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GaryK [48]
2 years ago
14

Thủy phân m gam mantozo, sau một thời gian thu được dung dịch X. Khi cho dung dịch X tác dụng gần hết với dung dịch AgNO3 trong

NH3 thì thu được tối đa 194,4 gam Ag. Biết hiêu suất quá trình thủy phân là 80%. Giá trị gần nhất của m là :
Chemistry
1 answer:
Westkost [7]2 years ago
4 0

Answer:

I dont undarsatnd 2gebwhsanKM<dwkdwndwkjdwnfwkjdnfkwnfwkf

Explanation:

wnkf mnf wnmd

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What we call "tin cans" are really iron cans coated with a thin layer of tin. The anode is a bar of tin and the cathode is the i
UNO [17]

Answer:

Fe (s) + Sn^{2+} (aq)\rightarrow Fe^{2+} (aq) + Sn (s)

Explanation:

Although the context is not clear, let's look at the oxidation and reduction processes that will take place in a Fe/Sn system.

The problem states that anode is a bar of thin. Anode is where the process of oxidation takes place. According to the abbreviation 'OILRIG', oxidation is loss, reduction is gain. Since oxidation occurs at anode, this is where loss of electrons takes place. That said, tin loses electrons to become tin cation:

Sn (s)\rightarrow Sn^{2+} (aq) + 2e^-

Similarly, iron is cathode. Cathode is where reduction takes place. Reduction is gain of electrons, this means iron cations gain electrons and produce iron metal:

Fe^{2+} (aq) + 2e^-\rightarrow Fe (s)

The net equation is then:

Sn (s) + Fe^{2+} (aq)\rightarrow Fe (s) + Sn^{2+} (aq)

However, this is not the case, as this is not a spontaneous reaction, as iron metal is more reactive than tin metal, and this is how the coating takes place. This implies that actually anode is iron and cathode is tin:

Actual anode half-equation:

Fe (s)\rightarrow Fe^{2+} (aq) + 2e^-

Actual cathode half-equation:

Sn^{2+} (aq) + 2e^-\rightarrow Sn (s)

Actual net reaction:

Fe (s) + Sn^{2+} (aq)\rightarrow Fe^{2+} (aq) + Sn (s)

6 0
3 years ago
26 grams of CO2 to moles
neonofarm [45]
First, find out how many grams are in one mole of CO2(the two oxygen atoms means you need to multiply oxygen’s amu by 2,then add whatever carbon’s amu is to that). Then divide 26 grams by that number and that will be your moles. There are only two significant figures, so round your answer correctly.
4 0
2 years ago
I Wanted to ask.. it's a weird question I know or maybe it's gonna get reported But
Temka [501]

Answer:

if hyper its probaly because the have ADHD which is a type of drug that helps people with behavior disorders . the might have ticks wich is something that people can not control.

Explanation:

5 0
2 years ago
6. In Reaction A, students are instructed to add no more than 0.25 mL of 15 M nitric acid. What volume of 15 M nitric acid is re
enyata [817]

The given question is incomplete, the complete question is:

In Reaction A, students are instructed to add no more than 0.25 mL of 15 M nitric acid. What volume of 15 M nitric acid is required to react with 0.030 g of copper metal? If 1.0 mL of acid contains approximately 20 drops, how many drops of nitric acid are needed?

Answer:

The correct answer is 0.0629 ml and 1.26 drops.

Explanation:

Based on the given question, the equation is:  

Cu + 2HNO₃ (aq) ⇒ Cu(NO₃)₂ + H₂

The mass of copper given is 0.030 grams.  

The molecular mass of copper is 63.55 gram per mole. The number of moles can be determined by using the formula, n = weight/molecular mass.  

Moles of Cu = 0.030 grams/63.55 grams per mole = 0.000472 moles

Based on the reaction, it is clear that 1 mole of Cu reacts with 2 moles of nitric acid, therefore, the number of moles of nitric acid needed will be,  

= 0.000472 mol Cu × 2 mol HNO₃ / 1 mole Cu

= 0.000944 mol HNO₃

The concentration of HNO₃ given is 15 M

Now the volume of HNO₃ required to react with 0.030 grams of copper metal will be,  

Volume = 0.000944 mol HNO₃ × 1L/15 mol HNO₃ × 1000 ml/ 1L

= 0.0629 ml.  

Based on the given information, if 1 ml of nitric acid comprise 20 drops, therefore, 0.0629 ml of the acid will require the drops,  

Number of drops of HNO₃ = 0.0629 ml × 20 drops / 1 ml  

= 1.26 drops.  

7 0
3 years ago
What is the metal that is grouped with metalloids
vlabodo [156]

Answer:

The post-transition metals cluster to the lower left of this line. Metalloids: The metalloids are boron (B), silicon (Si), germanium (Ge), arsenic (As), antimony (Sb), tellurium (Te) and polonium (Po). They form the staircase that represents the gradual transition from metals to nonmetals.

Explanation: i googled it

4 0
3 years ago
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