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oee [108]
3 years ago
9

26. Which waves have wavelengths longer than those of visible light? Give an example of how each kind

Chemistry
1 answer:
Kipish [7]3 years ago
8 0

Answer:

Infrared radiation lies between the visible and microwave portions of the electromagnetic spectrum. Infrared waves have wavelengths longer than visible and shorter than microwaves, and have frequencies which are lower than visible and higher than microwaves.

Explanation:

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decreases

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How Many Moles Of HCl Need To Be Added To 150.0 ML Of 0.50 M NaZ To Have A Solution With A PH Of 6.50
Aleks04 [339]

The number of mole of HCl needed for the solution is 1.035×10¯³ mole

<h3>How to determine the pKa</h3>

We'll begin by calculating the pKa of the solution. This can be obtained as follow:

  • Equilibrium constant (Ka) = 2.3×10¯⁵
  • pKa =?

pKa = –Log Ka

pKa = –Log 2.3×10¯⁵

pKa = 4.64

<h3>How to determine the molarity of HCl </h3>
  • pKa = 4.64
  • pH = 6.5
  • Molarity of salt [NaZ] = 0.5 M
  • Molarity of HCl [HCl] =?

pH = pKa + Log[salt]/[acid]

6.5 = 4.64 + Log[0.5]/[HCl]

Collect like terms

6.5 – 4.64 = Log[0.5]/[HCl]

1.86 = Log[0.5]/[HCl]

Take the anti-log

0.5 / [HCl] = anti-log 1.86

0.5 / [HCl] = 72.44

Cross multiply

0.5 = [HCl] × 72.44

Divide both side by 72.44

[HCl] = 0.5 / 72.4

[HCl] = 0.0069 M

<h3>How to determine the mole of HCl </h3>
  • Molarity of HCl = 0.0069 M
  • Volume = 150 mL = 150 / 1000 = 0.15 L
  • Mole of HCl =?

Mole = Molarity x Volume

Mole of HCl = 0.0069 × 0.15

Mole of HCl = 1.035×10¯³ mole

<h3>Complete question</h3>

How many moles of HCl need to be added to 150.0 mL of 0.50 M NaZ to have a solution with a pH of 6.50? (Ka of HZ is 2.3 x 10 -5 .) Assume negligible volume of the HCl

Learn more about pH of buffer:

brainly.com/question/21881762

7 0
2 years ago
How many formula units of sodium chloride (NaCl) may theoretically be produced from 13.0 g FeCl3?
algol [13]

Answer:

1.445\times 10^{23} formula units of sodium chloride can be formed from 13.0 gram of ferric chloride.

Explanation:

Mass of ferric chloride = 13.0 g

Moles of ferric chloride = \frac{13.0 g}{162.5 g/mol}=0.08 mol

1 mole of ferric chloride has three moles of chloride ions.Then 0.08 moles of ferric chloride will have :

3\times 0.08 mol=0.24 mol of chloride

Na^++Cl^-\rightarrow NaCl

1 mole of sodium ion reacts with 1 mole of chloride ion to form 1 mole of NaCl. Then 0.24 moles of chloride ion will give:

\frac{1}{1}\times 0.24 mol=0.24 mol of NaCl

1 mole = N_A=6.022\times 10^{23} molecules/ atoms

Number of NaCl molecules in 0.24 moles :

=6.022\times 10^{23}\times 0.24=1.445\times 10^{23} molecules

1.445\times 10^{23} formula units of sodium chloride can be formed from 13.0 gram of ferric chloride.

4 0
3 years ago
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