Hoá 10 || Kết nối tri thức || Bài 18: Ôn tập chương 5 || KNTT

Hello everyone, in this video, the teacher will guide us to the 18th chapter of the 5th chapter of the book. And the book he is teaching is the book The Connection of Knowledge of Life. If you have found this book, we will continue to follow this book. In addition to this book, the teacher also performs the book The Creation of the Universe and the Book of the Winged. Before conducting the teaching of the 3 chapters of the 5th chapter of the book, the teacher performs 2 types of videos. The first type of video is to guide the content of the book, summarize the content of the book as well as answer the questions of the 3 chapters of the book. The second type of video is to guide the tips and tricks of all books. In addition, the teacher also has a group of students called JALO. When you join the JALO group, you will receive documents for 1 year. This group includes a lot of exercises. The teacher has sent it to the group. When we join the group, we receive exercises for 1 year. Especially, the exercises have detailed instructions. There are a lot of exercises for us to learn and practice. Especially, when we do exercises that we don't understand, we will receive help from the teacher. Here are the exercises for 1 year. There are a lot of exercises. When you join the JALO group, contact the teacher. Now back to the issue of the 18th chapter of the 5th chapter of the book. The first is the knowledge system. Summarize quickly and quickly. Complete the missing content. Be sure to respond to the product. The content is already guided. Remember to use small and large. Large and small are required. DeltaH is small 0, heat is 0, heat is 0. Calculate the entropy according to the total energy. There are 2 calculations according to the total energy and the total energy connected. For the total energy, we use the total entropy. DeltaFH In the case of delta FH, we subtract from the product to calculate the total energy of the tank delta FH of the head That's how we calculate the entropy of the tank If we calculate the total energy of the tank the E and B of the head we subtract from the total energy of the product These are two ways to calculate If we calculate the entropy of the tank, we subtract from the product If we calculate the total energy of the tank, we subtract from the product In the answer to this question, I will guide you again Question 1 The reaction of the transition between the two elements of FH FH red to FH white This is the heat absorption If we want to change FH red to FH white we have to absorb 17.6 kJ This is the heat absorption This means that FH red is more elastic than FH white If we want to change FH red to FH white, we have to absorb the heat FH red is more elastic than FH white That's question 1 Question 2 Let me know the entropy of this reaction CO plus OC 1 mol CO plus 0.5 mol OC C2 H0 Let me know the heat of formation of CO Write it down here The entropy of the reaction If we calculate the heat of formation we take the product This is the product Take the product 5393.5 We subtract for the first element Subtract for CO Subtract for X Here And we subtract for 1 part 2 of OC OC is not We put this OC We can calculate this Then we subtract Then X is equal to 393.5 Subtract for this Plus this 393.5 Subtract for 2 8 3 We take the negative sign 110.5 That's question A We press this formula, we will get X 1110.5 Or press SHIFT SHOW This question is easy Question 3 5% glucose There is a separate volume This is C% This is a separate volume D The reaction of this 1 mol of oxygen The heat of formation is like this We ask a patient Give a bottle of glucose 500 ml Maximum energy How much does the patient receive? This is the reason When we are sick, we pour water into it Then we don't need to eat There is energy in here Already Now we calculate First we Calculate the volume Calculate the volume of glucose M glucose M by D D by V D is 1.02 V is 500 We calculate this 1.02 by 500 510 Then we calculate the