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2 years ago

__________ is one of many factors that can determine the rate of alcohol absorption.

1 answer:

6 0

Food consumption affects the rate of alcohol absorption in the bloodstream.

Explanation:

The type of food and therefore the amount of food that is a gift in your epithelial duct once you consume alcohol have the foremost direct impact on the speed of alcohol absorption.

When you consume alcohol on the associate empty abdomen, the alcohol is sometimes absorbed within the blood among fifteen minutes to two-and-a-half hours. If you have got a moderate quantity of food in your abdomen once you drink, that speed slows all the way down to thirty minutes to a few hours. If you’re drinking on a full abdomen, alcohol absorption ranges from 3 to 6 hours.

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Suppose you are studying the K sp of K C l O 3 , which has a molar mass of 122.5 g/mol, at multiple temperatures. You dissolve 4

Ghella [55]

Answer:

The K sp Value is $K_{sp}=7.40$

Explanation:

From the question we are told that

The of $KClO_3$ is = 122.5 g/ mol

The mass of $KClO_3$ dissolved is $m_s = 4.0g$

The volume of solution is $V_s = 12mL = 12*10^{-3}L$

The number of moles of $KClO_3$ is mathematically evaluated as

$No \ of \ moles \ = \frac{mass }{Molar \ mass}$

Substituting values

$= \frac{4}{122.5}$

$=0.0327\ moles$

Generally concentration is mathematically represented as

$concentration = \frac{No \ of \ moles}{volume }$

For $KClO_3$

$Z= \frac{0.0327}{12*10^{-3}}$

$=2.72 \ mol/L$

The dissociation reaction of $KClO_3$ is

$KClO_3 \ ----> K^{+}_{(aq)} + ClO_3^-_{(aq)}$

The solubility product constant is mathematically represented as

$K_{sp} = \frac{concentration of ionic product }{concentration of ionic reactant }$

Since there is no ionic reactant we have

$K_{sp} = [k^+] [ClO_3^-]$

$= Z^2$

$= 2.72^2$

$K_{sp}=7.40$

5 0

2 years ago

Read 2 more answers

When 10 g of diethyl ether is converted to vapor at its boiling point, about how much heat is absorbed? (c4h10o, δhvap = 15.7 kj

vagabundo [1.1K]

Answer:

ΔH= 3KJ

Explanation:

The total heat absorbed is the total energy in the process, and that is in form of entalpy.

ΔH = q + ΔHvap, where q is the heat necessary for elevate the temperature of dietil ether. Suppose the initial temperature is room temperature (25ºC=298 K), then

q= 10g x2.261 J/gK x(310 K - 298K)= 271.32 J= 0.3 kJ

Then

ΔHvap = 10g C4H10O x (1 mol C4H10O/74.12 g C4H10O) x( 15.7 KJ/ 1 mol C4H10O) = 2.12 KJ

ΔH= 2.5KJ ≈ 3KJ

5 0

2 years ago

Motohiro felt chilled by the rainy afternoon. He decided to make a cup of hot tea. He let the tea leaves steep too long and the

kati45 [8]

Answer: the correct one would be option B.

Explanation:

8 0

2 years ago

What would happen to the measured cell potentials if 30 mL solution was used in each half-cell instead of 25 mL

tatiyna

Answer:

The answer is " $\bold{\log \frac{[0] mole}{[R]mole}}$ "

Explanation:

$E_{cell} =E_{cell}^{\circ} - \frac{0.0591}{n}= \log\frac{[0]}{[R]}\\$

In the above-given equation, we can see from $E_{ceu}$ , of both oxidant $conc^n$ as well as the reactant were connected. however, weight decreases oxidant and reduction component concentration only with volume and the both of the half cells by the very same factor and each other suspend

$\to \log \frac{\frac{\text{oxidating moles}}{25 \ ml}}{\frac{\text{moles of reduction}}{25 ml}} \ \ = \ \ \log \frac{\frac{\text{oxidating moles}}{30 \ ml}}{\frac{\text{moles of reduction}}{30 ml}} \\\\\\$

$\to {\log \frac{[0] mole}{[R]mole}}$

3 0

2 years ago

What is the mass of a sample of iron if that sample lost 2300J of heat energy when it cooled from 80 oC to 30°C? The specific he

irakobra [83]

102 grams.
Equation:
Quantify of heat = mass x specific heat x difference in temperature
We have: quantity of heat : 2300J
specific heat: .449 J/g
difference in t: 80 - 30 = 50
Solve for mass: 2300 = mass x 0.449 x 50
mass = 102.449
2 sig-figs --> 102 grams

7 0

2 years ago