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bazaltina [42]
3 years ago
8

In a constant‑pressure calorimeter, 60.0 mL of 0.300 M Ba(OH)2 was added to 60.0 mL of 0.600 M HCl. The reaction caused the temp

erature of the solution to rise from 23.65 ∘C to 27.74 ∘C. If the solution has the same density and specific heat as water ( 1.00 g/mL and 4.184J/g⋅K,) respectively), what is Δ???? for this reaction (per mole H2O produced)? Assume that the total volume is the sum of the individual volumes.
Chemistry
1 answer:
Zepler [3.9K]3 years ago
3 0

Answer:

ΔH = 57.04 Kj/mole H₂O

Explanation:

60ml(0.300M Ba(OH)₂(aq) + 60ml(0.600M HCl(aq)

=> 0.06(0.3)mole Ba(OH)₂(aq) + 0.60(0.6)mole HCl(aq)

=> 0.018mole Ba(OH)₂(aq) + 0.036mole HCl(aq)

=> 100% conversion of reactants => 0.018mole BaCl₂(aq) + 0.036mole H₂O(l) + Heat

ΔH = mcΔT/moles H₂O <==> Heat Transfer / mole H₂O

=(120g)(4.0184j/g°C)(27.74°C - 23.65°C)/(0.036mole H₂O)

ΔH = 57,042 j/mole H₂O = 57.04 Kj/mole H₂O

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If a liquid substance is transferred to a different container, what can be predicted about the volume of the liquid in the new c
taurus [48]

The same

Explanation:

If a liquid substance is transferred to a different container, the volume of the liquid in the new container will remain the same.

The volumes of liquids are fixed and does not change. Wherever they are contained, just like solids, they maintain their constant space.

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8 0
2 years ago
Dilution question In many of the experiments, you will be asked to prepare a standard solution by diluting a stock solution. You
riadik2000 [5.3K]

Answer:

We could do two 1:50 dilutions and one 1:4 dilutions.

Explanation:

Hi there!

A solution that is 1000 ug/ ml  (or 1000 mg / l) is 1000 ppm.

Knowing that 1 ppm = 1000 ppb, 100 ppb is 0.1 ppm.

Then, we have to dilute the stock solution (1000 ppm / 0.1 ppm) 10000 times.

We could do two 1:50 dilutions and one 1:4 dilutions (50 · 50 · 4 = 10000). Since the first dilution is 1:50, you will use the smallest quantity of the stock solution (if we use the 10.00 ml flask):

First step (1:50 dilution):

Take 0.2 ml of the stock solution using the third dispenser (20 - 200 ul), and pour it in the 10.00 ml flask. Fill with water to the mark (concentration : 1000 ppm / 50 = 20 ppm).

Step 2 (1:50 dilution):

Take 0.2 ml of the solution made in step 1 and pour it in another 10.00 ml flask. Fill with water to the mark. Concentration 20 ppm/ 50 = 0.4 ppm)

Step 3 (1:4 dilution):

Take 2.5 ml of the solution made in step 3 (using the first dispenser 1 - 5 ml) and pour it in a 10.00 ml flask. Fill with water to the mark. Concentration 0.4 ppm / 4 = 0.1 ppm = 100 ppb.

6 0
3 years ago
There are two steps in the usual industrial preparation of acrylic acid, the immediate precursor of several useful plastics. In
adell [148]

Answer:

The net change in enthalpy for the formation of one mole of acrylic acid from calcium carbide, water and carbon dioxide is 523.2 kJ.

Explanation:

Step 1:

CaC_2(s) + 2H_2O(g)\rightarrow C_2H_2(g) + Ca(OH)_2(s),\Delta H_1=414.0 kJ...[1]

Step 2 :

6C_2H_2(g) + 3CO_2(g) + 4H_2O(g)\rightarrow 5CH_2CHCO_2H(g) \Delta H_2=132.0kJ..[2]

Adding 6 × [1] and [2]:

6CaC_2(s) + 12H_2O(g)\rightarrow 6C_2H_2(g) + 6Ca(OH)_2(s)

6C_2H_2(g)+3CO_2(g)+16H_2O(g)\rightarrow 5CH_2CHCO_2H(g)

we get :

6CaC_2(s) + 8H_2O(g)+3CO_2(g)\rightarrow 5CH_2CHCO_2H(g)+ 6Ca(OH)_2(s),\Delta H'=?

\Delta H'=6\times \Delta H_1+\Delta H_2

\Delta H'=6\times 414.0 kJ+132.0kJ

\Delta H'=2,626 kJ

Energy released on formation of 5 moles of acrylic acid = 2,626 kJ

Energy released on formation of 1 mole of acrylic acid:

\frac{ 2,626 kJ}{5 } = 523.2 kJ

7 0
3 years ago
A solution has a [oh−] of 1 x 10−12. what is the poh of this solution? 2 7 10 12
Colt1911 [192]
POH = - log [ OH-]

pOH = - log [ 1 x 10⁻¹²]

pOH = 12
8 0
3 years ago
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Luda [366]
Answer is: <span>The general rule of thumb is that the smaller a substance's atoms and the stronger the bonds, the harder the substance will be. 
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8 0
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