volume of glucose C% M glucose C% M glucose 510 215 215.5 g 25.5 g Okay, we have this, we will calculate the MONS of GLUCO See, SIGAMON, MONS GLUCO GLU is C6H12O6, M is greater than 180 So 25.5 divided by 180 Okay, let's see how many MONS Divide by 180 17 divided by 120 Okay, this is the MONS Now people say that this MONS transmits energy like this But how much energy is it? So it is DENTAI RH 17 divided by 120 We multiply by 1.2 I2803.0 2.803 Okay, we take this number and multiply by 2.803 Multiply by 2.803 We will get the answer 397.09 Yes, this is the A 397.09 This is the value of the energy that the patient can receive Okay, this is the whole formula The formula of the 9th class program M is divided by DNUV, C is divided by 100 Then continue to the 4th In the process of time, if you don't understand anything We can leave a comment for us to communicate with you If we want to receive more exercises Then we go to the dialog group Put a lot of exercises in there Then the 4th question The 4th question gives the average value of the connection Here, in these connections, this formula is also given People call it TINTOPY This is calculated according to the connection energy Now this formula Instead of writing C3H8, it is written like this We must know the connection to do it C3H8 123C H, H, this is C3H8 C3H8 The function is C4H4 Plus C2H4 C2H4 has a double connection We learned this in the link plus too much Here We have to do this So how to convert TINTOPY into DeltaRH Based on the connection energy We take the connection energy of the head Here we will see The connection energy of CH CH1, 2, 3, 4, 5, 6, 7 8 So there will be 8 times the connection energy of CH Plus CC There are 2 CC Plus 2 times the connection energy of CC Then we subtract Here 4 4 times the connection energy of CH We subtract 4 times the connection energy of CH And 1 times the connection energy of CC Then we substitute 8 times CH Plus 2 times CC 2, 3, 4, 6 4 times CH 4 times 4, 1, 8 4 times CH 4 times 4, 1, 8 Ok We add Here CH Subtract 2 times 1C to this pair 6, 1, 2 Now we can subtract these two These two are the same Then we will calculate the value This is this word 2 multiplied by 3 4, 6 2 multiplied by 3, 4, 6 We subtract for 6, 12 Then it will be equal to 80 It will be equal to 80 80 kilojoules Equal to 80 Here, the answer is C See 4C This is also easy Nothing difficult This is in part 3 of the instruction Continue to question number 5 Question number 5 Methane is formed in this direction Methane is a natural gas Methane is formed in this direction 2 of CO2 How many concentration of methane did it form density due to from This is the product Amp 393.5 Plus 2 times of water 2 times of water is Amp 285.8 Excluding the product, excluding the head The head is CH4, let's call CH4 as X ENTA-FH of CH4 Excluding the oxygen, oxygen is 0 Then we press this button, we will get Meta's constant energy, which is ENTA-FH Of CH4 Move this over here Then we take 8 Move to 8 90.3 We subtract for 3 93.5 Subtract for 2 Subtract for 285.8 Calculate Amp 74.8 Amp 74.8 Kilojoules We have the value This is the formula of CH4 This is question 5 Simple Applying gas Question 6 Question 6 requires us to compare Heat When burning 1 kg of copper and 1 kg of fatty acid Cheetorin fatty acid This process is possible When burning 1 mole 1 mole shows the energy is this 1 mole creates energy, so we calculate the mole Now the mole of Copper CH2H5OH Its mole is 1 Kilograms are easy to calculate No need to change Because it is easy to compare with the same unit Or change to 1000 grams Kilograms 1000 grams If we do it right We go by 1000 1 kg of copper by 1000 grams Then we divide by 46 This is the mole Then we calculate the thermal conductivity We calculate the thermal deformation Then we take 1000 Divide by 46 We multiply by 1365 Put it here Then Similarly, we calculate the mole of Fatty acid Cheetorin Cheetor 1 kilogram is also 1000 grams M is large, we add 890 890 Then We calculate the thermal conductivity Is 1000 Divide by 890 