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lorasvet [3.4K]
3 years ago
15

Draw an ether that contains exactly five carbon atoms and only single bonds

Chemistry
1 answer:
Eddi Din [679]3 years ago
8 0

<em>Answer:</em>

  •                                       H3CH2COCH2CH2CH3

<em>Ether:</em>

Ether are organic compounds that contain ether functional group , in which oxygen atom is connected with two alkyl or aryl group.

They have general formula as follow

  • R---O---R   or R'---O----R or R'---O---R'

            while  R = Alkyl

                       R' = Aryl

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Silver nitrate solution reacts with calcium chloride solution according to the equation: 2 AgNO3 + CaCl2 → Ca(NO3)2 + 2 AgCl All
HACTEHA [7]

Answer:

14.33 g

Explanation:

Solve this problem based on the stoichiometry of the reaction.

To do that we need the molecular weight of the masses involved and then calculate the number of moles, find the limiting reagent and  finally calculate the mass of AgCl.

                      2 AgNO₃   + CaCl₂    ⇒   Ca(NO₃)₂   + 2 AgCl

mass, g                  6.97        6.39                                     ?

MW ,g/mol         169.87      110.98                                  143.32

mol =m/MW           0.10         0.06                                    0.10

From the table above AgNO₃ is the limiting reagent and we will produce 0.10 mol AgCl which is a mass :

0.10 mol x 143.32 g/mol = 14.33 g

5 0
3 years ago
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8 0
3 years ago
3. A 4.00 gram sample of solid gold was heated from 274K to 314K. How much energy was involved?
MrMuchimi
Q = mcΔT = (4.00 g)(0.129 J/g•°C)(40.85 °C - 0.85 °C)
Q = 20.6 J of energy was involved (more specifically, 20.6 J of heat energy was absorbed from the surroundings by the sample of solid gold).
3 0
2 years ago
Calculate the energy required to heat 566.0mg of graphite from 5.2°C to 23.2°C. Assume the specific heat capacity of graphite un
maw [93]

Answer:

7.23 J

Explanation:

Step 1: Given data

  • Mass of graphite (m): 566.0 mg
  • Initial temperature: 5.2 °C
  • Final temperature: 23.2 °C
  • Specific heat capacity of graphite (c): 0.710J·g⁻¹K⁻¹

Step 2: Calculate the energy required (Q)

We will use the following expression.

Q = c × m × ΔT

Q = 0.710J·g⁻¹K⁻¹ × 0.5660 g × (23.2°C-5.2°C)

Q = 7.23 J

6 0
3 years ago
A 1.0l buffer solution contains 0.100 mol of hc2h3o2 and 0.100 mol of nac2h3o2. the value of ka for hc2h3o2 is 1.8×10−5. part a
Mandarinka [93]
There are two ways to solve this problem. We can use the ICE method which is tedious and lengthy or use the Henderson–Hasselbalch equation. This equation relates pH and the concentration of the ions in the solution. It is expressed as

pH = pKa + log [A]/[HA] 

 where pKa = - log [Ka]
[A] is the concentration of the conjugate base
[HA] is the concentration of the acid

Given:
Ka = 1.8x10^-5
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NaC2H3O2 = 0.1 mol

Solution:
pKa = - log ( 1.8x10^-5) = 4.74

[A] = 0.015 mol + 0.100 mol = .115 moles
[HA] = .1 - 0.015 = 0.085 moles

pH = 4.74 + log (.115/0.085)
pH = 4.87
6 0
3 years ago
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