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juin [17]
3 years ago
13

According to the following thermochemical equation, if 951.1 g of NO 2 is produced, how much heat is released at constant pressu

re? 2NO(g) + O2(g) 2NO 2(g); DH -114.4 kJ
Chemistry
1 answer:
nordsb [41]3 years ago
4 0

Answer:

The amount of released is 1,182 kJ.

Explanation:

When heat is released at constant pressure, this change in energy is known as enthalpy (ΔH°) of the reaction. Enthalpy is an extensive property, so it depends on the amount of reacting material. Let's take a look at the provided equation:

2 NO(g) + O₂ ⇄ 2 NO₂(g)   ΔH° = -114.4 kJ

Since this equation is balanced with 2 moles of NO₂(g), we can say that 114.4 kJ are released every 2 moles of NO₂(g) produced. By convention, when enthalpies are negative, it means that energy is <em>released</em> and the reaction is exothermic. Conversely, positive enthalpies mean energy is <em>absorbed</em> and the reaction is endothermic.

We can calculate the amount of energy released taking into account the previous relationship (-114.4 kJ/2 moles of NO₂(g)), the mass of NO₂(g) produced (951.1g) and its molar mass (46.00g/mol). The calculations would be:

951.1g.\frac{1molNO_{2} }{46.00g} .\frac{-114.4kJ}{2molesNO_{2} } =-1,182kJ

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Norma-Jean [14]

Explanation:

The given data is as follows.

   Current (I) = 3.50 amp,        Mass deposited = 100.0 g

  Molar mass of Cr = 52 g

It is known that 1 faraday of electricity will deposit 1 mole of chromium. As 1 faraday means 96500 C and 1 mole of Cr means 52 g.

Therefore, 100 g of Cr will be deposited by "z" grams of electricity.

                z \times 52 g = 96500 \times 100 g

                         z = \frac{96500 \times 100 g}{52 g}

                            = 185576.9 C

As we know that, Q = I × t

Hence, putting the given values into the above equation as follows.

                      Q = I × t

           185576.9 C = 3.50 amp \times t  

                      t = 53021.9 sec

Thus, we can conclude that 100 g of Cr will be deposited in 53021.9 sec.

3 0
3 years ago
An acid with molar mass 84.48 g/mol is titrated with 0.650 M KOH. What volume of KOH solution is needed to titrate 1.70 grams of
RSB [31]

Answer:

V=0.0310L=3.10mL

Explanation:

Hello.

In this case, since the acid is monoprotic and the KOH has one hydroxyl ion only, we can see that at the equivalence point the moles of both of them are the same:

n_{acid}=n_{KOH}

Thus, since we are given 1.70 g of the acid, we compute the moles of acid that were titrated:

n_{acid}=1.70g*\frac{1mol}{84.48g}=0.0201mol

Which equal the moles of KOH. In such a way, since the molarity is defined as moles over liters (M=n/V), the liters are moles over molarity (V=n/M), thus, the resulting volume is:

V=\frac{0.0201mol}{0.650mol/L}\\\\V=0.0310L=3.10mL

Best regards!

7 0
3 years ago
What is the purpose of sodium and chloride ions in the exoeruent?
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5 0
3 years ago
At sea level, where the pressure was 104 kPa and temperature 21.1 ºC, a certain mass of air occupies 2.0 m3 . To what volume wil
Romashka [77]

Answer:

The volume of air at where the pressure and temperature are  52 kPa, -5.0 ºC is 3.64 m^3.

Explanation:

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 104 kPa

P_2 = final pressure of gas = 52 kPa

V_1 = initial volume of gas = 2.0m^3

V_2 = final volume of gas = ?

T_1 = initial temperature of gas = 21.1^oC=273+21.1=294.1K

T_2 = final temperature of gas = -5.0^oC=273+(-5.0)=268 K

Now put all the given values in the above equation, we get:

\frac{104 kPa\times 2.0m^3}{294.1 K}=\frac{52 kPa\times V_2}{268 K}

V_2=3.64 m^3

The volume of air at where the pressure and temperature are  52 kPa, -5.0 ºC is 3.64 m^3.

3 0
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Using the fact that at 0.10 M, the absorbance was 0.357; and at 0.20 M, the absorbance was 0.714, the absorptivity (slope) is
olga_2 [115]

Answer:

The answer is "3.57 and 0.07".

Explanation:

Using the slop formula:

=\frac{n_2-n_1}{n_2-n_1}\\\\ =\frac{0.714-0.357}{0.20-0.10} \\\\ = ABSORPTIVITY, \ \ a=3.57 \\\\A= a\times b \times c

Given:

A=0.250

length path b=1  

from calibration it is found that

a=3.57\\\\consentation \ c=\frac{A}{a\times b} \\\\C=\frac{0.250}{3.57\times 1}

   =0.07 \ M

3 0
3 years ago
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