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Gelneren [198K]
3 years ago
12

3.15 mol of an unknown solid is placed into enough water to make 150.0 mL of solution. The solution's temperature increases by 1

9.2°C. Calculate ∆H for the dissolution of the unknown solid. (The specific heat of the solution is 4.18 J/g・°C and the density of the solution is 1.20 g/mL).
Chemistry
1 answer:
RUDIKE [14]3 years ago
7 0

Answer:

Enthalpy change for the dissolution of the unknown solid = 4.6 Kj/mole

Explanation:

Using Q = m x Cs x ΔT ................................(1)

where Q = Amount of heat absorbed

            m = Mass of solution

           Cs = Specific heat capacity of solution

          ΔT = Change in temperature

Given Density of solution = 1.20 g/ml

And volume of solution = 150 ml

Mass of solution = density x volume

                           = 1.2 x 150

                           = 180 g

From equation (1)

              Q = 180 x 4.18 x 19.2 = 14446.08 = 14.4 Kj

So, ΔH  of the dissolution of the unknown solid = \frac{Q}{n} = \frac{14.4}{3.15} = 4.6 kj/mole

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lisov135 [29]

Answer:

12.

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5 0
2 years ago
Match each substance with the correct designation for the equation HSO3- + CH3NH2 &lt;=&gt; SO32- + CH3NH3+ HSO3- CH3NH2 SO32- C
Zanzabum

Answer:

HSO_3^-: conjugate acid of SO_3^{2-}

CH_3NH_2 : conjugate base of CH_3NH_3^+

SO_3^{2-} : conjugate base of HSO_3^-

CH_3NH_3^+ : conjugate acid of CH_3NH_2

Explanation:

According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.

HSO_3^-+CH_3NH_2\rightleftharpoons SO_3^{2-}+CH_3NH_3^+

Here in forward reaction CH_3NH_2 is accepting a proton, thus it is considered as a base and after accepting a proton, it forms CH_3NH_3^+ which is a conjugate acid.

And HSO_3^-  is losing a proton, thus it is considered as an acid and after loosing a proton, it forms SO_3^{2-} which is a conjugate base.

Similarly in the backward reaction, CH_3NH_3^+ is loosing a proton, thus it is considered as a acid and after loosing a proton, it forms CH_3NH_2 which is a conjugate base.

And SO_3^{2-}  is accepting a proton, thus it is considered as a base and after accepting a proton, it forms HSO_3^{-} which is a conjugate acid.

4 0
3 years ago
Information related to a titration experiment is given in the balanced equation and table below.
Scorpion4ik [409]

Answer:

0.24M

Explanation:

The equation for the reaction is given below:

H2SO4 + 2KOH → K2SO4 + 2H2O

From the equation above, we obtained the following information:

nA (mole of acid) = 1

nB (mole of base) = 2

Data obtained from the question include:

Va (volume of the acid) = 12mL

Ca (concentration of the acid) =?

Vb (volume of the base) = 36mL

Cb (concentration of the base) = 0.16 M

The Ca (concentration of the acid) can be obtained as follow:

CaVa/CbVb = nA/nB

Ca x 12 / 0.16 x 36 = 1 /2

Cross multiply to express in linear form as shown below:

Ca x 12 x 2 = 0.16 x 36

Divide both side by 12 x 2

Ca = 0.16 x 36/ 12 x 2

Ca = 0.24M

Therefore, the concentration of the acid is 0.24M

6 0
3 years ago
What mass of NaCl is dissolved in 150 g of water in a .050<br> msolution?
mart [117]

Answer:

0.4383 g

Explanation:

Molality is defined as the moles of the solute present in 1 kg of the solvent.

It is represented by 'm'.

Thus,  

Molality\ (m)=\frac {Moles\ of\ the\ solute}{Mass\ of\ the\ solvent\ (kg)}

Given that:

Mass of solvent, water = 150 g = 0.15 kg ( 1 g = 0.001 g )

Molality = 0.050 m

So,

0.050=\frac {Moles\ of\ the\ solute}{0.15}

Moles = 0.050\times 0.15\ mol= 0.0075\ mol

Molar mass of NaCl = 58.44 g/mol

Mass = Moles*Molar mass = 0.0075\times 58.44\ g = 0.4383 g

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It makes it thinner to calcite the rhythm
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