To answer this question, you need to equalize the content of the cranberry for both solutions. The content could be count by multiplying the percentage with the volume of the cranberry juice. Then the calculation would be:
content1= content2
concentration1 * volume1= concentration2 * volume2
12%* 25 ml = 20%* volume2
volume2= 25ml * 12%/ 20%= 15ml
Answer: W
Explanation:
The cell in liquid W will look the smallest out of the four because of Osmosis.
Osmosis is a process by which water molecules move from an area of relatively higher concentration of water to an area with a relatively lower concentration through a selectively permeable membrane(cell membrane).
With W being saltier than the cell, water molecules will move from the cell to liquid W to balance the concentrations inside and outside the cell which will lead to the cell in W being smaller.
Answer:
The volume when the conditions were altered is 0.5109 L or 510.9 mL
Explanation:
Using the general gas equation,
P1 V1 / T1 = P2 V2 / T2
where;
P1 = 756 mmHg
V1 = 475 ml = 0.475 L
T1 = 23.5°C = 23.5 + 273K = 275.5 K
P2 = 722 mm Hg
T2 = 10°C = 10 + 273 K = 283 K
V2 = ?
Rearranging to make V2 the subject of the formula, we obtain:
V2 = P1 V1 T2 / P2 T1
V2 = 756 * 0.475 * 283 / 722 * 275.5
V2 = 101, 625.3 / 198911
V2 = 0.5109 L or 510.9 mL
Step one write the chemical equation for reaction
= Ag2O + 2(C10H10N4SO2)---> 2 ( AgC10H9N4SO2)
The reacting ratio of Ag2O to AgC10H9N4SO2 is 1:2 from the reaction above
step 2; find the number of moles of AgC10H9N4SO2
that is mass/molar mass
The molar mass of AgC10H9N4SO2 = 107.86 +(12 x10) + (1 x 9) + (4 x 14) +32 +(16 x2) =356.86g/mol
moles is therefore= 25g/356.86g/mol= 0.07moles
by use of mole ratio the moles of Ag2O= 0.0702=0.035moles
mass = moles x molar mass
the molar mass of Ag2O=231.72 g/mol
mass is therefore= 231.72g/mol x 0.035moles= 8.11grams
