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Anonymous users2024-02-09agno3 : mol
AgNO3+NaCl---AgCl+nano32AGno3+Na2CO3---2AG2CO3+2NANO3AG2CO3+2HNO3---2AGno3+CO2+H2O gas is: CO2
n(co2)=135ml/2240ml/mol=n(co2)=n(ag2co3)=
m(ag2co3)=n(ag2co3)*334g/mol=n(nacl)=n(agno3)-n(co2)m(nacl)=n(nacl)*
Mass fraction = m(NaCl) m total * 100%.
I haven't done it for a long time, hehe, if you don't do it well, you will be ha, just count it yourself.
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Anonymous users2024-02-08Relative molecular weight.
nano3--88
na2co3--106
agno3+nacl===agcl|+nano3x x2agno3+na2co3===2ag2co3|+2nano3ag2co3+2hno3===2agno3|+co2|+H2O produces CO2 N(CO2) = 135ml 22400ml mol = AMOL (A, you can use a calculator to calculate accurately).
n(co2)=n(ag2co3)=n（na2co3）n（agno3）=2 n（na2co3）
n(agno3)=( *1mol/l=
x=m(nacl)=b*
Mass fraction = [m(NaCl) m total]*100%=(CG.)
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Anonymous users2024-02-07ag+..hcl 2ag+..CO2 knows the molar mass of the gas, and the molar mass of the silver ion is not difficult to calculate the molar mass of Cl-, and then calculate the result.
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Anonymous users2024-02-06God, I'm still a college student, damn it, I can't even understand the questions.
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Anonymous users2024-02-051) The solution set is; Poor Bridge (2); (7) Any number of the celebration modulus.
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Anonymous users2024-02-04Solution: (4) find the vector am multiplication vector an=am film xan membrane xcos, an film is known = root number (4 2 ten 2 2) = 2 root number 5, this problem is transformed into the maximum value of vector am multiplication cos, that is, the projection of am on an is the largest, and it is easy to get the maximum when m is a square c point. There are two ways to require this maximum, first, to establish a plane Cartesian coordinate system with a as the origin, then the vector am=(4,4) and the vector an=(4,2), then the product is =4x4, ten, 2x4=24
Law. 2. In the triangle amn, use the cosine theorem to find cos=(am 2 deca 2-mn 2) (2an am) = (32 dec 20-4) (2x2 root number 5x4 root number 2).
So the vector am multiplied by the vector an=4 root number 2x2 root number 5xcos=2412, because x,y is positive, so 2 x times 2 y=2 (x ten y)>=2 (2 root number(xy))=16
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