Solve this whole experiment,

A common function of proteins and carbohydrates in the plasma membrane is...

Vika [28.1K]

I believe your answer is B.

8 0

3 years ago

What is the definition of relief in science terms

SOVA2 [1]

The apparent topography exhibited by minerals in thin section as a consequence of refractive index.

6 0

3 years ago

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

Consider 80.0-g samples of two different compounds consisting of only carbon and oxygen. One of the compounds consists of 21.8 g

goldenfox [79]

Answer:

<u><em>Ratio of the mass carbon that combines with 1.00 g of oxygen in compound 2 to the mass of carbon that combines with 1.00 g of oxygen in compound 1 = 2</em></u>

Explanation:

First, detemine the mass of oxygen in the two samples by difference:

mass of oxygen = mass of sample - mass of carbon

Item Compound 1 Compound 2

Sample 80.0 g 80.0 g

Carbon 21.8 g 34.3 g

Oxygen: 80.0 g - 21.8g = 58.2 g 80.0 g - 34.3 g = 45.7 g

Second, determine the ratios of the masses of carbon that combine with 1.00 g of oxygen:

For each sample, divide the mass of carbon by the mass of oxygen determined above:

Sample Mass of carbon that combines with 1.00 g of oxygen

Compound 1 21.8 g / 58.2 g = 0.375

Compound 2 34.3 g / 45.7 g = 0.751

Third, determine the ratio of the masses of carbon between the two compounds.

Divide the greater number by the smaller number:

Ratio = 0.751 / 0.375 = 2.00 which in whole numbers is 2

6 0

3 years ago

What is discriminate?

vodka [1.7K]

Answer:

Explanation:

To make a distinction in favor of or against a person or thing on basic or group, class or category, to which the person or thing(s) belong

7 0

3 years ago