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kirill115 [55]
3 years ago
7

Viết các đồng phân cấu tạo mạch hở của C4H6O2 cùng nhóm chức axit

Chemistry
1 answer:
Aleksandr-060686 [28]3 years ago
8 0

Answer:

+ axit

CH2=CH-CH2-COOH,

CH3-CH=CH-COOH (tính cả đồng phân hình học)

CH2=C(CH3)-COOH.

+ este

HCOOCH=CH-CH3 (tính cả đồng phân hình học)

HCOO-CH2-CH=CH2,

HCOOC(CH3)=CH2.

CH3COOCH=CH2

CH2=CH-COOCH3

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The reducing agent in this catalytic hydrogenation reaction was molecular hydrogen (H2), which was produced in situ (in the reac
creativ13 [48]

Answer:

a)  After the balloon inflated after 440 uL of dropwise due to the reaction of 1-Decene and the solution in the conical vial. b) 4NaBH_{4} + 2HCl + 7H_{2}O ⇒ 16H_{2(g)}+ 2NaCl + Na_{2}B_{4}O_{7} c) No H_{2} was not the limiting reactant.

Explanation:

Generally, hydrogenation is the chemical reaction between a compound or element and molecular hydrogen in the presence of catalysts such as platinum.

a) After the balloon inflated after 440 uL of dropwise 1-Decene solution was added due to the reaction between 1-Decene and the solution in the conical vial.

b)  4NaBH_{4} + 2HCl + 7H_{2}O ⇒ 16H_{2(g)}+ 2NaCl + Na_{2}B_{4}O_{7}

c) H_{2} was not the limiting reactant based on the mol to mol ratio of H_{2} and decane which is 1:1. Therefore, if 0.8 mol of decane was produced then 0.8 mol of H_{2} would also be produced.

4 0
3 years ago
Question 12 A chemistry student must write down in her lab notebook the concentration of a solution of sodium thiosulfate. The c
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Weight of sodium thiosulfate   = 76.148 - 8.2

= 67.948 g.

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3 0
3 years ago
I NEED HELP PLEASE!!!! CHEMISTRY QUESTION: If 38 g of Li3P and 15 grams of Al2O3 are reacted, what total mass of products will r
maksim [4K]

Answer:

21.5 g.

Explanation:

Hello!

In this case, since the reaction between the given compounds is:

2Li_3P+Al_2O_3\rightarrow 3Li_2O+2AlP

We can see that according to the law of conservation of mass, which states that matter is neither created nor destroyed during a chemical reaction, the total mass of products equals the total mass of reactants based on the stoichiometric proportions; in such a way, we first need to compute the reacted moles of Li3P as shown below:

n_{Li_3P}^{reacted}=38gLi_3P*\frac{1molLi_3P}{51.8gLi_3P}=0.73molLi_3P

Now, the moles of Li3P consumed by 15 g of Al2O3:

n_{Li_3P}^{consumed \ by \ Al_2O_3}=15gAl_2O_3*\frac{1molAl_2O_3}{101.96gAl_2O_3} *\frac{2molLi_3P}{1molAl_2O_3} =0.29molLi_3P

Thus, we infer that just 0.29 moles of 0.73 react to form products; which means that the mass of formed products is:

m_{Li_2O}=0.29molLi_3P*\frac{3molLi_2O}{2molLi_3P} *\frac{29.88gLi_2O}{1molLi_2O} =13gLi_2O\\\\m_{AlP}=0.29molLi_3P*\frac{2molAlP}{2molLi_3P} *\frac{57.95gAlP}{1molAlP} =8.5gAlP

Therefore, the total mass of products is:

m_{products}=13g+8.5g\\\\m_{products}=21.5g

Which is not the same to the reactants (53 g) because there is an excess of Li₃P.

Best Regards!

7 0
3 years ago
At a particular temperature, 12.0 moles of so3 is placed into a 3.0-l rigid container, and the so3 dissociates by the reaction 2
saul85 [17]
            2 SO₃ --> 2 SO₂ + O₂
I             12             0          0
C           -2x           +2x      +x
---------------------------------------------
E         12-2x          2x         x

Since the moles of SO₂ at equilibrium is 3 mol, 2x = 3. Then, x = 1.5 mol. So, the amounts at equilibrium is:
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SO₂: 2(1.5) = 3
O₂: 1.5

The formula for K basing on the stoichiometric reaction is:
K = [SO₂]²[O₂]/[SO₃]² 
where the unit used is conc in mol/L.

K = [3 mol/3 L]²[1.5 mol/3 L]/[9 mol/3 L]²
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how often should gas appliances be checked for proper operation in order to prevent carbon monoxide poisoning?
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You should be checking your gas appliances every year, it should always be a qualified technician.
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