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RideAnS [48]
3 years ago
6

Water will always move from ________ areas to ________ areas, if the membrane is permeable to water. isotonic, hypotonic hyperto

nic, hypotonic hypertonic, isotonic hypotonic, hypertonic
Chemistry
2 answers:
ryzh [129]3 years ago
4 0

Answer:

From hypotonic to hypertonic

Explanation:

Water diffusion is a phenomenon that occurs when a solute (eg. a salt) is present in different concentrations in different areas. Because the <em>concentration is inversely proportional to volume </em>(meaning that the higher the volume, the lower the concentration), water will move from areas with lower concentration <u>(hypotonic)</u> to areas with higher concentration <u>(hypertonic)</u>, so as to match the concentrations.

docker41 [41]3 years ago
4 0

Answer:

Water will move from hypotonic to hypertonic solutions

Explanation:

When a hypotonic solution is separated by a permeable membrane from another solution with more solute (hypertonic), water will move across the membrane until both solutions have the same concentration: when water comes into the compartment with hypertonic solution, this solute dissolves.

Water will not move from hypertonic to hypotonic without energy addition.

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3 years ago
A 40.0 mL sample of 0.18 M HCI is titrated with 0.36 M CoHsNH2. Dctermine the pH at these points: At the beginning (before base
Whitepunk [10]

Answer:

at the beginning:

pH = 0.745

Explanation:

HCl is a strong acid, so:

  • HCl + H2O  → H3O+  +  Cl-

       0.18 M             0.18        0.18.....equilibrium

before base is added:

∴ [ H3O+ ] ≅ <em>C </em>HCl = 0.18 M

⇒ pH = - Log [ H3O+ ] = - Log ( 0.18 )

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8 0
3 years ago
It has been suggested that the surface melting of ice plays a role in enabling speed skaters to achieve peak performance. Carry
vesna_86 [32]

Explanation:

Relation between pressure, latent heat of fusion, and change in volume is as follows.

          \frac{dP}{dT} = \frac{L}{T \times \Delta V}

Also, \frac{L}{T} = \Delta S^{fusion}_{m}

where, \Delta V^{fusion}_{m} is the difference in specific volumes.

Hence,    \frac{dP}{dT} = \frac{\Delta S^{fusion}_{m}}{\Delta V^{fusion}_{m}}

As, \Delta S^{fusion}_{m} = \frac{L}{T} = \frac{6010}{273.15} = 22.0 J/mol K

And,   \Delta V^{fusion}_{m} = \frac{M}{d_{H_{2}O}} - \frac{M}{d_{ice}} ...... (1)

where,    d_{H_{2}O} = density of water

              d_{ice} = density of ice

             M = molar mass of water = 18.02 \times 10^{-3} kg

Therefore, using formula in equation (1) we will calculate the volume of fusion as follows.

        \Delta V^{fusion}_{m} = \frac{M}{d_{H_{2}O}} - \frac{M}{d_{ice}}

                       = \frac{18.02 \times 10^{-3}}{997} - \frac{18.02 \times 10^{-3}}{920}  

                       = -1.51 \times 10^{-6}        

Therefore, calculate the required pressure as follows.

              \frac{dP}{dT} = \frac{22}{-1.51 \times 10^{-6}}

                              = 1.45 \times 10^{7} Pa/K

or,                           = 145 bar/K

Hence, for change of 1 degree pressure the decrease is 145 bar  and for 4.7 degree change dP = 145 \times 4.7 bar

                              = 681.5 bar

Thus, we can conclude that pressure should be increased by 681.5 bar to cause 4.7 degree change in melting point.

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If u add u will get ur answer
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Read 2 more answers
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