5 M of NaCl is equivalent to 292,2 grams.
5 % NaCl is equal to 5 g NaCl in 100 g of a material.
0,16 mg NaCl is equivalent to 2,7.10-5 mol.
5 x 0.25 = 1.25 moles
The concentration is the same !
A solution of NaCl 1 M.
i = isotonic molar [glucose] / isotonic molar [NaCl] i = 14 M / 7 M = 2 i = isotonic molar [glucose] / isotonic molar [NaCl] i = 14 M / 7 M = 2 i = isotonic molar [glucose] / isotonic molar [NaCl] i = 14 M / 7 M = 2 i = isotonic molar [glucose] / isotonic molar [NaCl] i = 14 M / 7 M = 2
To prepare 1000 ml of 0.02 M NaCl solution, you would need 40 ml of 5 M NaCl solution, which you can calculate using the formula C1V1 = C2V2, where C1 is the concentration of the stock solution, V1 is the volume of the stock solution needed, C2 is the desired concentration, and V2 is the final volume. The dilution factor in this case would be 25, as you are diluting the 5 M solution 25 times to achieve the desired 0.02 M concentration.
The momentum of 5 kg m/s is equivalent to 25 joules.
To find the volume of 0.075 M NaCl solution that can be made, you can use the formula C1V1 = C2V2. Plugging in the values, we get (9.0 M)(450 mL) = (0.075 M)(V2). Solving for V2 gives V2 = 1.1 L. Therefore, 1.1 L of 0.075 M NaCl solution can be made by diluting 450 mL of 9.0 M NaCl.
2M NaCl is a equivalent to a solution with 116,88 g NaCl in 1 L water.
Since 1 m is equivalent to 100 cm, 5 m is equal to 500 cm.
Pure water will have the lowest boiling point because it does not contain any solute particles to elevate the boiling point. As the concentration of NaCl increases, the boiling point also increases due to an increase in the number of solute particles that disrupt the formation of water vapor. Therefore, 0.5 M NaCl will have a higher boiling point than pure water, followed by 1.0 M NaCl, and finally 2.0 M NaCl will have the highest boiling point.