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uysha [10]
3 years ago
10

If the molecule C6H12 does not contain a double bond, and there are no branches in it, what will its structure look like?

Chemistry
1 answer:
NARA [144]3 years ago
5 0
I have attached a photo of the structure. 
You can get better at solving problems like this by practicing a lot! 




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A sample of hydrogen gas at a pressure of 0.926 atm and a temperature of 29.5 C, occupies a volume of 457 mL. If the gas is allo
klio [65]

Answer:

V₂ = 946.72 mL

Explanation:

Given data;

Initial pressure = 0.926 atm

Initial volume = 457 mL

Temperature = constant = 29.5°C

Final pressure = 0.447 atm

Final volume = ?

Solution:

The given problem will be solved through the Boyle's law,

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume  

by putting values,

P₁V₁ = P₂V₂

0.926 atm × 457 mL = 0.447 atm × V₂

V₂ = 423.18 atm. mL/ 0.447 atm

V₂ = 946.72 mL

4 0
2 years ago
Which statement best describes the process that the diagrams show?
TEA [102]

B. This looks like ice turning to water then to steam

7 0
3 years ago
Read 2 more answers
Using the law of conservation of energy what is the kinetic energy at D
seraphim [82]

Answer:

The law of conservation of energy states that energy can neither be created nor destroyed - only converted from one form of energy to another.

Explanation:

This means that a system always has the same amount of energy, unless it's added from the outside.

3 0
2 years ago
A gas takes up a volume of 35 L, and has a pressure of 4.8 atm. What is the new pressure of
joja [24]

Answer:

7.3 atm

Explanation:

- Use the formula P1V1 = P2V2

- Rearrange formula and then plug in values.

- Hope this helped! Let me know if you need more help or a further explanation.

3 0
3 years ago
The formation of ammonia is represented by the equation N2(g) + 3H2(g) ⇌ 2NH3(g). Determine the enthalpy of formation of ammonia
Savatey [412]

Answer:

\Delta _fH_{NH_3}=-66\frac{kJ}{mol}

Explanation:

Hello!

In this case, since the study of the bond energy allows us to compute the enthalpies of some reactions, for this combination reaction by which ammonia is yielded, we understand the enthalpy of reaction equals the enthalpy of formation of ammonia, and, in terms of the bonds energy we can write:

\Delta _fH_{NH_3}=Delta _rH=\Sigma \Delta H(bonds \ broken)-\Sigma \Delta H(bonds \ formed)

Whereas the bonds enthalpy of those bonds that get broken cover the N≡N and the three H-H bonds at the reactants side and the enthalpy of those bonds that are formed cover the six N-H bonds at the products; which means we obtain:

\Delta _fH_{NH_3}=942\frac{kJ}{mol} +3*436\frac{kJ}{mol}-6*386\frac{kJ}{mol}\\\\\Delta _fH_{NH_3}=-66\frac{kJ}{mol}

Which differs from the theoretical value that is -46 kJ/mol.

Best regards!

7 0
3 years ago
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