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beks73 [17]
4 years ago
10

Procaine hydrochloride ( = 272.77 g/mol) is used as a local anesthetic. Calculate the molarity of a 4.666 m solution which has a

density of 1.1066 g/ml.
Chemistry
2 answers:
AleksAgata [21]4 years ago
8 0

Answer : The molarity of solution will be, 2.27 mole/L

Explanation :

The relation between the molarity, molality and the density of the solution is,

d=M[\frac{1}{m}+\frac{M_b}{1000}]

where,

d = density of solution  = 1.1066 g/ml

m = molality of solution  = 4.666 m

M = molarity of solution  = ?

M_b = molar mass of solute (Procaine hydrochloride) = 272.77 g/mole

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

1.1066=M[\frac{1}{4.66}+\frac{272.77}{1000}]

M=2.27mole/L

Therefore, the molarity of solution will be, 2.27 mole/L

Mkey [24]4 years ago
5 0

Procaine hydrochloride ( = 272.77 g/mol) is used as a local anesthetic. Calculate the molarity of a 4.666 m solution which has a density of 1.1066 g/ml.

molarity = Moles of solute /  volume of solution

          Molarity = m d / [ 1 + (mW / 1000)]

Molarity = 4.666 X 1.1066 / [ 1 + (4.666 X 272.77 / 1000)]

Molarity =  5.16 / 2.272= 2.271 M

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Consider the following reaction:
iren [92.7K]

Answer:

A. ΔG° = 132.5 kJ

B. ΔG° = 13.69 kJ

C. ΔG° = -58.59 kJ

Explanation:

Let's consider the following reaction.

CaCO₃(s) → CaO(s) + CO₂(g)

We can calculate the standard enthalpy of the reaction (ΔH°) using the following expression.

ΔH° = ∑np . ΔH°f(p) - ∑nr . ΔH°f(r)

where,

n: moles

ΔH°f: standard enthalpy of formation

ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g)) - 1 mol × ΔH°f(CaCO₃(s))

ΔH° = 1 mol × (-635.1 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1206.9 kJ/mol)

ΔH° = 178.3 kJ

We can calculate the standard entropy of the reaction (ΔS°) using the following expression.

ΔS° = ∑np . S°p - ∑nr . S°r

where,

S: standard entropy

ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g)) - 1 mol × S°(CaCO₃(s))

ΔS° = 1 mol × (39.75 J/K.mol) + 1 mol × (213.74 J/K.mol) - 1 mol × (92.9 J/K.mol)

ΔS° = 160.6 J/K. = 0.1606 kJ/K.

We can calculate the standard Gibbs free energy of the reaction (ΔG°) using the following expression.

ΔG° = ΔH° - T.ΔS°

where,

T: absolute temperature

<h3>A. 285 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 285K × 0.1606 kJ/K = 132.5 kJ

<h3>B. 1025 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 1025K × 0.1606 kJ/K = 13.69 kJ

<h3>C. 1475 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 1475K × 0.1606 kJ/K = -58.59 kJ

5 0
4 years ago
Water is a pure substance. Which of the following is true about water? (4 points) Its compounds can only be physically separated
jonny [76]

Answer:

Water/H20 is a compound made of hydrogen and oxygen. Although water is the most abundant substance on earth, it is rarely found naturally in its pure form. Most of the time, pure water has to be created. Pure water is called distilled water or deionized water.

Explanation:

7 0
3 years ago
A 0.75M solution of CH3OH is prepared in 0.500 kg of water. How many moles of CH3OH are needed?
4vir4ik [10]

Answer:

We need 0.375 mol of CH3OH to prepare the solution

Explanation:

For the problem they give us the following data:

Solution concentration 0,75 M

Mass of Solvent is 0,5Kg

knowing that the density of water is 1g / mL,  we find the volume of water:

                           d = \frac{g}{mL} \\\\ V= \frac{g}{d}  = \frac{500g}{1 \frac{g}{mL} } = 500mL = 0,5 L

Now, find moles of CH_{3} OH are needed using the molarity equation:

                           M = \frac{ moles }{ V (L)} \\\\\\molesCH_{3}OH  = M . V(L) = 0,75 M . 0,5 L\\\\molesCH_{3}OH = 0,375 mol

therefore the solution is prepared using 0.5 L of H2O and 0.375 moles of CH3OH,  resulting in a concentration of 0,75M

5 0
4 years ago
Chemistry! help! please!
VARVARA [1.3K]
It would be 187 75 Re since 76 - 1 is 75 and number 2 would be beta particle and a positron.
7 0
4 years ago
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What is the volume of 8.7 * 10 ^ 23 molecules of chlorine gas (Cl 2 ) ?
Mariana [72]

Answer:

32.3 dm³

Explanation:

Data given:

no. of molecules of Cl₂ = 8.7 x 10²³

Volume of chlorine gas (Cl₂) = ?

Solution:

First we have to find number of moles

For this formula used

    no. of moles = no. of molecules / Avogadros number

    no. of moles = 8.7 x 10²³ / 6.022 x 10²³

    no. of moles = 1.44 moles

Now we have to find volume of the gas

for this formula used

                      no. of moles = volume of gas / molar volume

molar volume = 22.4 dm³/mol

Put values in above equation

                 1.44 moles = volume of Cl₂ gas / 22.4 dm³/mol

rearrange the above equation

                 volume of Cl₂ gas = 1.44 moles x 22.4 dm³/mol

                volume of Cl₂ gas =  32.3 dm³

8 0
3 years ago
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