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A hypertonic solution has a higher solute concentration compared to another solution, typically the cytoplasm of a cell. When a cell is placed in a hypertonic solution, water moves out of the cell to balance the solute concentrations, causing the cell to shrink. This process is known as osmosis and can lead to cell dehydration and impaired function.
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What is hysotonic?
"Hypotonic" refers to a solution that has a lower concentration of solutes compared to another solution. In biology, a hypotonic solution has a lower concentration of solutes outside of a cell compared to inside, causing water to move into the cell and potentially cause it to swell and burst.
What is the molarity of 4.50 moles of Na-Br dissolved in water to make 3.00 L of solution?
To find the molarity (M) of a solution, you use the formula: M = moles of solute / liters of solution. Here, you have 4.50 moles of NaBr and 3.00 L of solution. Therefore, the molarity is M = 4.50 moles / 3.00 L = 1.50 M. Thus, the molarity of the NaBr solution is 1.50 M.
What is of 25 m mol solution?
A 25 millimole (m mol) solution means there are 25 millimoles of solute in every liter of solution. It is a unit used to express the concentration of a solute in a solution.
What is the molarity of a solution which contains 0.120 moles of K2CrO4 in 5.0L of solution?
The Molarity is equal to the number of moles divided by the liters. M=mol/L M=0.120 mol/5.0 L M=0.024 so the molarity is 0.024 M
How many milliliters of 2 M NaCl solutions are required to make 1 liternof 0.4 M NaCl solution?
To prepare 1 liter of a 0.4 M NaCl solution from a 2 M NaCl solution, you can use the dilution formula, (C_1V_1 = C_2V_2), where (C_1) is the concentration of the stock solution, (V_1) is the volume of the stock solution needed, (C_2) is the concentration of the diluted solution, and (V_2) is the final volume of the diluted solution. Plugging in the values: (2 , \text{M} \times V_1 = 0.4 , \text{M} \times 1000 , \text{mL}). Solving for (V_1) gives (V_1 = \frac{0.4 \times 1000}{2} = 200 , \text{mL}). Thus, 200 mL of the 2 M NaCl solution is required.
What is hysotonic?
"Hypotonic" refers to a solution that has a lower concentration of solutes compared to another solution. In biology, a hypotonic solution has a lower concentration of solutes outside of a cell compared to inside, causing water to move into the cell and potentially cause it to swell and burst.
What is the m/v of the solution in question?
The m/v of a solution is the mass of the solute divided by the volume of the solution, multiplied by 100.
Will a 0.5 M solution have a higher freezing point than a 0.75 M solution of the same substance?
No, the 0.75 M solution will have a higher freezing point. The freezing point depression is directly proportional to the molality of the solution, so a higher concentration solution will have a greater effect on lowering the freezing point.
preparation of 0.1 m of NaCL ML SOLUTION AND ITS LAP REPORT?
A solution of NaCl 1 M.
What is the formula for calculating the molality (m) of a solution?
The formula for calculating the molality (m) of a solution is: molality (m) moles of solute / kilograms of solvent.
How do you prepare 1 M solution from 6 M solution?
It's simple to dilute a solution, just add more water. If you have one liter of 6 molar solution, and add another 5 liters of water, it becomes a 1 molar solution. You can also use M1V1 = M2V2 formula for dilution. For example, if you want to prepare 1 M of solution in 1 litre of water, then how much volume you need from the initial 6 M solution? Simply use the equation, (6 M) x (V1) = (1 M) x (1 litre) ---> V1 = (1 M) x (1 litre)/(6 M) = 0.167 L or 167 ml
What volume of a 2.5 M stock solution of acetic acid (HCHÀO) is required to prepare 100.0 milliliters of a 0.50 M acetic acid solution?
To prepare a 0.50 M acetic acid solution, you would need to dilute the 2.5 M stock solution. By using the formula M1V1 = M2V2, you can calculate the volume of the stock solution needed as: (0.5 M)(100.0 mL) = (2.5 M)(V2), where V2 is the volume of the stock solution needed. So, V2 = (0.5 M x 100.0 mL) / 2.5 M = 20.0 mL. Therefore, you would need 20.0 mL of the 2.5 M stock solution to prepare the desired 100.0 mL of 0.50 M acetic acid solution.
How much 1 M NaHCO3 solution does it take to fully react with 5 ml of a 1 M solution of HCl?
First , write down the BALANCED reaction eq'n HCl + NaHCO3 = NaCl + H2O + CO2 We note the molar ratios are all 1:1 :: 1:1:1 So using the eq'n moles = [conc] X vol(mL) / 1000 NB The '1000' is needed to convert 'mL' to 'L' to agree with 'M' (mol/L). So calculate the moles(HCl) used. mol(HCl) = 1.0 M X 5mL / 1000 = 0.005 moles (This is equivalent to ;1; ratio above). On an equivlance basis only 0.005 moles (NaHCO3) will react. So 0.005mol(NaHCO3) = 1.o M X vol(mL) / 1000 Algebraically rearrange vol(mL) = 0.005 mol X 1000 / 1.0 M vol(mL) = 5 mL ( of NaHCO3 is needed).
How many milliliters of a 0.266 m lino3 solution are required to make 150.0 ml of 0.075 m lino3 solution?
To find out how many milliliters of the 0.266 M LiNO3 solution are needed, you can use the formula C1V1 = C2V2, where C1 is the concentration of the first solution, V1 is the volume of the first solution, C2 is the concentration of the second solution, and V2 is the volume of the second solution. Plugging in the values, you can solve for V1, which will give you the volume of the 0.266 M LiNO3 solution needed to make 150.0 ml of 0.075 M LiNO3 solution.
What is the molarity of 4.50 moles of Na-Br dissolved in water to make 3.00 L of solution?
To find the molarity (M) of a solution, you use the formula: M = moles of solute / liters of solution. Here, you have 4.50 moles of NaBr and 3.00 L of solution. Therefore, the molarity is M = 4.50 moles / 3.00 L = 1.50 M. Thus, the molarity of the NaBr solution is 1.50 M.
What is of 25 m mol solution?
A 25 millimole (m mol) solution means there are 25 millimoles of solute in every liter of solution. It is a unit used to express the concentration of a solute in a solution.
A 0.100 M solution of K 2SO 4 would contain the same total ion concentration as what solutions?
It would contain the same total ion concentration as 0.0750 M Na3PO4.
