1. A
2. base
3. strong ones have more h30 weak ones have more OH
4. strong ones have more OH weak ones have more H30
5. <span>One of the main differences between acids and bases is that acids have a pH that is less than 7 and bases have a pH that is greater than 7. When dissolved in water, acids are substances that will cause the concentration of hydrogen ions (H+) to increase. Bases, when dissolved in water, will instead cause the number of hydroxide ions (OH-) to increase.6.Idk </span>
Answer:

Explanation:
Hello!
In this case, according to the Avogadro's number, it is possible to compute the atoms of Kr in 2.00 moles as shown below:

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For all three questions, we will use the fact that
- molarity = (moles of solute)/(liters of solution)
1) For 175 mL of solution at 0.203 M, this means that:
- 0.203 = (moles of solute)/0.175
- moles of solute = 0.035523 mol
Considering the hydrochloric acid solution, if we have 0.035523 mol, then:
- 6.00 = 0.035523/(liters of solution)
- liters of solution = 0.035523/6.00 = 0.0059205 = <u>5.92 mL (to 3 sf)</u>
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2) If there is 20.3 mL = 0.0203 L, then:
- 8.20 = (moles of solute)/0.0203
- moles of solute = 0.16646 mol
This means that the molarity of the diluted solution is:
- 0.16646/(0.200) = <u>0.832 M (to 3 sf)</u>
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3) If we need 1.50 L of 0.700 M solution, then:
- 0.700 = (moles of solute)/1.50
- moles of solute = 1.05 mol
Considering the 9.36 M acid solution, from which we need 1.05 mol of perchloric acid from,
- 9.36 = 1.05/(liters of solution)
- liters of solution = 1.05/9.36, which is 0.11217948717949 L, or <u>112 mL (to 3 sf)</u>
Answer:
Molarity =5.32 M
Explanation:
Given data:
Mass of glucose = 239 g
Volume = 250 mL (250 /1000 = 0.25 L)
Molarity = ?
Solution;
Formula:
Molarity = number of moles / volume in litter
Number of moles:
Number of moles = mass/ molar mass
Number of moles = 239 g / 180.2 g/mol
Number of moles = 1.33 mol
Molarity:
Molarity = number of moles / volume in litter
Molarity = 1.33 mol / 0.25 L
Molarity =5.32 M
Answer:
You would get 19.969 moles
Explanation: