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maria [59]
2 years ago
12

Draw the four structures of the compounds with molecular formula C5H10O that contain a carbon-carbon double bond, an unbranched

carbon chain, and a hydroxyl group one carbon from the end of the chain.

Chemistry
1 answer:
kondaur [170]2 years ago
3 0

Explanation:

Molecular formula of the compound =

The carbon chain in the molecule will look like :

1. H_2C=CH-CH_2-CH_2-CH_2-OH

2. H_3C-CH=CH-CH_2-CH_2-OH

3. H_3C-CH-CH=CH-CH_2-OH

4. H_3C-CH_2-CH_2-CH=CH-OH

No branching is present, so that means the valency of the carbon will be fulfilled by 10 hydrogen atoms and 1 oxygen atom

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Which of the following statements is true about the following reaction?
MAVERICK [17]

The correct reaction equation is:

3NaHCO_{3} (aq) + C_{6}H_{8}O_{7} (aq) \rightarrow 3CO_{2} (g) + 3H_{2}O (l) + Na_{3}C_{6}H_{5}O_{7} (aq)

Answer:

b) 1 mole of water is produced for every mole of carbon dioxide produced.

Explanation: <u>CONVERT EVERYTHING TO MOLES OR VOLUME, THEN COMPARE IT WITH THE COMPOUND'S STOICHIOMETRY IN CHEMICAL EQUATION.</u>

a) <u>22.4 L of CO_{2} gas</u> is produced only when <u>\frac{22.4}{3} L of  C_{6}H_{8}O_{7}</u> is reacted with 22.4 L of NaHCO_{3}. So it is wrong.

b) Since in the chemical equation the stoichiometric coefficient of CO_{2}  and H_{2}O are same so the number of moles or volume of each of them will be same whatever the amount of reactants taken. <u>Therefore it is correct option.</u>

c)  6.02\times 10^{23} molecules is equal 1 mole of Na_{3}C_{6}H_{5}O_{7} if produced then 3 moles of NaHCO_{3} is required, which is not given in the option. So it is wrong.

d) 54 g of water or 3 moles of H_{2}O (<em>Molecular Weight of water is 18 g</em>) is produced when 3 moles of NaHCO_{3} is used but in this option only one mole of NaHCO_{3} is given. So it is wrong.

8 0
2 years ago
Read 2 more answers
4.8g of calcium is added to 3.6g of water. The following reaction occurs
notka56 [123]
Q1)
the number of moles can be calculated as follows
number of moles = mass present / molar mass
number of moles is the amount of substance.
4.8 g of Ca was added therefore mass present of Ca is 4.8 g
molar mass of Ca is 40 g/mol 
molar mass is the mass of 1 mol of Ca
therefore if we substitute these values in the equation 
number of moles of Ca = 4.8 g / 40 g/mol = 0.12 mol
0.12 mol of Ca is present 

q2)
next we are asked to calculate the number of moles of water present 
again we can use the same equation to find the number of moles of water
number of moles = mass present / molar mass
3.6 g of water is present 

sum of the products of the molar masses of the individual elements by the number of atoms 
H - 1 g/mol and O - 16 g/mol 
molar mass of water = (1 g/mol x 2 ) + 16 g/mol = 18 g/mol 
molar mass of H₂O is 18 g/mol 
therefore number of moles of water  = 3.6 g / 18 g/mol = 0.2 mol 
0.2 mol of water is present 
8 0
2 years ago
Lithium chloride forms three hydrates. They are LiCl.H2O, LiCl.2H2O and LiCl.3H2O.
Stels [109]

Answer:

The answer is LiCl.2H2O

Explanation:

Li=7

Cl=35.5

O=16

LiCl.H2O

7+35.5+16+2

60.5

%comp=60.5/78.5×100

22.9

LiCl.2H20

7+35.5+2(2+16)

42.5+36

78.5

%comp=36/78.5×100

45.9

LiCl.3H20

7+35.5+3(2+16)

42.5+54

96.5

54/96.5×100

56.0

7 0
2 years ago
The target diol is synthesized in one step from 1-methylcyclopentene, but your lab partner exhausted the supply of that alkene.
andrezito [222]

Answer:

The reagents are CH_{3}CH_{3}O^{-},OsO_{4},NaHSO_{3}and H_{2}O.

Explanation:

1-Methylenecyclopentene is treated with HBr form 1-bromo-1-methylcyclopentane, which is treated with strong base ethoxide ion and forms 1-methylcyclopent-1-ene.

This alkene is treated with osmium tetraoxide in the presence of sodium bisulfite to form target product.

The chemical reaction is as follows.

4 0
2 years ago
5.00 g of hydrogen gas and 50.0g of oxygen gas are introduced into an otherwise empty 9.00L steel cylinder, and the hydrogen is
GenaCL600 [577]
1) Balanced chemical reaction:

2H2 + O2 -> 2H20

Sotoichiometry: 2 moles H2: 1 mol O2 : 2 moles H2O

2) Reactant quantities converted to moles

H2: 5.00 g / 2 g/mol = 2.5 mol

O2: 50.0 g / 32 g/mol = 1.5625 mol

Limitant reactant: H2 (because as per the stoichiometry it will be consumed with 1.25 mol of O2).

3) Products

H2 totally consumed -> 0 mol at the end

O2 = 1.25 mol consumed -> 1.5625 mol - 1.25 mol = 0.3125 mol at the end

H2O: 2.5 mol H2 produces 2.5 mol H2O -> 2.5 mol at the end.

Total number of moles: 0.3125mol + 2.5 mol = 2.8125 mol

4) Pressure

Use pV = nRT
n = 2.8125
V= 9 liters
R = 0.082 atm*lit/K*mol
T = 35 C + 273.15 = 308.15K

p = nRT/V  = 7.9 atm
3 0
2 years ago
Read 2 more answers
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