Answer:
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Explanation:
9.) 10-2.76 =0.0174 [H30+]= 1.74*10-3 M
10.)10-3.65=0.00224 [H3O+] =2.24*10-2 M
11.)10-3.65=0.00224 [OH-]= 2.224*10-4M
12.)10-6.87=0.00000135 [OH-]= 1.35*10-7M
Answer:
M is increased loudness
O is decreased loudness
N is decreased pitch
P is increased pitch
Louder soundwaves ten to have longer and deeper waves visa versa.
Increased pitch tends to have shorter soundwaves. visa versa
The molality of the solution = 17.93 m
<h3>Further explanation</h3>
Given
6.00 L water with 6.00 L of ethylene glycol(ρ=1.1132 g/cm³= 1.1132 kg/L)
Required
The molality
Solution
molality = mol of solute/ 1 kg solvent
mol of solute = mol of ethylene glycol
- mass of ethylene glycol :
= volume x density
= 6 L x 1.1132 kg/L
= 6.6792 kg
= 6679.2 g
- mol of ethylene glycol (MW=62.07 g/mol)
=mass : MW
=6679.2 : 62.07
=107.608
6 L water = 6 kg water(ρ= 1 kg/L)