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An aqueous solution containing 6.06 g of lead(II) nitrate is added to an aqueous solution containing 6.58 g of potassium chlorid

sertanlavr [38]

Answer:

Pb(NO₃)₂ is the limiting reactant.

The equation is: Pb(NO₃)₂ (aq) + 2KCl (aq) → PbCl₂ (s) ↓ + 2KNO₃ (aq)

Explanation:

We identify our reactants:

Pb(NO₃)₂ → Lead (II) nitrate

KCl → Potassium chloride

Our reaction is:

Pb(NO₃)₂ (aq) + 2KCl (aq) → PbCl₂ (s) ↓ + 2KNO₃ (aq)

1 mol of Lead (II) nitrate reacts to 2 moles of KCl, in order to produce 2 moles of potassium nitrate and 1 mol of slid Lead (II) chloride.

We determine the moles of the reactants:

6.06 g . 1mol /331.2 g = 0.0183 moles of Lead (II) nitrate

6.58 g . 1mol / 74.55g = 0.0882 moles of KCl

2 moles of KCl react to 1 mol of Pb(NO₃)₂

Then, 0.0882 moles of KCl may react to (0.0882 . 1) /2 = 0.0441 moles

We have 0.0183 moles of Pb(NO₃)₂ and we need 0.441 moles. Then, the

Pb(NO₃)₂ is our limiting reactant.

4 0

3 years ago

Which of the following statements about the nuclear envelope is false?

SVETLANKA909090 [29]

Answer:

The statement about the nuclear envelope which is false is D - The nuclear envelope is continuos with the Golgi apparatus.

Explanation:

The nuclear envelope is composed of two membranes pierced by pores. They are distributed throughout the entire surface of the envelope and are much more than perforations in which both external and internal membranes are continued. Inside the pores a set of proteins form a structure called NPC where the smallest molecules pass, since the larger ones require energy input. The nuclear envelope that is reinforced by filaments, its two bilayers are continued with the endoplasmic reticulum membrane system.

4 0

4 years ago

A 1.00-L flask was filled with 2.00 moles of gaseous and 2.00 moles of gaseous and heated. After equilibrium was reached, it was

Bogdan [553]

Answer:

2.35

Explanation:

A 1.00-L flask was filled with 2.00 moles of gaseous SO₂ and 2.00 moles of gaseous NO₂ and heated. After equilibrium was reached, it was found that 1.21 moles of gaseous NO was present. Assume that the reaction

SO₂ + NO₂ ⇌ SO₃ + NO

occurs under these conditions. Calculate the value of the equilibrium constant, K, for this reaction.

Step 1: Calculate the molar concentrations

Since the reaction takes place in a 1.00 L flask, the molar concentrations are:

[SO₂]i = 2.00 M

[NO₂]i = 2.00 M

[NO]eq = 1.21 M

Step 2: Make an ICE chart

SO₂ + NO₂ ⇌ SO₃ + NO

I 2.00 2.00 0 0

C -x -x +x +x

E 2.00-x 2.00-x x x

Step 3: Calculate the concentrations at equilibrium

The concentration of NO at equilibrium is 1.21 M. Then, x = 1.21.

[SO₂]eq = 2.00-1.21 = 0.79 M

[NO₂]eq = 2.00-1.21 = 0.79 M

[SO₃]eq = x = 1.21 M

[NO]eq = x = 1.21 M

Step 4: Calculate the value of the equilibrium constant, K, for this reaction

K = [SO₃] × [NO]/[SO₂] × [NO₂]

K = 1.21²/0.79² = 2.35

6 0

3 years ago

Petrochemicals are used in some cosmetics. True or False

Hunter-Best [27]

True. Any cosmetic containing paraffin wax, benzene, mineral oil, PEG, DEA, MEA, the suffix ethyl- or propyl-, etc. About 95% of fragrances contain petrochemicals.

4 0

3 years ago

Nuclear energy is released during

Zepler [3.9K]

fusion and fission

Explanation:

the nuclear energy is released by fusion when the nuclei of two atoms are combined while the fission is process when the nucei of two atoms split apart

3 0

3 years ago