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Rufina [12.5K]
2 years ago
10

Which of these facts best illustrates why regulation of alcohol consumption is necessary

Chemistry
1 answer:
exis [7]2 years ago
4 0

So that people don’t break laws or drive under the influence.

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A gas is contained in a thick-walled balloon. When the pressure changes from 319 mm Hg to 215 mm Hg, th volume changes from 0.55
Dmitry_Shevchenko [17]
In this question we need to find the new volume of the gas. Since we have been given the pressure and temperature change, we can used to combined gas law equation.
\frac{P1V1}{T1} =  \frac{P2V2}{T2}
the parameters for 1st instance are given on the left side and parameters for the second instance are given on the right side of the equation
(319 mmHg x 0.558 L)/ 115 K = (215 mmHg x V)/387 K
V = 2.79 L
7 0
2 years ago
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Water is dissolved into n-butanol (a polar liquid). Which is the second step at the molecular level as water dissolves into n-bu
sergeinik [125]
<span>The steps of solubility of water in N-butanol is as follows:1. N-butanol molecules are attracted to the surface of the water, 2. N-butanol molecules surround water molecules, 3. Butanol mixes with water and 4. Water molecules are carried into N-butanol.</span>
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84. Heavy water, D2O (molar mass = 20.03 g mol–1), can be separated from ordinary water, H2O (molar mass = 18.01), as a result o
ipn [44]

Answer:

relative rate of diffusion is 1.05

Explanation:

According to Graham's law of difussion:

Rate of diffusion is inversely proportional to the square root of molecular weight of a molecule.

For two given molecules:

\frac{(rate)_{1}}{(rate)_{2}}=\frac{\sqrt{M_{2}} }{\sqrt{M_{1}}}

The given molecules are

Water = 18.01

Heavy water =20.03

Thus the relative rate of diffusion will be:

\frac{(rate)_{water}}{(rate)_{heavywater}}=\sqrt{ \frac{20.03}{18.01}}=1.05

8 0
2 years ago
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Phantasy [73]
1.
cost in cents

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miles = c + 2
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14. What is the pH of a 0.24 M solution of sodium propionate, NaC3H502, at 25°C? (For
Murrr4er [49]

Answer:

9.1

Explanation:

Step 1: Calculate the basic dissociation constant of propionate ion (Kb)

Sodium propionate is a strong electrolyte that dissociates according to the following equation.

NaC₃H₅O₂ ⇒ Na⁺ + C₃H₅O₂⁻

Propionate is the conjugate base of propionic acid according to the following equation.

C₃H₅O₂⁻ + H₂O ⇄ HC₃H₅O₂ + OH⁻

We can calculate Kb for propionate using the following expression.

Ka × Kb = Kw

Kb = Kw/Ka = 1.0 × 10⁻¹⁴/1.3 × 10⁻⁵ = 7.7 × 10⁻¹⁰

Step 2: Calculate the concentration of OH⁻

The concentration of the base (Cb) is 0.24 M. We can calculate [OH⁻] using the following expression.

[OH⁻] = √(Kb × Cb) = √(7.7 × 10⁻¹⁰ × 0.24) = 1.4 × 10⁻⁵ M

Step 3: Calculate the concentration of H⁺

We will use the following expression.

Kw = [H⁺] × [OH⁻]

[H⁺] = Kw/[OH⁻] = 1.0 × 10⁻¹⁴/1.4 × 10⁻⁵ = 7.1 × 10⁻¹⁰ M

Step 4: Calculate the pH of the solution

We will use the definition of pH.

pH = -log [H⁺] = -log 7.1 × 10⁻¹⁰ = 9.1

5 0
2 years ago
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