D...
W...
m...
A student runs two experiments with a constant-volume "bomb" calorimeter
containing 1400. g of water (see sketch at right).
sam....
First, a 5.000 g tablet of benzoic acid (C6H5CO₂H) is put into the "bomb" and
burned completely in an excess of oxygen. (Benzoic acid is known to have a heat of
combustion of 26.454 kJ/g.) The temperature of the water is observed to rise from
13.00 °C to 32.28 °C over a time of 10.7 minutes.
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an Am...
Next, 4.900 g of ethane (C₂H6) are put into the "bomb" and similarly completely
burned in an excess of oxygen. This time the temperature of the water rises from
13.00 °C to 44.56 °C.
Use this information, and any other information you need from the ALEKS Data
resource, to answer the questions below about this reaction:
-
stirrer
Is this reaction exothermic, endothermic, or neither?
ก
thermometer
chemical reaction
"bomb"
A "bomb" calorimeter.
If you said the reaction was exothermic or endothermic, calculate the amount of heat that was
released or absorbed by the reaction in the second experiment.
2C₂H,(g) + 70₂(g) 4CO₂(g) + 6H₂O(g)
Be sure any of your answers that are calculated from measured data are rounded to the correct number of significant digits.
Note for advanced students: it's possible the student did not do these experiments sufficiently carefully, and the values you
calculate may not exactly match published values for this reaction.
exothermic
Oendothermic
neither
water
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The pH of the resulting solution is 2.82.

pH is the measure of acidity and alkalinity of a solution , It has a range of 1 to 14.

HCl          +                   NaOH      →         NaCl       +         H₂O

0.0040                         .0025                 0                       0

0.0040 -0.0025             0                   0.0025          0.0025

Moles of HCl = 0.0015

As 0.0025 moles of NaOH reacts with 0.0025 moles of HCl leaving behind 0.0015 moles of HCl

Concentration of HCl = 0.0015/ 1 = 0.0015 M

[ H⁺] = 0.0015 M

pH = -log [ H⁺]

pH = -log [ 0.0015]

pH = 2.82

Therefore the pH of the resulting solution is 2.82.

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Answer:

1.46 moles NaCl

Explanation:

To find the moles of NaCl, you need to multiply the given value (0.731 moles BaCl₂) by the mole-to-mole ratio of the two relevant molecules. This ratio is made up of the coefficients in the balanced reaction. The numerator should contain moles NaCl to allow for the cancellation of moles BaCl₂. The final answer should have 3 sig figs to reflect the given value.

1 BaCl₂(aq) + Na₂SO₄(aq) ---> BaSO₄(s) + 2 NaCl(aq)

0.731 mole BaCl₂           2 moles NaCl
---------------------------  x  -------------------------  = 1.46 moles NaCl
                                         1 mole BaCl₂

If 0.731 mol of BaCl₂ is allowed to react with an excess of Na₂SO₄, 1.462  moles of NaCl will be produced.

The balanced chemical equation is:

BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq)

The mole ratio of BaCl₂  to NaCl is 1:2. This means that 1 mole of BaCl2 will produce 2 moles of NaCl.

Moles of BaCl₂  = 0.731 mol

Mole ratio BaCl₂:NaCl = 1:2

Moles of NaCl = 0.731 mol * 2 = 1.462 mol

If we have 0.731 moles of BaCl₂, then we will produce 2 * 0.731 = 1.462 moles of NaCl.

Therefore, 1.462 moles of NaCl will be produced.

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Answer:

The Chinese civilization really has an exact chronology, and the historical period began in the first year of the Western Zhou Dynasty in 841 BC. Before that, only the lineage records of the kings were recorded. Therefore, the earliest origin of Chinese civilization has not yet been finalized. Although there are many records about the Xia Dynasty in the traditional literature, due to the late completion of the book, there is no recognized evidence of the existence of the Xia Dynasty and its previous history. Therefore, some people in the modern and modern historical circles have always questioned the existence of the Xia Dynasty. sex.

Explanation:

Answer:

9.487 L

Explanation:

argon  mole wt = 39.948 gm/mole

             14.3 gm  / 39.948 gm/mole   = .357965 moles of argon

Use Ideal Gas Law:

PV = n RT     where   R = .082057  L-atm / K-mol

V =  .357965 * .082057 * 436 / 1.35  = 9.487 L