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irina [24]
3 years ago
11

The burning of a sample of propane generated 1 04.6 kJ of heat. All of this heat was used to heat 500.0 g of water that had an i

nitial temperature of 20.0/C. What was the final temperature of the water?
Chemistry
1 answer:
Paul [167]3 years ago
3 0

Answer: 70.0°C

Explanation:

Quantity of heat = Mass * Specific heat * Change in temperature

Quantity of heat = 104.6 KJ

Mass = 500.0 g

Specific heat of water is 4.18 J/g°C

Change in temperature assuming final temperature is x = x - 20

Units should be in grams and joules:

104,600 = 500 * 4.18 * (x - 20)

104,600 = 2,090 * (x - 20)

x - 20 = 104,600/2,090

x = 104,600/2,090 + 20

x = 69.8

= 70.0°C

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Infrared (IR) spectroscopy is used to identify functional groups within a molecule. Click on the peak that corresponds to the st
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Complete Question

The complete question is shown on the first uploaded image

Answer:

The correct place to click on any of the  double peak showing between  2500 and 3000cm^{-1} on the uploaded question

Explanation:

The group which is being highlighted is an Aldehyde  functional group denoted by this structure (=C-H)

This groups stretch on the infrared spectrometer gives two medium intensities peaks

Its stretch comes at  2820 - 2850 cm^{-1}  peak 1

and at  2720 - 2750 cm^{-1}  peak 2

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2 years ago
the burning of wood is a chemical reaction between cellulose molecules in the wood and oxygen the products are carbon dioxide an
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Put the <em>wood and excess oxygen in a closed contai</em>ner that contains some device that can create a spark (to start the reaction).

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3 years ago
1. Using the Slater rule, determine the effective nuclear charge of platinum.
AleksandrR [38]

Answer:

Z* = 3.55

Explanation:

Slater rule says that:

Z*= Z - S

Z* be the nuclear effective charge

Z is the nuclear charge

S is the shielding constant

First we write the electronic configuration of platinum:1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{10} 4s^{2} 4p^{6} 4d^{10} 5s^{2} 5p^{6} 4f^{14} 5d^{9} 6s^{1}

The first Slater rule says that we need to group:

(1s^{2}) (2s, 2p)^{8} (3s, 3p)^{8} (3d^{10}) (4s, 4p)^{8} (4d^{10}) (5s, 5p)^{8} (4f^{14}) (5d^{9}) (6s^{1})

The second rule says that the electrons to the right are not shielding, but we are going to solve the exercise for the last level (6s), so we don't have electrons to the right.

For the third rule we have two considerations, if is ns or np and if is nd or nf:

For our case, we have an electro that is in ns, so the rule says that

-electrons within same group shield 0.35, except the 1s which shield 0.30

-electrons within the n-1 group shield 0.85

-electrons within the n-2 or lower groups shield 1.00

Now we can proceed with the calculation:

The first consideration in the third rule does not apply as we only have one electron on this level.

The second consideration will be as follow for the level 5, where we have 17 electrons.

Finally the third consideration will be for levels 1, 2, 3 and 4, where we have 14 for 4f, 10 for 4d, 8 for 4s and 4p, 10 for 3d, 8 for 3s and 3p, 8 for 2s and 2p and finally 2 for 1s, which gives 60 electrons.

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3 years ago
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