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2 years ago

What is the compound name of NO3?

Chemistry

2 answers:

Ganezh [65]2 years ago

8 0

Answer:

NITRATE ION

Explanation:

OlgaM077 [116]2 years ago

5 0

Nitrate is the compound Nitrogen Oxygen Oxygen Oxygen

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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

A 400. 0 g sample of liquid water is at 30. 0 ºc. how many joules of energy are required to raise the temperature of the water t

Agata [3.3K]

Energy required to raise the temperature from 35°C - 45 °C= 25116 J.

specific heat, the quantity of warmth required to raise the temperature of one gram of a substance by means of one Celsius degree. The units of precise warmth are generally energy or joules consistent with gram according to Celsius diploma. for instance, the unique warmth of water is 1 calorie (or 4.186 joules) according to gram in step with Celsius degree.

solving,

Sample of liquid = 400. 0 g

temperature = 30. 0 ºc

joules of energy are required to raise the temperature of the water to 45. 0 ºc

therefore rise in temperature 45 - 30 = 15°C

Specific heat capacity = 4.186 J/g m °C

In kelvin = 273 + 15 = 288

= ∴ energy required = Q = m s ( t final - t initial)

= 400*4.186 * 15

= 25116 joule

Learn more about specific heat here:-brainly.com/question/21406849

#SPJ4

7 0

1 year ago

Is anyone else's brainly not working or is mine the only one that is tweaking?

omeli [17]

Mine tweaking as well

5 0

3 years ago

Uranium-238 (U-238) decays into lead-206 (Pb-206). Which of the following is a true statement?

Doss [256]

C is the correct answer

4 0

3 years ago

What pressure is exerted by 932.3 g of CH4 in a 0.560 L steel container at 136.2 K?

Roman55 [17]

Molar mass CH4 = 16.0 g/mol

* number of moles:

932.3 / 16 => 58.26875 moles

T = 136.2 K

V = 0.560 L

P = ?

R = 0.082

Use the clapeyron equation:

P x V = n x R x T

P x 0.560 = 58.26875 x 0.082 x 136.2

P x 0.560 = 650.76

P = 650.76 / 0.560

P = 1162.07 atm

5 0

3 years ago