The term which is described as a long, narrow depression in the ocean floor would be ocean trench. They <span>are hemispheric-scale long but narrow topographic depressions of the sea floor. They are also the deepest parts of the ocean floor. Hope this answers the question.</span>

8 0

2 years ago

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The vapor pressure of water at 25.0°C is 23.8 torr. Determine the mass of glucose (molar mass = 180 g/mol) needed to add to 500.

mrs_skeptik [129]

According to Raoult's law the relative lowering of vapour pressure of a solution made by dissolving non volatile solute is equal to the mole fraction of the non volatile solute dissolved.

the relative lowering of vapour pressure is the ratio of lowering of vapour pressure and vapour pressure of pure solvent

$\frac{p^{0-}p}{p^{0}}=x_{B}$

Where

xB = mole fraction of solute=?

$p^{0}=23.8torr$

p = 22.8 torr

$x_{B}=\frac{23.8-22.8}{23.8}=0.042$

mole fraction is ratio of moles of solute and total moles of solute and solvent

moles of solvent = mass / molar mass = 500 /18 = 27.78 moles

putting the values

$molefraction=\frac{molessolute}{molesolute+molessolvent}$

$0.042=\frac{molessolute}{27.78+molessolute}$

$1.167+0.042(molesolute)=molessolute$

$molessolute=1.218$

mass of glucose = moles X molar mass = 1.218 X 180 = 219.24 grams

3 0

3 years ago

Given the reaction:CO2 + 6 H2O==> C6H12O6 + 6 O2

kari74 [83]

For every 1 mole of C6H12O6, you need 6 moles of water. Multiply the 2.5 moles you are trying to make by the 6 of water you need, and 4) 15 is your answer.

5 0

3 years ago

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