Wiki User
∙ 13y agoDid you mean 250 mL of 0.15 M Na2CO3? (It's impossible to make a 15 M Na2CO3 solution, as Na2CO3 is not that soluble.) Yes, you can assume that this will be an aqueous solution.
Steps.
1. Calculate the mass of solid Na2CO3 needed.
2. Place this mass of Na2CO3 in the volumetric flask.
3. Add some water and swirl to dissolve the Na2CO3.
4. Carefully add more water until the total volume of solution is 250 mL, as indicated by the line etched on the neck of the volumetric flask.
250 mL x 1 L x 0.15 mol Na2CO3 x 105.99 g Na2CO3 = 4.0 g Na2CO3 needed
........... 1000 mL ......... 1 L ..................1 mol Na2CO3
Wiki User
∙ 13y agoTo prepare a 250ml 15M NaCO3 solution in a volumetric flask without using the flask to dilute, you would first calculate the mass of NaCO3 needed based on its molar mass. Then, dissolve this mass in a small volume of water in a separate container and transfer it quantitatively to the 250ml volumetric flask, and then add enough water to make up to the mark. Finally, mix thoroughly to ensure homogeneous distribution.
If the color change is too dark, it may be difficult to accurately determine the endpoint in a titration. This can result in an overestimation or underestimation of the NaOH concentration. To address this, dilute the solution or use an indicator that changes color at a more suitable endpoint for the titration.
The pH of the weak acids is lower than 7 (neutral) but higher than strong acids (which can be around 1). This is because pH is a measure of hydrogen ion concentration and weak acids only partially dissociate. It is a logarithmic scale, so pH=-log[H]+ so a larger concentration of hydrogen means a lower pH and a smaller concentration, a higher pH. Thus, when water is added to a weak acid, the concentration of hydrogen ions will decrease and the pH will become greater.
When a sample of zinc powder is added to a solution of sulfuric acid (1 mole solution for a reaction that can be observed instantly), zinc sulfate and hydrogen gas are produced. Immediately, hydrogen can be observed effervescing from the solution. The chemical equation for this reaction is as follows: H2SO4(aq) + Zn(s) = ZnSO4(aq) + H2(g) (aq) = aqueous When a sample of zinc powder is added to a solution of sulfuric acid (1 mole solution for a reaction that can be observed instantly), zinc sulfate and hydrogen gas are produced. Immediately, hydrogen can be observed effervescing from the solution. The chemical equation for this reaction is as follows: H2SO4(aq) + Zn(s) = ZnSO4(aq) + H2(g) (aq) = aqueous Best answer: Zn(s)+H2SO4(aq)------- ZnSO4(aq)+H2(g)
Potassium 31 is potassium's normal state, the way you would find it on the Periodic Table. I presume potassium 41 is an isotope of potassium. (An isotope is the same atom, but it has a different number of neutrons, so the mass differs.)
Water is the most common liquid in the world, covering about 70% of the Earth's surface. It is essential for life and is found in oceans, lakes, rivers, and even underground.
because they ran out of gold, i presume
Presume is an action. Therefore the word is already a verb.Presumes, presuming and presumed are other verbs, depending on the tense.
I presume that she will arrive on time for the meeting.
I presume that you were talking about Presume. Not persume , as it is incorrect.
Because of its looks, it looks sad. Like it is crying. That is what i presume
Even though she knew him well, she did not want to presume how he felt about the situation. Presume is a verb.
I presume that the word "presume" is a verb.
Even though she knew him well, she was careful not to presume what he was thinking. He didn't want to presume he knew what would happen next.
I would presume America simply because it's an American show.
The noun forms of the verb to presume are presumption and the gerund, presuming.
I presume from category that the question ask about the heat of dissolution. Assume the solvent and solution is at thermal equilibrium. Adding more solvent would yield change in interaction for non ideal solution and thus it could yield increase or decrease of temperature depend on the infinite heat of solution of the solute that we interested in. Adding more solvent would not yield temperature change for ideal solution.
I presume from category that the question ask about the heat of dissolution. Assume the solvent and solution is at thermal equilibrium. Adding more solvent would yield change in interaction for non ideal solution and thus it could yield increase or decrease of temperature depend on the infinite heat of solution of the solute that we interested in. Adding more solvent would not yield temperature change for ideal solution.