The molar mass of glucose is 180.16g/mol. So to convert grams of glucose into moles of glucose, divided the mass in grams by the molar mass, and grams will cancel out leaving you with moles. 120g / 180.16 g/mol = 0.666 moles
Answer:
Land and sea breeze
Explanation:
This uneven warming of the land during the day results in the phenomenon called land and sea breeze.
- During the day, the land heats up faster than the nearby ocean.
- Ocean is made up of water with a high specific heat capacity.
- This derives pressurized air near the surface of the land to the ocean.
- At night, the ocean does not easily lose heat after the land has lost its own.
- Pressurized air moves from the ocean to the land.
- This causes the warming of the land during the night.
To find how many atoms are in the compound, we first have to find how many moles there are. The numbers in the subscript tell you how many moles of each element are present in the compound. Therefore, to find the number of total moles, we can add up each of the numbers in the subscript:
Cu₂Mg₁H₁O₁ *when there is no number under an element, it is implied that there is 1 mole of the element
2+1+1+1=5 moles
We can then use Avogadro's number and dimensional analysis to find how many atoms are in the compound (Avogadro's number is 6.02x10^23):
5 moles (6.02x10^23/1 mole)= 3.01x10^24 atoms
<u>Answer:</u> The volume of NaOH solution required to reach the half-equivalence point is 0.09 mL
<u>Explanation:</u>
The chemical equation for the dissociation of butanoic acid follows:
The expression of for above equation follows:
We are given:
Putting values in above expression, we get:
Neglecting the negative value because concentration cannot be negative
To calculate the volume of base, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is butanoic acid
are the n-factor, molarity and volume of base which is NaOH.
We are given:
Putting values in above equation, we get:
Hence, the volume of NaOH solution required to reach the half-equivalence point is 0.09 mL