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beks73 [17]
2 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]2 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]2 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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Octane (C8H18) undergoes combustion according to the following thermochemical equation. 2C8H18(l) + 25O2(g) → 16CO2(g) + 18H2O(l
Zepler [3.9K]

Answer: The standard enthalpy of formation of liquid octane is -250.2 kJ/mol

Explanation:

The given balanced chemical reaction is,

2C_8H_{18}(l)+25O_2(g)\rightarrow 16CO_2(g)+18H_2O(l)

First we have to calculate the enthalpy of reaction (\Delta H^o).

\Delta H^o=H_f_{product}-H_f_{reactant}

\Delta H^o=[n_{O_2}\times \Delta H_f^0_{(O_2)}+n_{H_2O}\times \Delta H_f^0_{(H_2O)}]-[n_{C_8H_{18}}\times \Delta H_f^0_{(C_8H_{18})+n_{O_2}\times \Delta H_f^0_{(O_2)}]

where,

We are given:

\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(C_8H_{18}(l))}=?kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.8kJ/mol

Putting values in above equation, we get:

-1.0940\times 10^4=[(16\times -393.5)+(18\times -285.8)]-[(25\times 0)+(2\times \Delat H_f{C_8H_{18}(l)}]

\Delta H^o_f_{(C_8H_{18}(l))}=-250.2kJ/mol

Thus the standard enthalpy of formation of liquid octane is -250.2 kJ/mol

4 0
2 years ago
Ron and Hermione begin with 1.50 g of the hydrate copper(II)sulfate ∙ x-hydrate (CuSO4 ∙ xH2O), where x is an integer. Part of t
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Answer

5

Explanation:

We can go about this using the percentage compositions.

First, we calculate the percentage composition of the copper sulphate. This is obtainable by using the mass.

0.96/1.5 * 100 = 64%

Hence the percentage by mass of the water present is 36%

The molar mass of the anhydrous sulphate is 64 + 32 +4(16) = 160g/mol

The molar mass of the water is 2(1) + 16 = 18g/mol

Not forgetting that it is in multiples of x, the total molar mass of the water is 18x moles

The total mass of the copper sulphate hydrate is 160+ 18x

Now how do we get x? Like it is said earlier, the percentage composition is constant.

Hence, 64/100 * (160 + 18x) = 160

16000 = 64(160 + 18x)

16000 = 10,240 + 1152x

16,000 - 10,240 = 1152x

1152x = 5760

x = 5760/1152

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7 0
2 years ago
In the PhET simulation, make sure that the checkbox Stable/Unstable in the bottom right is checked. Using the PhET simulation as
lord [1]

Answer:

kindly check the EXPLANATION SECTION

Explanation:

In order to be able to answer this question one has to consider the neutron proton ratio. Considering this ratio will allow us to determine the stability of a nuclei. The most important rule that helps us in determination of stability is that when the Neutron- Proton ratio  of any nuclei ranges from  to 1 to 1.5, then we say the nuclei is STABLE.

Also, we need to understand that when the  Neutron- Proton ratio is LESS THAN 1 or GREATER THYAN 1.5, then we say the nuclei is UNSTABLE.

So, let us check which is stable and which is unstable:

a. 4 protons and 5 neutrons =  Neutron- proton ratio = N/P = 5/4= stable.

b. 7 protons and 7 neutrons  =  Neutron- proton ratio = N/P = 7/7= 1 = stable.

c. 2 protons and 3 neutrons  =  Neutron- proton ratio = N/P = 3/5 =0.6 =unstable.

d. 3 protons and 0 neutrons  =  Neutron- proton ratio = N/P = 0/3= 0= unstable.

e. 6 protons and 5 neutrons  =  Neutron- proton ratio = N/P = 5/6= 0.83 = unstable.

f. 9 protons and 9 neutrons  =  Neutron- proton ratio = N/P = 9/9 = 1 = stable.

g. 8 protons and 7 neutrons  =  Neutron- proton ratio = N/P =  7/8 =0.875 = unstable.

h. 1 proton and 0 neutrons =  Neutron- proton ratio = N/P = 0/1 =0 = unstable

6 0
2 years ago
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