Answer: is neither an acid nor a base
Explanation: Water is a universal solvent which means it can dissolve most of the substances in it.
Water has high thermal heat capacity , which means large heat is required to heat the water.
Water is not always pure as it gets contaminated by various pollutants present in the atmosphere such as gases, bacteria and suspended matter.
Water is an amphoteric substance which can act as both acid and base, thus can donate and acept [texH^+[/tex] ions.Thus it is neither an acid nor a base.
Here water is accepting a proton, thus it acts as base.
Here water is donating a proton, thus it acts as acid.
Observe to to gather facts, by paying close attention towards what you are working on.
meanwhile, reference is the act or process on reaching to your conclusion, based on facts you already know .
In a food chain we arrange the energy in the form of a pyramid.
The producers are on the base of pyramid and then consumers are towards peak.
in the given food chain grass is being eaten by grasshopper which are food of birds.
Grasshoppers are also eaten up by Hawks. so both brids and hawks are feeding upon grasshoppers thus the amount of energy transferred from the grass to the grasshopper is the same as the amount of energy transferred from the grasshopper to the bird.
<u>Answer:</u> The number of moles of weak acid is 
moles.
<u>Explanation:</u>
To calculate the moles of KOH, we use the equation:
We are given:
Volume of solution = 43.81 mL = 0.04381 L (Conversion factor: 1L = 1000 mL)
Molarity of the solution = 0.0969 moles/ L
Putting values in above equation, we get:
The chemical reaction of weak monoprotic acid and KOH follows the equation:
By Stoichiometry of the reaction:
1 mole of KOH reacts with 1 mole of weak monoprotic acid.
So, 
of KOH will react with = 
of weak monoprotic acid.
Hence, the number of moles of weak acid is moles.
Answer:
V KOH = 41 mL
Explanation:
for neutralization:
- ( V×<em>C </em>)acid = ( V×<em>C </em>)base
∴ <em>C </em>H2SO4 = 0.0050 M = 0.0050 mol/L
∴ V H2SO4 = 41 mL = 0.041 L
∴ <em>C</em> KOH = 0.0050 N = 0.0050 eq-g/L
∴ E KOH = 1 eq-g/mol
⇒ <em>C</em> KOH = (0.0050 eq-g/L)×(mol KOH/1 eq-g) = 0.0050 mol/L
⇒ V KOH = ( V×<em>C </em>) acid / <em>C </em>KOH
⇒ V KOH = (0.041 L)(0.0050 mol/L) / (0.0050 mol/L)
⇒ V KOH = 0.041 L
