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lora16 [44]
2 years ago
8

How many joules are required to heat 2 l of water in a pot from 20 ◦c to the boiling point of water? the specific heat of water

is 4.18 j g · ◦ c and the density of water is 1 g/ml?
Chemistry
1 answer:
miss Akunina [59]2 years ago
4 0
  The  joules  required  to   heat 2L  of  water  in  a  pot  from  20  c  to  the  boiling  point  of  water  is  calculated     using the following  formula

  Q=    MC  delta  T
M  = mass  =   density  x   volume(  2  x 1000= 2000ml)
M  =     1g/ml  x2000  ml  =  2000g
C  =  specific  heat   capacity =  4.18  g/c
delta  T  =  change  in  temperature  =  100 c (  boiling  point  of  water)   -  20 c =  80 c 
Q  is therefore  =   2000 g x 4.18 j/g c  x  80c  =  668800j
 


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Answer:

Kp = \frac{P(NH_{3}) ^{4} P(O_{2}) ^{5}}{P(NO) ^{4} P(H_{2}O)^{6}}

Explanation:

First, we have to write the balanced chemical equation for the reaction. Nitrogen monoxide (NO) reacts with water (H₂O) to give ammonia (NH₃) and oxygen (O₂), according to the following:

NO(g) + H₂O(g) → NH₃(g) + O₂(g)

To balance the equation, we add the stoichiometric coefficients (4 for NH₃ and NO to balance N atoms, then 6 for H₂O to balance H atoms and then 5 for O₂ to balance O atoms):

4 NO(g) + 6 H₂O(g) → 4 NH₃(g) + 5 O₂(g)

All reactants and products are in the gaseous phase, so the equilibrium constant is expressed in terms of partial pressures (P) and is denoted as Kp. The Kp is expressed as the product of the reaction products (NH₃ and O₃) raised by their stoichiometric coefficients (4 and 5, respectively) divided into the product of the reaction reagents (NO and H₂O) raised by their stoichiometric coefficients (4 and 6, respectively). So, the pressure equilibrium constant expression is written as follows:

Kp = \frac{P(NH_{3}) ^{4} P(O_{2}) ^{5}}{P(NO) ^{4} P(H_{2}O)^{6}}

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The diagram below shows a gas sample being measured with an open-end mercury (Hg) manometer. If a barometer reads 730. 1 torr, w
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The pressure of the gas in the flask (in atm) when Δh = 5.89 cm is 1.04 atm

<h3>Data obtained from the question</h3>

The following data were obtained from the question:

  • Atmospheric pressure (Pa) = 730.1 torr = 730.1 mmHg
  • Change in height (Δh) = 5.89 cm
  • Pressure due to Δh (PΔh) = 5.89 cmHg = 5.89 × 10 = 58.9 mmHg
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The pressure of the gas can be obtained as illustrated below:

P = Pa + PΔh

P = 730.1 + 58.9

P = 789 mmHg

Divide by 760 to express in atm

P = 789 / 760

P = 1.04 atm

Thus, the pressure of the gas when Δh = 5.89 cm is 1.04 atm

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In an experiment, you measure a solution absorbance of 0.2 with a path length of 1cm. If the molar absorptivity coefficient is 5
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A solution has an absorbance of 0.2 with a path length of 1 cm. Given the molar absorptivity coefficient is 59 cm⁻¹ M⁻¹, the molarity is 0.003 M.

<h3>What does Beer-Lambert law state?</h3>

The Beer-Lambert law states that for a given material sample, path length and concentration of the sample are directly proportional to the absorbance of the light.

A solution has an absorbance of 0.2 with a path length of 1 cm. Given the molar absorptivity coefficient is 59 cm⁻¹ M⁻¹, we can calculate the molarity of the solution using the following expression.

A = ε × b × c

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where,

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A solution has an absorbance of 0.2 with a path length of 1 cm. Given the molar absorptivity coefficient is 59 cm⁻¹ M⁻¹, the molarity is 0.003 M.

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The energy released by the system is 12.4J

Work done on the surrounding is 4.2J

Follow the steps below to find the change in energy

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Recall, that exothermic reaction is a type of reaction in which heat is released to the surroundings. Hence, change in enthalpy is negative

Step 1; Write the formula for calculating change in energy

\Delta E\text{ }=\text{ q }+\text{ w}

Since heat is released to the surrounding, then q = -12J

Recall, that work done by the system on the surroundings is always negative

Hence, w = -4.2J

Step 2; Substitute the given data into the formula in step 1

\begin{gathered} \text{  }\Delta E\text{ = q + w} \\ \text{  }\Delta E\text{ }=\text{ -12.4 }+\text{ \lparen-4.2\rparen} \\ \text{  }\Delta E\text{ = -12.4 - 4.2} \\ \text{  }\Delta E\text{ }=\text{ -16.6J} \end{gathered}

Therefore, the change i

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