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2 years ago

Draw the structural formula of 4,5-diisopropylnonane.

Chemistry

1 answer:

mamaluj [8]2 years ago

3 0

Following the IUPAC rules, the longest carbon chain should consist of 9 carbons because of the word 'nonane'. From there, attach two isopropyl (3-carbon chains) to the 4th and 5th carbon starting either from the left or right of the parent carbon chain. The structural formula is shown in the picture attached.

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A 2.80 g sample of Al reacts with 4.15 g sample of Cl2 according to the equation shown below.

solong [7]

Answer:

Mass = 5.33 g

Explanation:

Given data:

Mass of Al = 2.80 g

Mass of Cl₂ = 4.15 g

Theoretical yield of AlCl₃ = ?

Solution:

Chemical equation:

2Al + 3Cl₂ → 2AlCl₃

Number of moles of Al:

Number of moles = mass/molar mass

Number of moles = 2.80 g/ 27 g/mol

Number of moles = 0.10 mol

Number of moles of Cl₂:

Number of moles = mass/molar mass

Number of moles = 4.15 g/71 g/mol

Number of moles = 0.06 mol

Now we will compare the moles of AlCl₃ with Al and Cl₂.

Cl₂ : AlCl₃

3 : 2

0.06 : 2/3×0.06 = 0.04

Al : AlCl₃

2 : 2

0.10 : 0.10

Number of moles of AlCl₃ produced by chlorine are less so it will be limiting reactant.

Mass of AlCl₃:Theoretical yield

Mass = number of moles ×molar mass

Mass = 0.04 mol × 133.34 g/mol

Mass = 5.33 g

4 0

1 year ago

An archeologist finds a 1.62 kg goblet that she believes to be made of pure gold. She determines that the volume of the goblet i

katovenus [111]

The answer is

The density (D) is quotient of mass (m) and volume (V):
 $D= \frac{m}{V}$
The unit is g/cm³

It is given:
m = 1.62 kg = 1620 g
V = 205 mL = 205 cm³
D = ?

Thus:
$D= \frac{m}{V} = \frac{1620g}{205cm ^{3} } = 7.90 g/cm ^{3}$

The density of the goblet is 7.90 g/cm³.

5 0

2 years ago

Based on the results of this lab, write a short paragraph that summarizes how to distinguish physical changes from chemical chan

Kaylis [27]

Physical changes occur when the properties of a substance are retained and/or the materials can be recovered after the change. Chemical changes involve the formation of a new substance. Formation of a gas, solid, light, or heat are possible evidence of chemical change.

6 0

1 year ago

Read 2 more answers

A pan containing 20.0 grams of water was allowed to cool from a temperature of 95.0 °C. If the amount of heat released is 1,200

Sedbober [7]

Answer:

81°C.

Explanation:

To solve this problem, we can use the relation:

Q = m.c.ΔT,

where, Q is the amount of heat released from water (Q = - 1200 J).

m is the mass of the water (m = 20.0 g).

c is the specific heat capacity of water (c of water = 4.186 J/g.°C).

ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = final T - 95.0°C).

∵ Q = m.c.ΔT

∴ (- 1200 J) = (20.0 g)(4.186 J/g.°C)(final T - 95.0°C ).

(- 1200 J) = 83.72 final T - 7953.

∴ final T = (- 1200 J + 7953)/83.72 = 80.67°C ≅ 81.0°C.

So, the right choice is: 81°C.

7 0

2 years ago

A 0.500 g sample of C7H5N2O6 is burned in a calorimeter containing 600. g of water at 20.0∘C. If the heat capacity of the bomb c

Nata [24]

Answer:

22.7

Explanation:

First, find the energy released by the mass of the sample. The heat of combustion is the heat per mole of the fuel:

ΔHC=qrxnn

We can rearrange the equation to solve for qrxn, remembering to convert the mass of sample into moles:

qrxn=ΔHrxn×n=−3374 kJ/mol×(0.500 g×1 mol213.125 g)=−7.916 kJ=−7916 J

The heat released by the reaction must be equal to the sum of the heat absorbed by the water and the calorimeter itself:

qrxn=−(qwater+qbomb)

The heat absorbed by the water can be calculated using the specific heat of water:

qwater=mcΔT

The heat absorbed by the calorimeter can be calculated from the heat capacity of the calorimeter:

qbomb=CΔT

Combine both equations into the first equation and substitute the known values, with ΔT=Tfinal−20.0∘C:

−7916 J=−[(4.184 Jg ∘C)(600. g)(Tfinal–20.0∘C)+(420. J∘C)(Tfinal–20.0∘C)]

Distribute the terms of each multiplication and simplify:

−7916 J=−[(2510.4 J∘C×Tfinal)–(2510.4 J∘C×20.0∘C)+(420. J∘C×Tfinal)–(420. J∘C×20.0∘C)]=−[(2510.4 J∘C×Tfinal)–50208 J+(420. J∘C×Tfinal)–8400 J]

Add the like terms and simplify:

−7916 J=−2930.4 J∘C×Tfinal+58608 J

Finally, solve for Tfinal:

−66524 J=−2930.4 J∘C×Tfinal

Tfinal=22.701∘C

The answer should have three significant figures, so round to 22.7∘C.

8 0

2 years ago