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Nadusha1986 [10]
3 years ago
8

"which compound would be expected to show intense ir absorption at both 3500 and 1733cm-1?"

Chemistry
1 answer:
erik [133]3 years ago
4 0

The compound suppose to contain a carbonyl group which shows peak about 1735 Cm^-1 and it also contians an primary amino group for peak of 3500.

Carbonyl group is present in carboxyl group, ketones or ester.

Amino group in presence of carbonyl produces amide group

Thus, the compound must contain CONH_2 group


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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
Which type of fat is a solid at room temperature and does not have double bonds between carbon
melamori03 [73]

Answer:

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Saturated fats are all of the above.

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7 0
3 years ago
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galina1969 [7]

Answer:

a. True

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6 0
3 years ago
Please help, I really don’t understand this!!!
kap26 [50]

<u>Analysing the Question:</u>

We are given the balanced equation:

C₆H₁₂O₆ + 6O₂→ 6CO₂ + 6H₂O

from this equation, we can say that: <em>for every 1 mole of Glucose, we need 6 moles of Oxygen</em>

<u>Moles of Glucose used in the reaction:</u>

Molar mass of Glucose = 180 grams / mol

Given mass of Glucose = 1 gram

Mole of Glucose = Given mass / Molar mass

Moles of Glucose = 1 / 180 moles

<u>Mass of Oxygen required:</u>

We know that for every mole of Glucose, we need 6 moles of Oxygen

So, for 1/180 moles of Glucose, we need 6 / 180 = 1 / 30 moles of Oxygen

Mass of 1 / 30 moles of Oxygen:

Mass = Molar mass * number of moles

Mass of Oxygen = 32 * 1/30

Mass of Oxygen  = 32 / 30

Mass of Oxygen = 1.06 grams

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3 years ago
Which of the following techniques can be used to separate a heterogeneous mixture into its component parts? (3 points)
avanturin [10]

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Separation by density

Hope this helps

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3 years ago
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