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4 years ago

Why is the actual dihedral angle of gauche butane different from 60degrees?

Chemistry

1 answer:

tatyana61 [14]4 years ago

5 0

Answer:The actual dihedral angle of gauche butane differs because of a phenomenon known as steric effect.

Explanation: Steric effect is phenomenon that influences the shape and reactivity of ions and molecules caused by nonbonding interactions.

Steric effects goes with electronic effects, which usually tells the shape and reactivity.

Steric effects is a result of repulsive forces between overlapping electron clouds activities. These steric effects will even distort angles and molecular geometries.

In the context of butane, because presence of two bulky methyl groups (-CH3), the steric interference occurs which causes the methyl groups to be some what far apart hence, increasing the dihedral angle between the two methyl groups which in turn, affects the dihedral angles of other groups present in the molecules.

Therefore, the dihedral angles between the hydrogens are not exactly 60 or 180 degrees.

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Para formar bronce, se mezclan 150g de cobre a 1100°C y 35g de estaño a 560°C. Determine la temperatura final del sistema.

jek_recluse [69]

Answer:

La temperatura final del sistema es 1029,346 °C.

Explanation:

Asumamos que el sistema conformado por el cobre y el estaño no tiene interacciones con sus alrededores. Por la Primera Ley de la Termodinámica, el cobre cede calor al estaño con tal de alcanzar el equilibrio térmico. El cobre se encuentra inicialmente en su punto de fusión, mientras que el estaño está por encima de ese punto, de modo que la transferencia de calor es esencialmente sensible:

$m_{Cu}\cdot c_{Cu}\cdot (T-T_{Cu}) = m_{Sn}\cdot c_{Sn}\cdot (T_{Sn}-T)$

$(m_{Cu}\cdot c_{Cu} + m_{Sn}\cdot c_{Sn})\cdot T = m_{Sn}\cdot c_{Sn}\cdot T_{Sn} + m_{Cu}\cdot c_{Cu}\cdot T_{Cu}$

$T = \frac{m_{Sn}\cdot c_{Sn}\cdot T_{Sn}+m_{Cu}\cdot c_{Cu}\cdot T_{Cu}}{m_{Cu}\cdot c_{Cu}+m_{Sn}\cdot c_{Sn}}$ (1)

Donde:

$m_{Sn}$ - Masa del estaño, en gramos.

$m_{Cu}$ - Masa del cobre, en gramos.

$c_{Sn}$ - Calor específico del estaño, en calorías por gramo-grados Celsius.

$c_{Cu}$ - Calor específico del cobre, en calorías por gramo-grados Celsius.

$T_{Sn}$ - Temperatura inicial del estaño, en grados Celsius.

$T_{Cu}$ - Temperatura inicial del cobre, en grados Celsius.

Si sabemos que $m_{Cu} = 150\,g$ , $m_{Sn} = 35\,g$ , $c_{Cu} = 0,093\,\frac{cal}{g\cdot ^{\circ}C}$ , $c_{Sn} = 0,060\,\frac{cal}{g\cdot ^{\circ}C}$ , $T_{Sn} = 560\,^{\circ}C$ y $T_{Cu} = 1100\,^{\circ}C$ , entonces la temperatura final del sistema es:

$T = \frac{(35\,g)\cdot \left(0,060\,\frac{cal}{g\cdot ^{\circ}C} \right)\cdot (560\,^{\circ}C)+(150\,g)\cdot \left(0,093\,\frac{cal}{g\cdot ^{\circ}C} \right)\cdot (1100\,^{\circ}C)}{(35\,g)\cdot \left(0,060\,\frac{cal}{g\cdot ^{\circ}C} \right)+(150\,g)\cdot \left(0,093\,\frac{cal}{g\cdot ^{\circ}C} \right)}$

$T = 1029,346\,^{\circ}C$

La temperatura final del sistema es 1029,346 °C.

3 0

3 years ago

When an atom gains an electron, the resulting particle is calledA) a proton.B) an anion.C) a cation.D) an isotope.E) none of the

MAVERICK [17]

Answer:

B) an anion

Explanation:

During bonding atoms may share electron or give out electron . In the case of electrovalent or ionic bonding, electron(s) is/are transferred from one atom to another . The atom that donate the electron are known as donor atom and the atom that receives the electron are known as receivers.

The atom that donate the electron becomes a cation(positively charged) as it lose it electron(s) while the receiver gains electron to become an anion which is a negatively charge ion.

An example of such interaction is between a sodium atom and a chlorine atom. The Na atom lose it electron to become positively charged(cation) while the chlorine atom gains electron to become negatively charged(anion). Both atom bond together losing and gaining electron to fulfill the octet rule.

Na+ + Cl- → NaCl

5 0

3 years ago

How many times higher is the concentration of H+ in the Hubbard Brook sample than in unpolluted rainwater?

Anna [14]

Answer:

1. 7 (a neutral solution)

Answer: 10-7= 0.0000001 moles per liter

2. 5.6 (unpolluted rainwater)

Answer: 10-5.6 = 0.0000025 moles per liter

3. 3.7 (first acid rain sample in North America)

Answer: 10-3.7 = 0.00020 moles per liter

The concentration of H+ in the Hubbard Brook sample is 0.00020/0.0000025, which is 80 times higher than the H+ concentration in unpolluted rainwater.

Explanation:

8 0

3 years ago

Consider the following reactions. (Note: (s) = solid, (l) = liquid, and (g) = gas.) ½H2(g) + ½I2(g) → HI(g), ΔH = +6.2 kcal/mole

BARSIC [14]

½H2(g) + ½I2(g) → HI(g) ΔH = +6.2 kcal/mol
or...
½H2(g) + ½I2(g) + 6,2kcal/mole → HI(g)
________
21.0 kcal/mole + C(s) + 2S(s) → CS2(l)
or...
C(s) + 2S(s) → CS2(l) ΔH = +2,1 kcal/mole
_________
ΔH > 0 ----------->>> ENDOTHERMIC REACTIONS

4 0

3 years ago

Read 2 more answers

A scientist is working in a lab and accidentally combines two liquids that quickly form a solution. Which process could be used

PIT_PIT [208]

Answer:

Distillation

Explanation:

The method of distillation can be used to separate the two liquids, if their boiling point is known. The liquid with lower boiling point will be evaporated and its vapours will be captured, while the liquid with higher boiling point will remail in the container in the liquid state.

7 0

3 years ago