All categories

3 years ago

How many methylene groups are present in 2,4-dimethylhexane?

Chemistry

2 answers:

stepan [7]3 years ago

8 0

There are zero number of methylene groups in the substance 2,4-dimethylhexane. Methylene groups are the part of a molecule that contains a carbon atom that has a double bond or =CH. Looking at the structure of 2,4-dimethylhexane, you cannot see any carbon atom that has a double bond.

dusya [7]3 years ago

4 0

Answer:

Zero methylene groups

Explanation:

Hi, on one side you have the 2.4-dimethylhexane molecule as shown in the figure.

On the other side, you have the methylene group with the following structure:

=CH2

As can be seen there isn't any methylene group in the 2.4-dimethylhexane

Important: don't confuse the methylene group with the methyl group. Methyl group is -CH3 and there are two of them in this molecule.

Read more on moles here:

brainly.com/question/15356425

Hope it helps

7 0

2 years ago

What is the entropy change in the environment when 5.0 MJ of energy is transferred thermally from a reservoir at 1000 K to one a

Leni [432]

Answer:

The entropy change in the environment is 3.62x10²⁶.

Explanation:

The entropy change can be calculated using the following equation:

$\Delta S = \frac{Q}{k_{B}}(\frac{1}{T_{f}} - \frac{1}{T_{i}})$

Where:

Q: is the energy transferred = 5.0 MJ

$k_{B}$ : is the Boltzmann constant = 1.38x10⁻²³ J/K

$T_{i}$ : is the initial temperature = 1000 K

$T_{f}$ : is the final temperature = 500 K

Hence, the entropy change is:

$\Delta S = \frac{5.0 \cdot 10^{6} J}{1.38 \cdot 10^{-23} J/K}(\frac{1}{500 K} - \frac{1}{1000 K}) = 3.62 \cdot 10^{26}$

Therefore, the entropy change in the environment is 3.62x10²⁶.

I hope it helps you!

7 0

3 years ago

Question 19 of 26

Alona [7]

Answer:

3Mg(NO3)2(aq)+2Na3PO4(aq)⇒Mg3(PO4)2(s)+6NaNO3(aq)

Explanation:

4 0

3 years ago

The atomic mass of gallium is 69.72 . The density of iron is 7.87 . The atomic mass of iron is 55.847 . Calculate the number of

laiz [17]

Answer:

the atomic mass of any elemet contains avogardo numberof atoms

In case of Gallium,

69.72 gram is atomic mass and it cotnains around 6.023*10^23 atoms of Gallium

but, 2000 punds = 907184.7 grams

907184.7 gram of gallium contains= 6.023*10^23* 907184/69.72

= 79 *10^26 atoms

Explanation:

8 0

3 years ago

Assuming steam to be an ideal gas, calculate its specific volume and density at a pressure of 90 lb/in2 and a temperature of 650

Vitek1552 [10]

Answer:

1) Sv = 0.4584 m³/Kg...assuming steam as an ideal gas

% deviation from the values in the steam tables

⇒ % dev = 45 %

2) mass air = 272.617 Kg; assuming air to be an ideal gas

Explanation:

ideal gas:

PV = RTn

molar volume:

⇒ V/n = RT/P

∴ P = 90 psi * ( 0.06895 bar/psi ) = 6.2055 bar

∴ T = 650 F = 343.33 °C = 616.33 K

∴ R = 0.08314 bar.L/mol.K

⇒ V/n = (( 0.08314 )*(616.33 K )) / 6.2055 bar

⇒ V/n = 8.2574 L/mol * ( m³/1000L ) = 8.2574 E-3 m³/mol

specific volume ( Sv ):

∴ Mw = 18.01528 g/mol

⇒ Sv = 8.2574 E-3 m³/mol * ( mol / 18.01528 g ) * ( 1000 g/Kg )

⇒ Sv = 0.4584 m³/Kg

steam table:

∴ P = 6.2055 bar ≅ 6 bar → Sv = 0.3157 m³/Kg

⇒ % deviation = (( 0.4584 - 0.3157 ) / 0.3157) * 100

⇒ % dev = 45.2 %; significant value, assuming steam to be a ideal gas

2) mass air, assuming ideal gas:

∴ V = 20ft * 35ft * 10ft = 7000ft³ * ( 28.3168 L/ft³ ) = 198217.6 L

∴ P = 17 psi * ( 0.06895 bar/psi ) = 1.172 bar

∴ T = 75 °F = 23.89 °C = 296.89 K

∴ R = 0.08314 bar.L/K.mol

⇒ n air = PV/RT = (( 1.172 )*( 198217.6 )) / (( 0.08314 )*( 296.89 ))

⇒ n air = 9411.616 mol air

∴ Mw air = 28.966 g/mol

⇒ mass air = 9411.616 mol * ( 28.966 g/mol ) = 272616.892 g = 272.617 Kg

7 0

3 years ago