+

1 answer:

5 0

Equivalents express the combining power of an element for example what mass of an element can combine with or displace one gram of hydrogen or eight grams of oxygen.

Since Mg is divalent, one gram of Mg2+ can combine with 2 grams of hydrogen

Therefore, 0.6 mole of Mg2+ can combine with 0.6 x 2 = 1.2 moles of hydrogen

Hence there are 1.2 equivalents in 0.60 mole of Mg2+

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This formula equation is unbalanced. Pb(NO3)2(aq) + Li2SO4(aq)--> PbSO4(s) + LiNO3(aq). Which coefficient should appear in fr

Temka [501]

Balancing is making sure there are the same number of atoms on either side of the reaction.

Pb(NO3)2 + Li2SO4--> PbSO4 + LiNO3

There are 2 NO3 groups and 2 Li on the right side, need 2 on the left side.
Need a coefficient of 2 for LiNO3

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2 years ago

Read 2 more answers

(b) A compound X contains carbon, hydrogen and oxygen only

faust18 [17]

Answer:

C2H4O

Explanation:

C H O

54.54/12 9.09/1 36.37/16

4.545 9.09 2.27

4.545/2.27 9.09/2.27 2.27/2.27

2.00 4.00 1

Empirical formula is C2H4O

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2 years ago

If the symbol X represents a central atom, Y represents outer atoms, and Z represents lone pairs on the central atom, the struct

Inessa05 [86]

Answer:

The correct option is: bent 109°

Explanation:

Covalent molecules are the molecules in which the atoms are linked by covalent bonds. The electrons involved in the formation of a covalent bond are known as shared pair or bond pair of electrons.

The three-dimensional arrangement of the atoms of a molecule in space is known its molecular structure or geometry.

Given molecule: XY₂, having two lone pairs around the central atom X.

Since the molecule XY₂ has two lone pairs and two bond pairs of electrons. Therefore according to the VSEPR theory, the given molecule has a bent molecular geometry with 109° bond angle.

7 0

3 years ago

How many moles of carbon are in 3.7 moles of C 8 H11 NO2

Dafna1 [17]

In every molecule of $C_{8} H_{11}NO_2$ there is 8 atoms of Carbon.

IF we have 3.7 moles of $C_{8} H_{11}NO_2$ to find the number of moles of Carbon, just multiply by 8

3.7 * 8 = 29.6 mol Carbon

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2 years ago

Sulfuric acid is produced in larger amounts by weight than any other chemical. It is used in manufacturing fertilizers, oil refi

Fed [463]

Answer:

A. -166.6 kJ/mol

B. -127.7 kJ/mol

C. -133.9 kJ/mol

Explanation:

Let's consider the oxidation of sulfur dioxide.

2 SO₂(g) + O₂(g) → 2 SO₃(g) ΔG° = -141.8 kJ

The Gibbs free energy (ΔG) can be calculated using the following expression:

ΔG = ΔG° + R.T.lnQ

where,

ΔG° is the standard Gibbs free energy

R is the ideal gas constant

T is the absolute temperature (25 + 273.15 = 298.15 K)

Q is the reaction quotient

The molar concentration of each gas ([]) can be calculated from its pressure (P) using the following expression:

$[]=\frac{P}{R.T}$

Calculate ΔG at 25°C given the following sets of partial pressures.

Part A 130atm SO₂, 130atm O₂, 2.0atm SO₃. Express your answer using four significant figures.

$[SO_{2}]=[O_{2}]=\frac{130atm}{(0.08206atm.L/mol.K).298K} =5.32M$

$[SO_{3}]=\frac{2.0atm}{(0.08206atm.L/mol.K).298K} =0.0818M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0818^{2} }{5.32^{3} } =4.44 \times 10^{-5}$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln (4.44 × 10⁻⁵) = -166.6 kJ/mol

Part B 5.0atm SO₂, 3.0atm O₂, 30atm SO₃ Express your answer using four significant figures.

$[SO_{2}]=\frac{5.0atm}{(0.08206atm.L/mol.K).298K}=0.204M$

$[O_{2}]=\frac{3.0atm}{(0.08206atm.L/mol.K).298K}=0.123M$

$[SO_{3}]=\frac{30atm}{(0.08206atm.L/mol.K).298K}=1.23M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{1.23^{2} }{0.204^{2}.0.123 } =296$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 296 = -127.7 kJ/mol

Part C Each reactant and product at a partial pressure of 1.0 atm. Express your answer using four significant figures.

$[SO_{2}]=[O_{2}]=[SO_{3}]=\frac{1.0atm}{(0.08206atm.L/mol.K).298K}=0.0409M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0409^{2} }{0.0409^{3}} =24.4$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 24.4 = -133.9 kJ/mol

7 0

2 years ago