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alexdok [17]
2 years ago
9

phillip morris cigaretteshave 5mg tar and 0.4 mg nicotine per cigarette and 20 cigarettes per pack how many packs of cigarettes

would have to be smoked to coat your lungs with 4.0 oz (1/4 lb) of tar? How many packs of cigarettes would have to be smoked to introduce your lungs to one gram of nicotine?
Chemistry
1 answer:
Shalnov [3]2 years ago
8 0

Answer: A solution is made by dissolving 27.8 g of nicotine (Molar Mass = 160 g/mol) in 145 g of cyclohexane (C6H12) to form 142 mL of solution. Calculate the mole fraction of the solute and the molarity, respectively, of this solution.

Explanation:

You might be interested in
Determain the number of moles in 2.24l of ch4 at stp
valkas [14]

Answer:

0.1 mole of CH₄

Explanation:

From the question given above, the following data were obtained:

Volume of CH₄ = 2.24 L

Number of mole of CH₄ =?

The number of mole of CH₄ can be obtained as follow:

Recall:

1 mole of a gas occupy 22.4 L at stp. This implies that 1 mole of CH₄ occupies 22.4 L at stp.

22.4 L = 1 mole of CH₄

Therefore,

2.24 L = 2.24 × 1 mole of CH₄ / 22.4

2.24 L = 0.1 mole of CH₄.

6 0
3 years ago
Phenolphthalein has a pKa of 9.7 and is colorless in its acid form and pink in its basic form. calculate [In-}/{HIn} for the fol
zvonat [6]

Answer:

1.58x10⁻⁵

2.51x10⁻⁸

0.0126

63.10

Explanation:

Phenolphthalein acts like a weak acid, so in aqueous solution, it has an acid form HIn, and the conjugate base In-, and the pH of it can be calculated by the Handerson-Halsebach equation:

pH = pKa + log[In-]/[HIn]

pKa = -logKa, and Ka is the equilibrium constant of the dissociation of the acid. [X] is the concentrantion of X. Thus,

i) pH = 4.9

4.9 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = - 4.8

[In-]/[HIn] = 10^{-4.8}

[In-]/[HIn] = 1.58x10⁻⁵

ii) pH = 2.1

2.1 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = -7.6

[In-]/[HIn] = 10^{-7.6}

[In-]/[HIn] = 2.51x10⁻⁸

iii) pH = 7.8

7.8 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = -1.9

[In-]/[HIn] = 10^{-1.9}

[In-]/[HIn] = 0.0126

iv) pH = 11.5

11.5 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = 1.8

[In-]/[HIn] = 10^{1.8}

[In-]/[HIn] = 63.10

6 0
3 years ago
C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l), ΔH = –1.37 × 103 kJ For the combustion of ethyl alcohol as described in the above equati
babymother [125]

Answer:

The true statements are: I. The reaction is exothermic.

II. The enthalpy change would be different if gaseous water was produced.

Explanation:

The given chemical reaction: C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l),  ΔH= -1.37×10³kJ

1. In an exothermic reaction, heat or energy is released from the system to the surrounding. Thus for an exothermic process the change in enthalpy is less than 0 or negative (ΔH < 0) .

Since the enthalpy change for a combustion reaction is negative. <u>Therefore, the given reaction is exothermic.</u>

2. The change in enthalpy (ΔH) of a reaction is equal to difference of the sum of standard enthalpy of formation (ΔHf°) of the products and the reactants.

ΔHr° = ∑ n.ΔHf°(products) − ∑ n.ΔHf°(reactants)

As the value of ΔHf° of water in gaseous state and liquid state is not the same.

<u>Therefore, the enthalpy change of the reaction will be different, if gaseous water was present instead of liquid water.</u>

3. An oxidation-reduction reaction or a redox reaction involves simultaneous reduction and oxidation processes.

The given chemical reaction, represents the combustion reaction of ethanol.

Since combustion reactions are redox reactions. <u>Therefore, the given combustion reaction is an oxidation-reduction reaction.</u>

4. According to the ideal gas equation: P.V =n.R.T

Volume (V) ∝ n (number of moles of gas)

Since the number of moles (n) of gaseous reactants is 3 and number of moles of gaseous (n) products is 2.

<u>Therefore, the volume occupied by 3 moles of the reactant gaseous molecules will be more than 2 moles product gaseous molecules.</u>

3 0
3 years ago
Ratio of atoms in Hydrochloric acid
Lesechka [4]

One to one ratio

The formula of hydrochloric acid is HCl so there is one atomic of hydrogen and one atom of oxygen

7 0
3 years ago
what is the limiting reactant when 1.50g of lithium and 1.50 g of nitrogen combine to form lithium nitride, a component of advan
PSYCHO15rus [73]
<h3>Answer:</h3>

Limiting reactant is Lithium

<h3>Explanation:</h3>

<u>We are given;</u>

  1. Mass of Lithium as 1.50 g
  2. Mass of nitrogen is 1.50 g

We are required to determine the rate limiting reagent.

  • First, we write the balanced equation for the reaction

6Li(s) + N₂(g) → 2Li₃N

From the equation, 6 moles of Lithium reacts with 1 mole of nitrogen.

  • Second, we determine moles of Lithium and nitrogen given.

Moles = Mass ÷ Molar mass

Moles of Lithium

Molar mass of Li = 6.941 g/mol

Moles of Li = 1.50 g ÷ 6.941 g/mol

                   = 0.216 moles

Moles of nitrogen gas

Molar mass of Nitrogen gas is 28.0 g/mol

Moles of nitrogen gas = 1.50 g ÷ 28.0 g/mol

                                     = 0.054 moles

  • According to the equation, 6 moles of Lithium reacts with 1 mole of nitrogen.
  • Therefore, 0.216 moles of lithium will require 0.036 moles (0.216 moles ÷6) of nitrogen gas.
  • On the other hand, 0.054 moles of nitrogen, would require 0.324 moles of Lithium.

Thus, Lithium is the limiting reagent while nitrogen is in excess.

7 0
3 years ago
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