When Nitrogen gas gets cold enough it can form a liquid. This liquid nitrogen is an

calculate δg o for each reaction using δg o f values: (a) h2(g) i2(s) → 2hi(g) 2.6 kj (b) mno2(s) 2co(g) → mn(s) 2co2(g) kj (c)

Reika [66]

(a)The change in Gibbs free energy for the reaction has been 2.6 kJ/mol.

(b) The change in Gibbs free energy for the reaction has been -49.3 kJ/mol.

(c) The change in Gibbs free energy for the reaction has been 91.38 kJ/mol.

6 0

2 years ago

A reaction starts with 20.0 grams of lithium hydroxide (LiOH) and produces 31.0 grams of lithium chloride (LiCl), what is the pe

castortr0y [4]

Answer:

87.6%

Explanation:

Step 1:

The balanced equation for the reaction.

LiOH + HCl —> LiCl + H2O

Step 2:

Determination of the mass of LiOH that reacted and the mass of LiCl produced from the balanced equation.

This is illustrated below:

Molar mass of LiOH = 7 + 16 + 1 = 24g/mol

Mass of LiOH from the balanced equation = 1 x 24 = 24g

Molar Mass of LiCl = 7 + 35.5 = 42.5g/mol

Mass of LiCl from the balanced equation = 1 x 42.5 = 42.5g

Thus, from the balanced equation, 24g of LiOH reacted to produce 42.5g of LiCl.

Step 3:

Determination of the theoretical yield of LiCl.

This is illustrated below:

From the balanced equation above,

24g of LiOH reacted to produce 42.5g of LiCl.

Therefore, 20g of LiOH will react to produce = (20 x 42.5)/24 = 35.4g

Therefore, the theoretical yield of LiCl is 35.4g.

Step 4:

Determination of the percentage yield of LiCl. This is illustrated below:

Actual yield of LiCl = 31g

Theoretical yield of LiCl = 35.4g

Percentage yield of LiCl =.?

Percentage yield = Actual yield /Theoretical yield x 100

Percentage yield = 31/35.4 x 100

Percentage yield of LiCl = 87.6%

8 0

3 years ago

How many protons electrons and neutrons does the following isotopes contain 1H+

kap26 [50]

Answer:

H+ contains 1 proton, 0 neutrons, 0 electrons.

Explanation:

7 0

3 years ago

1.How many mL of 0.523 M HBr are needed to dissolve 8.60 g of CaCO3?

prohojiy [21]

Answer:

The answer to your question is:

Explanation:

1.-

HBr = 0.523 M V = ?

CaCO3 = 8.6 g

2HBr(aq) + CaCO₃(s) ⇒ CaBr₂(aq) + H₂O(l) + CO₂(g)

MW CaCO₃ = 40 + 12 + 48 = 100 g

MW HBr = 80 + 1 = 81 g

Molarity = moles / volume

100 g of CaCO₃ ---------------- 1 mol

8.6 g ---------------- x

x = (8.6 x 1) / 100

x = 0.086 moles

2 moles of HBr ----------------- 1 mol of CaCO₃

x ----------------- 0.086 moles

x = (0.086 x 2) / 1 = 0.172 moles of HBr

Volume = moles / molarity

Volume = 0.172/ 0.523 = 0.323 l or 323 ml of HBr

2.-

V = ? ml NaOH 0.487 M

V = 101 ml of 0.628 M MnSO₄

MnSO₄(aq) + 2NaOH(aq) ⇒ Mn(OH)₂(s) + Na₂SO₄(aq)

MW MnSO₄ = 55 g

MW NaOH = NaOH = 40 g

Moles = Molarity x volume

Moles = (0.628) x (0.101)

Moles = 0.065 moles of MnSO₄

1 mol of MnSO₄ ------------------ 2 moles of NaOH

0.065 ----------------- x

x = (0.065x 2) / 1

x = 0.131 moles of NaOH

Volume = moles / molarity

Volume = 0.131 / 0.487

Volume = 0.268 l or 268 ml of NaOH

4 0

3 years ago

Perform the following

exis [7]

Answer:

665.22247

Explanation:

Esta ves la respuesta

7 0

3 years ago