A chemistry question Sodium of Alkali Metals, and a multiple choice question of Sodium Alkali Metals

1.The mass of sodium that produces H2 is m

.2na + 2h2o = 2naoh + h2...m...

m=2.Quality of Sodium Oxide:

3.The sodium that is oxidized is the mass of sodium contained in the sodium oxide: grams.

The selection is C2Na+2H2O=2NaOH+H2|x...

46/x=2/

Get x = oxidized as:

Explain that there is that there is, that is.

So it turned out to be oxidized. 1g

Summary. Answer, the amount of sodium is the amount of the substance, the calculation process: the molar mass of sodium is 23g mol, and 1mol of sodium produces 1mol NaOH then generated.

Some metal sodium reacts with water, and sodium reacts completely to find: + (1) the amount of sodium + (2) to be generated.

Hello, classmate, the question is incomplete, please send a screenshot of the original question, so that you can solve the problem quickly and effectively<>

Answer, the amount of sodium substance is, the calculation process: the molar mass of sodium liquid potato is 23g mol, and 1mol of sodium produces 1mol NaOH from acorns, as discussed then generated.

Answer, the amount of sodium substance is, the calculation process: the molar mass of sodium liquid potato is 23g mol, and 1mol of sodium produces 1mol NaOH from acorns, as discussed then generated.

2molna

2molna produces 1mol of hydrogen, then the volume of hydrogen produced is.

Ca and B are both phenomena of the reaction of Na and water, so it is feasible to observe. At the same time, the NaOH generated by the reaction increased the concentration of OH- in the solution, resulting in the precipitation of Ca(OH)2 from saturated clarified lime water.

Na cannot displace any metal in aqueous solution, and the activity of Ca is in front of Na, and Na cannot displace Ca2.

Sodium C reacts with water and may melt into small balls and swim around on the liquid surface and form gases, while it is impossible to displace any metal, but the increase in the concentration of Oh(-) leads to the precipitation of Ca(OH)2 from saturated clarified lime water.

Sodium C reacts with water first, there is phenomenon A, hydrogen and NaOH are obtained, and there is phenomenon B

Since the water is reacted, Ca(OH) is precipitated, and there is phenomenon D

NaOH then reacts with Ca(OH).

b Sodium reacts more violently with table salt than with water – that's wrong.

Reasons for choosing B are as follows:

A is correct because the sodium hydroxide produced is a base and turns red when it encounters phenolphthalein b is incorrect There is no reason for sodium to react faster than water in salt water, b is pure nonsense c correct The density of sodium is smaller than that of water.

D is correct It can be seen from 2 perspectives: 1. The concentration of Na+ in the solution increases, and from the perspective of KSP, NaCl precipitate is generated (please don't think that NaCl cannot have precipitate!) Whether any substance precipitates or not is to be judged according to KSP) 2. The reaction consumes water, and salt precipitation in saturated salt water is turbid.

A, 2Na+2H O=2NaOH+H, NaOH is a strong base, which will make the phenolphthalein reagent red.

B, false, first of all, because NaCl is dissolved in the salt water, that is to say, the same volume of salt water and water, the concentration of water is compared: salt water, and Na is reactive with water, so less water, it is slow. It can also be said that there will be impurity ions Na+ and Cl - in the salt water, which will collide with Na atoms like HO molecules, while they do not react with Na, so the collision between water and Na is relatively less.

C, because 2Na + 2H O = 2NaOH + H this reaction exothermic reaction (you can touch the test tube to feel the temperature), and the melting point of Na is very low, so it will become a liquid state, due to the action of surface tension, Na is not infiltrated by water, so it becomes a small ball, the reason for swimming is because the reaction releases hydrogen, just like a balloon will fly everywhere when it is deflated, the released hydrogen will push the ball, and then swim while reacting with the water, while becoming smaller, and finally disappearing.

d I think it's also wrong, because it should be a white solid, not cloudy, but not as wrong as b.

Sodium and water react to form sodium hydroxide and hydrogen gas.

A correct because sodium hydroxide is a base that turns red when it encounters phenolphthalein.

b Incorrect Sodium and table salt do not react.

c Correct sodium is less dense than water.

d. Correct, the reaction consumes water, and salt precipitation in saturated salt water is so turbid.

b Sodium reacts more violently with table salt than with water, and emits heat, and a hissing sound is found.

