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1 year ago

Describe a method to investigate how the temperature changes when different masses of ammonium

Chemistry

1 answer:

rewona [7]1 year ago

4 0

Answer:

Decrease temperature with increase in ammonium nitrate concentration.

Explanation:

The temperature changes when different masses of ammonium nitrate are dissolved in water because the ammonium nitrate takes energy from the surrounding water environment for the breaking of ionic bonds as a result the temperature of water decreases. When more concentration of ammonium nitrate is added to water, more decrease occur in temperature of water and the water becomes cold.

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The periodic table _____.

zhuklara [117]

Explanation:

eto po yung pic thank me later☺️

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1 year ago

Read 2 more answers

Which of the following is NOT a clue that a chemical change has occurred? * 1 point Sugar turns blackish-brown as it is heated.

Vladimir [108]

Answer:

The answer is Steam is produced when water is heated. The color does not change, a precipice is not formed, and the odor does not change!

5 0

2 years ago

A 0.150-kg sample of a metal alloy is heated at 540 Celsius an then plunged into a 0.400-kg of water at 10.0 Celsius, which is c

Zarrin [17]

Answer:

$C_{alloy}=0.497\frac{J}{g\°C}$

Explanation:

Hello there!

In this case, according to this calorimetry problem on equilibrium temperature, it is possible for us to infer that the heat released by the metal allow is absorbed by the water for us to write:

$Q_{allow}=-(Q_{water}+Q_{Al})$

Thus, by writing the aforementioned in terms of mass, specific heat and temperature, we have:

$m_{alloy}C_{alloy}(T_{eq}-T_{alloy})=-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})$

Then, we solve for specific heat of the metallic alloy to obtain:

$C_{alloy}=\frac{-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})}{m_{alloy}(T_{eq}-T_{alloy})}$

Thereby, we plug in the given data to obtain:

$C_{alloy}=\frac{-(400g*4.184\frac{J}{g\°C} (30.5\°C-10.0\°C)+200g*0.900\frac{J}{g\°C}(30.5\°C-10.0\°C)}{150g(30.5\°C-540\°C)} \\\\C_{alloy}=0.497\frac{J}{g\°C}$

Regards!

3 0

2 years ago

A mixture of 454 kg of applesauce at 10 degrees Celsius is heated in a heat exchanger by adding 121300 kJ. Calculate the outlet

stira [4]

Explanation:

The given data is as follows.

Mass of apple sauce mixture = 454 kg

Heat added (Q) = 121300 kJ

Heat capacity ( $C_{p}$ ) of apple sauce at $32.8^{o}C$ = 4.0177 $kJ/kg^{o}C$

So, Heat given by heat exchanger = heat taken by apple sauce

Q = $mC_{p} \Delta T$

or, Q = $mC_{p} (T_{f} - T_{i})$

Putting the given values into the above formula as follows.

Q = $mC_{p} (T_{f} - T_{i})$

121300 kJ = $454 kg \times 4.0177 kJ/kg^{o}C \times (T_{f} - 10)$

$T_{f}$ = $76.5^{o}C$

Thus, we can conclude that outlet temperature of the apple sauce is $76.5^{o}C$ .

3 0

2 years ago

En un experimento hacemos reaccionar 12 g de carbono con 32 g de oxígeno para formar dióxido de carbono. Razona si podemos saber

max2010maxim [7]

Answer:

La masa de óxido de carbono iv formado es 44 g.

Explanation:

En esta pregunta, se nos pide calcular la masa de óxido de carbono iv formado a partir de la reacción de masas dadas de carbono y oxígeno.

En primer lugar, necesitamos escribir una ecuación química equilibrada.

C + O2 → CO2

De la ecuación, 1 mol de carbono reaccionó con 1 mol de oxígeno para dar 1 mol de óxido de carbono iv.

Ahora, si marca las masas en la pregunta, verá que corresponde a la masa atómica y la masa molar de la molécula de carbono y oxígeno, respectivamente. ¿Qué indica esto?

Como tenemos una relación molar de 1: 1 en todo momento, lo que esto significa es que la masa de óxido de carbono iv producida también es la misma que la masa molar de óxido de carbono iv.

Por lo tanto, procedemos a calcular la masa molar de óxido de carbono iv Esto es igual a 12 + 2 (16) = 12 + 32 = 44 g Por lo tanto, la masa de óxido de carbono iv formado es 44 g

5 0

2 years ago