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2 years ago

PLEASE HELP DUE TONIGHT THANK YOU

Chemistry

1 answer:

Sav [38]2 years ago

3 0

Answer:

check the attached image. hope it helps

Explanation:

a. What does mol/L mean? mol/L means Molar Concentration

b. Describe in your own words how mol/L compares to grams/liter. (similarities and differences)

Similarities: mol/L and grams/liter both are units of concentration.

Differences:

mol/L concentration says how many moles of solute are present in 1L of solution.

grams/liter concentration says how many grams of solute are present in 1L of solution.

3. Describe at least 2 ways in the simulation to change each of the parameters:

a. Volume of solution= evaporating or adding more water

b. amount of solute=reducing the water or evaporating more water

c. Concentration of solute in solution = evaporating or adding more solute

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3. Which of the following is NOT a major ingredient in weather?

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Explanation: B Sunshine

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When a student mixes 50 mL of 1.0 M HCl and 50 mL of 1.0 M NaOH in a coffee-cup calorimeter, the temperature of the resultant so

zmey [24]

Answer: 54.4 kJ/mol

Explanation:

First we have to calculate the moles of HCl and NaOH.

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=1.0M\times 0.05=0.05mole$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=1.0\times 0.05L=0.05mole$

The balanced chemical reaction will be,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.05 mole of HCl neutralizes by 0.05 mole of NaOH

Thus, the number of neutralized moles = 0.05 mole

Now we have to calculate the mass of water:

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = $50ml+50ml=100ml$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 100ml=100g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

$c$ = specific heat of water = $4.18J/g^oC$

m = mass of water = 100 g

$T_{final}$ = final temperature of water = $27.5^0C$

$T_{initial}$ = initial temperature of metal = $21.0^0C$

Now put all the given values in the above formula, we get:

$q=100g\times 4.18J/g^oC\times (27.5-21.0)^0C$

$q=2719.6J=2.72kJ$

Thus, the heat released during the neutralization = 2.72 KJ

Now we have to calculate the enthalpy of neutralization per mole of $HCl$ :

0.05 moles of $HCl$ releases heat = 2.72 KJ

1 mole of $HCl$ releases heat = $\frac{2.72}{0.05}\times 1=54.4KJ$

Thus the enthalpy change for the reaction in kJ per mol of HCl is 54.4 kJ

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3 years ago

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A 7.27-gram sample of a compound is dissolved in 250. grams of benzene. The freezing point of this solution is 1.02°C below that

UkoKoshka [18]

Answer:

The correct answer is 146 g/mol

Explanation:

Freezing point depression is a colligative property related to the number of particles of solute dissolved in a solvent. It is given by:

ΔTf = Kf x m

Where ΔTf is the freezing point depression (in ºC), Kf is a constant for the solvent and m is the molality of solution. From the problem, we know the following data:

ΔTf = 1.02ºC

Kf = 5.12ºC/m

From this, we can calculate the molality:

m = ΔTf/Kf = 1.02ºC/(5.12ºC/m)= 0.199 m

The molality of a solution is defined as the moles of solute per kg of solvent. Thus, we can multiply the molality by the mass of solvent in kg (250 g= 0.25 kg) to obtain the moles of solute:

0.199 mol/kg benzene x 0.25 kg = 0.0498 moles solute

There are 0.0498 moles of solute dissolved in the solution. To calculate the molar mass of the solute, we divide the mass (7.27 g) into the moles:

molar mass = mass/mol = 7.27 g/(0.0498 mol) = 145.9 g/mol ≅ 146 g/mol

Therefore, the molar mass of the compound is 146 g/mol

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3 years ago

The pKa of the α‑carboxyl group of serine is 2.21 , and the pKa of its α‑amino group is 9.15 . Calculate the average net charge

sleet_krkn [62]

Answer:

Net charge in serine at pH equal to 8.30 is "0"

Explanation:

At pH > $pK_{a}$ , carboxyl group exists as $-CO_{2}^{-}$ (charged)

At pH < $pK_{a}$ , carboxyl group exists as -COOH (neutral)
At pH > $pK_{a}$ , amino group exists as $-NH_{2}$ (neutral)
At pH < $pK_{a}$ , amino group exists as $-NH_{3}^{+}$ (charged)
So, at pH = 8.30, both carboxyl and amino group exists in charged state.
Net charge in serine at pH equal to 8.30 is "0".
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Answer:

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Explanation:

number 2

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