D is not true. According to the experimental phenomenon BC is right. The equation for the reaction of Na and water can know that the product has NaOH and all are bases. The alkali can turn phenolphthalein red. There is no precipitation and no turbidity.

d, the first option, sodium and water to produce sodium hydroxide, alkaline makes phenolphthalein red, c is given in the book, specifically I will help you look at it We should also have it in our professional book Sorry to answer incorrectly You should choose B

There is a place upstairs that is not right, the Na acid preferentially reacts with the H+ ion to form H2, rather than reacting with the water and then with the H ion of the acid. Because in the acid solution, the H ions dissociated by the sensitive hall acid are more dissolved in the solution, and the activity is greater than that of the H+ dissociated by hydrolysis.

In the salt solution, Na preferentially reacts with H ions dissociated from H2O to form NaOH. At this time, if there is a metal cation and it can react with OH ions to form a precipitate, NaOH reacts with metal ions to form a precipitate (that is, neoalkali). Common Shenbi lakes such as Fe(OH)3, Al(OH)3, Cu(OH)2, MG (Qiaohuiyin OH)2

For example: FeCl3+3Na(OH)==Fe(OH)3 +3NaCl

Sodium metal. First, it reacts violently with water to produce sodium hydroxide.

and hydrogen, sodium hydroxide dissolves in water, hydrogen is partially burned and part escapes.

In acids (such as hydrochloric acid), sodium hydroxide and acids (such as hydrochloric acid) react to form jujube salts (such as sodium chloride).

and water. In salt, do not continue the reaction.

In alkalis, there is generally no reaction, but there are exceptions, such as the stool bend in aluminum hydroxide solution to continue the reaction. As follows:

1 Na2O2 2 Na2O undergoes the following reaction: 2Na2O2+2H2O=4NaOH+O2, Na2O+H2O=2NaOH

It can be seen that the substance is formed by reaction and consumes a certain amount of water.

Na2CO3, on the other hand, does not react, because CO32- will be partially hydrolyzed, CO32- +H2O = HCO3- +OH-

The two-step hydrolysis is very weak, do not need to consider, at this time it can be seen that the hydrolysis of carbonate occurs, and the number of anions produced by hydrolysis is higher than that of non-hydrolysis, that is, the final anion is greater than, but will not be reached.

NaCl, on the other hand, does not undergo hydrolysis and is completely ionized in solution, that is, ionized anions, and C 1 = 2> 3 > 4 should be selected

2Na2O2+2H2O=4Naon+O2 is 2Na2O+2H2O=4NaOH+O2 after dissolved in water, and Na2CO3 is soluble in water.

NaCl is present after dissolving in water.

Because the volume of the solution remains the same, D should be chosen, right?

Both 1 and 2 react with water to form NaOH, and the amount of water consumed when reacting with water is equal, and the product is also equal, so 1=2, both are anions;

3 will hydrolyze a part of the carbonate ions, because the hydrolyzed anions are -1 valence, so the total amount of anions is increased, greater than, and 4 is the anion.

Select C, the solute in 1 is Naoh, there are sodium ions in the solution, and the anion hydroxide in 2 is the solute NaOH hydroxide.

3 If carbonate hydrolysis is not considered, the anions should be, but one carbonate hydrolysis will produce two anions, so it is greater than.

4 Needless to say, chloride concentration.

The volume of the solution does not change, so the conclusion is C.

Na2O2+2NaHCO3=2Na2CO3+Na2O2+NaHCO3=Na2CO3+NaOH+ The rest is just a good idea.

To be precise, it is greater than or equal to 1 3 and less than 1 2

First, the mass of Na in the original mixture (2Na + 2H2O = = 2NaOH + H2) is calculated from the hydrogen generated

To find the quality of na, reuse, is the quality of na2o.

i.e. the mass of the NA that is oxidized to Na2O is.

The mass of the NA before being oxidized is.

According to the equation, it can be seen that the amount of substance that produces 1molnaohna from 1molna is.

The amount of substance of Naoh is too.

Quality Score:

According to case B, the first 50ml does not generate CO2, indicating that H+ and CO3 - generate HCO-, because there is 50ml of HCl, and the solubility of HCl is, indicating that the H+ sum is consumed, and CO3- is obtained by the reaction of NaOH with CO2, and the volume of CO3- solution taken is 20ml, indicating that there is in 20ml, then there is 100ml, which can be obtained from the chemical equation you wrote, and the volume of the modified solution is 100ml, So the solubility of NaOH solution is 1mol l, which is not easy to type.