Then we multiply by 35807 Then we press these two values ​​and compare which one is larger Above is 1000 Divide by 46 Multiply by 1365 Then the value will be 2967 1000 29674 Let's make it round 29674 Kilojoules Then below 1000 divided by 890 890 Multiply by 35807 Divide by 40 40 2 32 Then People ask us to compare Compare So when Burn 1 kilogram Then Here The energy produced is 29,000 29.6 Here More than 40,000 At this time we clearly see The Heat produced when burning A kilogram of fatty acid will be larger Larger That's it Then question 7 Question 7 is based on connected energy Connected energy Then people give this acid From the data Of version 12.2 Then we calculate the connected energy of this Then now Question 7 Connected energy is the head of the product Version 12.2 Here is the value, look at the table Here is the connected energy, here is the group So The TNTP of the reaction is equal to 12456 There are 6 Connection CH And 2 CC connections 2 Connected energy of CC And 1 CO double connection 1 Connected energy CO double connection Plus for OC OC is plus 4 OC OC is OO double connection Then minus 3 connected energy Of CO2 CO2 is CO But at this time there are 2 CO It becomes 6 Minus for water, water is HO So 6 HO OK We put it in Calculate 6 connected energy of CH CH414 6x414 Plus for 2 CC 347 2x347 Plus for 1 CO double connection CO CO double connection 736 Plus for 4 OC O2 is 498 498 4x498 Minus for 6 CO 6 CO, CO of CO2 is 799 CO2 is different 6x799 Minus for 6 HO, HO we find HO only HO, OH here, 464 4, 4, then we press this button and it's done Simple Mainly to see if the board is correct, we can calculate Then this sentence we press We press it ourselves, the value is already pressed Then It will be 1, 2, 3, 8 Kilojoules There This is the value of the 7th question Show a volume of 1,238 Kilojoules There Then the last sentence Here, the last sentence Then a bottle of gas A bottle of gas Too loose is a mixture of Roban and Butane Dilimon 1, 2 We ask to determine The amount of gas emitted when burning 12 Kilo gas gas above Here Then we ask after a day to use up a bottle of gas 12 kg Know this family Need to burn 10,000 Kilojoules of heat In 1 day 80% consumption is considered to be 20% Now we calculate The first two are Roban and Butane 1, 2 So Roban is C3H8 Butane is C4H10, this is grade 11 Then the ratio is 1, 2 Now I call this guy X Above is 2X Then I will get 12 kg C3H8 is 12 x 3 Plus 8 44, so 44X Plus this is 58 58 Multiply by 2X Will be equal to 12 44 44 plus 2 multiplied by 58 160 12 divided by 160 Then X will be equal to 0.075 X will be equal to 0.075 And 2X Will be equal to multiply by 2 0.15 This is the calculation in grams 12 kg of grams We multiply by 1,000 0.075 Multiply by 1,000 75 This is 75 moles Multiply by 2 2 multiply by 75 150 moles Now let's continue A guy 75 This guy is 75 moles This guy is 150 moles Each mole creates this much Now 75 moles is how much Delta Rh Equal to 75 Multiply by 2, 2, 2, 0 Delta Rh of Phu Du Equal to 150 Multiply by 2, 8, 7, 4 Then we calculate how much We add these two So in total Delta Rh Of this In 12 kg Then 75 Multiply by 2, 2, 2, 0 Plus 150 Multiply by 2, 2, 8, 7, 4 So Delta Rh 5, 9, 7, 6, 0, 0 597,600 This is the heat output of the gas tank 12 kg The total heat output of the gas tank is 12 kg Now Average 1 day Now let's find the number of days Let's write it straight Now the number of days Number of days Now let's take this 5, 9, 7, 6, 0, 0 Absorb only 80%, 80% is 0.8 We divide it by 1 day to use 10,000 Then we will get the number of days Use up the gas tank 12 kg 5 9, 7, 6, 0, 0 Multiply by 0.8 Divide by 10,000 1 day to use 10,000 It will be 47.808 days It is almost equal to 48 days If the heat ratio We use the average If the heat ratio we use the average 1 day 10,000 kJ We use Almost 48 days to run out of gas tank 12 kg That's it I repeat To do more exercises In the YOLO group, we receive more exercises There are a lot of exercises About this The enthalpy change We receive and receive I will end the video here