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AURORKA [14]
3 years ago
14

What is largest value for the third quantum number

Chemistry
1 answer:
Kryger [21]3 years ago
4 0

Answer:

18.

Explanation:

3s^2 3p^6 d^10

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How many mL of 0.100 M NaCl would be required to make a 0.0365 M solution of NaCl when diluted to 150.0 mL with water?
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Answer:

54.75 mL

Explanation:

First calculate the number of moles of NaCl in the 150mL solution of NaCl

0.0365 moles should be present on 1000cm3 or 1dm3 of water.

1L = 1 dm3

1 mL = 1 / 1000 dm3

150 mL = 150/1000 dm3 = 0.15 dm3

If x moles are present in 0.15 dm3,

x/ 0.15 = 0.0365

We get x= 0.0365 × 0.15 mol

Now x amount of moles should be taken from the initial 0.100 M NaCl solution

So 0.1 moldm-3 = 0.0365× 0.15 mol / V

we get V = 0.05475 dm3

V= 0.05475 L

V= 54.75 mL

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A sample of hexane (C6H14) has a mass of 0.580 g. The sample is burned in a bomb calorimeter that has a mass of 1.900 kg and a s
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We first calculate the heat released:
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Heat of combustion = 27.7 / 0.0067 kJ/mol

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