answer: dispersed from the liquid so cold air can take its place
Answer:
We need 8.11 grams of glucose for this solution
Explanation:
Step 1: Data given
Molarity of the glucose solution = 0.300 M
Total volume = 0.150 L
The molecular weight of glucose = 180.16 g/mol
Step 2: Calculate moles of glucose in the solution
Moles glucose = molarity solution * volume
Moles glucose = 0.300 M * 0.150 L
Moles glucose = 0.045 moles glucose
Step 3: Calculate mass of glucose
MAss glucose = moles glucose* molecular weight of glucose
MAss glucose = 0.045 moles * 180.16 g/mol
MAss glucose = 8.11 grams
We need 8.11 grams of glucose for this solution
Answer:

Explanation:
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In this case, since the reaction between phosphorous and oxygen to form diphosphorous pentoxide is:

Thus, since phosphorous is in excess and oxygen and diphosphorous pentoxide are in a 5/2:1 mole ratio, we can compute the maximum moles of product as shown below:

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If H+ would have to dissociate from a substance and must be the only positive ion in the solution that substance must be an acid.
CH3CHO is an aldehyde called acetaldehyde.
CH3CH2OH is an alcohol called ethanol
CH3OCH3 is an ether called dimethyl ether.
and finally, the answer CH3COOH is a weak acid called acetic acid.
Answer:
Molar mass of Ca = 40 g / mol , given 123 g Ca is 123/40= 3.075 moles,
1 mole = 6.022 * 10^23 atoms, so 3.075 moles Ca= 18.51*10^23 atoms
Explanation: