The chemical reaction equation of Si and NaOH is Si and O2... Cl2 and NaBr

The chemical reaction equation of Si and NaOH is Si and O2... Cl2 and NaBr

Si + 2 NaOH + H2O = Na2SiO3 + 2 H2↑
Si + O2 = SiO2 (condition: heating)
Cl2 + 2 NaBr = 2 NaCl + Br2

Can Si react with NaOH

Si + 2naoh + H2O = heating = Na2SiO3 + 2h2

The essence of the reaction between Si and NaOH is Si + 3H2O = H2SiO3 + 2h2
H2SiO3+2NaOH=Na2SiO3+2H2O
Or Si + 2H2O = SiO2 + 2h2
SiO2+2NaOH=Na2SiO3+H2O

Strictly speaking, silicon can react with water, but the reaction is too weak. It is generally regarded as non reaction, which is proved by the equation
Si+2NaOH+H2O=Na2SiO3+2H2
It's from the equation
Si+4H2O=H4SiO4+2H2↑
H4SiO4=H2SiO3+H2O
H2SiO3+2NaOH=Na2SiO3+2H2O
Synthetic

How to turn the mathematical problem solving method into equation train travel problem

You can set the distance of a certain place in the train to be x, and then find out the equation relationship in the question, and then enumerate the equation. However, in enumerating the equation, you should pay attention to the equation listed by yourself. In application problems, the equation often has a formula, for example: distance △ time = speed, generally, A multiplication equation can be changed into two divisions. But this answer is too abstract for me. Can you tell me what kind of question it is?

How to deal with the pursuit of junior one?
A truck has a speed of 60 kilometers per hour from place a to place B. one hour later, a car starts from place a to catch up with the truck. The speed of the truck is 80 kilometers per hour. How many hours does it take for the car to start from place a to catch up with the truck?

It took x hours
60+60x=80x
20x=60
x=3
A: it took three hours for the car to catch up with the truck

A passenger car and a freight car run in the same direction on parallel tracks. The length of the passenger car is 200 meters, and the length of the freight car is 310 meters. The speed ratio of the passenger car and the freight car is 4:3. If the passenger car chases the freight car from the rear, the time from the front to the rear is more than 2 minutes. The speed of the two trains can be calculated

2 minutes = 120 seconds, so the speed difference between the two cars is (200 + 310) △ 120 = 4.25 M / s. suppose that the speed of passenger cars is 4x M / s and that of freight cars is 3x M / s. according to the meaning of the question, we can get the equation: 4x-3x = 4.25, & nbsp; & nbsp; & nbsp; X = 4.25, then the speed of passenger car is: 4 × 4.25 = 17 m / s, the speed of freight car is: 3 × 4.25 = 12.75 M / s, a: the speed of passenger car is 17 m / s, the speed of freight car is 12.75 M / s

On the pursuit of junior high school students
Xiao Ming's mother looked at her watch when she went out in the morning. The hour hand and the minute hand coincided between 7:00 and 8:00. When she came back, she looked at her watch again. The hour hand and the minute hand coincided between 8:00 and 9:00. How long did Xiao Ming's mother go out?

Suppose Xiaoming's mother goes out at 7:x in the morning
X / 60 = (x-35) / 5'60 minutes is an hour, 35 is the place where the needle reaches 7, 5 is five minutes for a word
The solution is as follows
X=38.1818
That is to say, Xiaoming's mother went out at 7:38.1818 in the morning
Let's suppose that when Xiaoming's mother comes back, it's 8:00 y
Y/60=(Y-40)/5
The solution is as follows
Y=43.6363
So Xiaoming's mother went out at 8:43.6363-7:38.1818 = 5.4545 in an hour
At twelve o'clock the hour and minute hands coincide
The angular velocity of the hour hand is 30 degrees per hour
The angular velocity of minute hand is 360 degrees per hour
When the needles and minute needles coincide, the minute needles travel n more circles than the hour hand (n = 1, 2, 3, 4,...)
If the minute hand coincides with the hour hand, then
The solution of 360t-30t = 360n is obtained
t=12N/11
Take n7-8 = 7 and n8-9 = 8 for the two time periods respectively
The time interval is
T = (12 * n8-9-12 * n7-8) / 11 = 12 / 11 (hours)
Minute hand speed: 6 degrees / minute
Clockwise speed: 0.5 deg / min
Let's say it takes X minutes to get out
210/(6-0.5)+x=60+240/(6-0.5)
X = 65.455 minutes
It took 65.455 minutes to buy vegetables

The distance between the enemy and our army is 25 kilometers. The enemy runs away at the speed of 5 kilometers per hour. Our army pursues at the speed of 8 kilometers per hour at the same time, and fights at the distance of 1 kilometer. How many hours did the fighting take place?

Set the time as X hours, so that the enemy's escape distance is 5x kilometers, and our pursuit distance is 8x. According to the title, we can get: 8x-5x = 25-1, sort out: 3x = 24, and solve: x = 8, that is, the battle took place eight hours after the beginning of pursuit

The equivalence relation in encounter problem and pursuit problem: () + () = original distance () - () = original distance

The equivalence relations in the encounter problem and the pursuit problem are as follows
Encounter problem: (the distance of line a) + (the distance of line B) = the original distance
Tracing problem: (the distance of line a) - (the distance of line B) = the original distance

When a car meets a red light at the intersection, when the green light is on, the car starts to accelerate at an acceleration of 4m / S2. Just at this time, a motorcycle comes at a speed of 10m / s in the same direction as the car, and the car chases the motorcycle
(1) What is the maximum distance between the two cars before catching up with the motorcycle when the car accelerates from the intersection?
(2) How long does it take for the car to catch up with the motorcycle
2: On a plane highway, a car starts from a standstill somewhere. At this time, a truck passes by the car at a speed of 15mgs and rushes forward. As a result, the car just catches up with the truck when it is 225m away from the starting point. Let the car do a uniform acceleration, and try to find the acceleration a of the car and the maximum speed X of catching up with the first two cars
3: When the speed of the car is 18m / s, the car makes a uniform deceleration linear motion after the emergency brake, and its acceleration is 6m / S2?
4: Car a and car B move to the intersection along the mutually perpendicular track. Car a is 16m away from the intersection, and the acceleration is 1mgs2. Car B is 12m away from the intersection, and moves to the intersection at a uniform speed of 7m / S. in order to avoid collision, car B brakes and let car a pass through the intersection first. How much is the acceleration when car B brakes? (two cars are considered as mass points)
5: A car is in a static state. There is a person at the beginning of 25m behind the car. When the car starts to move at a constant acceleration of 1mgs2, the person starts to overtake at a constant speed of 6m / s. can he catch up? If not, what is the minimum distance between the person and the workshop?

First question,
Automobile s = 1 / 2 * a * T ^ 2 = 2 * T ^ 2, motorcycle s = 10t, maximum distance between two vehicles motorcycle s-automobile s = 10t-2 * T ^ 2,
When t = 2.5, s = 12.5
How long does it take for a car to catch up with a motorcycle? S-car s = 10t-2 * T ^ 2 = 0, t = 5
Second question
The acceleration of car a 225 = 1 / 2 * a * T ^ 2, where t = 225 / 15 = 15, the solution is a = 2
It seems that there is a problem with the maximum speed X of the first two cars, because the car is at a constant speed, the car's speed is 0, then x = 10
The third question
That is s = vt-1 / 2 * a * T ^ 2, where V = 18, t = 6, a = 6, the solution is s = 0
After calculation, when t = 3, v = 0, the process is that the car moves forward with uniform deceleration, and the speed decreases to 0 after 3 seconds, and then the car goes back to the origin after 3 seconds
The fourth question
S = VT + 1 / 2 * a * T ^ 2, where V = 2, s = 16, a = 1, the solution is t = 4 ', that is, a reaches the focus in 4 seconds
S = vt-1 / 2 * a * T ^ 2, where V = 7, s = 12, t = 4, the solution is a = 2, that is, when a = 2, two cars collide, so a should be less than 2
The fifth question
People s = 6T, cars s = 1 / 2 * a * T ^ 2 = 0.5T ^ 2, people s-cars s = 6t-0.5t ^ 2, the same as the first problem, the solution is t = 6, s = 18, that is, the maximum distance between the two is 18, the actual distance between the two is 25, so it can't catch up