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DiKsa [7]
2 years ago
8

Lol nevermind i figured it out

Chemistry
1 answer:
Scorpion4ik [409]2 years ago
6 0
Hehehehe hi,
If someone else answers please please pleaseeeee mark me as brainliest. Thank youuu
You might be interested in
He following balanced equation shows the formation of water. 2H2 + O2 mc020-1.jpg 2H2O How many moles of oxygen (O2) are require
mrs_skeptik [129]

Answer : The correct option is, 13.7 mole

Solution : Given,

Moles of H_2 = 27.4 moles

The given balanced chemical reaction is,

2H_2+O_2\rightarrow 2H_2O

From the balanced chemical reaction, we conclude that

As, 2 moles of H_2 react with 1 moles of O_2

So, 27.4 moles of H_2 react with \frac{27.4}{2}=13.7 moles of O_2

Therefore, the number of moles of oxygen O_2 required are, 13.7 moles

7 0
2 years ago
Read 2 more answers
A sample of pure NO2 is heated to 335 ∘C at which temperature it partially dissociates according to the equation 2NO2(g)⇌2NO(g)+
vodka [1.7K]

The equilibrium constant for the reaction is 0.00662

Explanation:

The balanced chemical equation is :

2NO2(g)⇌2NO(g)+O2(g

At t=t  1-2x ⇔ 2x + x moles

The ideal gas law equation will be used here

PV=nRT

here n= \frac{w}{W} = \frac{w}{V}= density

P = \frac{density RT}{M}       density is 0.525g/L, temperature= 608.15 K, P = 0.750 atm

putting the values in reaction

0.75 = \frac{0.525 x 0.0821 x 608.15 }{M}

  M    = 34.61

         

to calculate the Kc

Kc=\frac{ [NO] [O2]}{NO2}

  \frac{1-2x}{1+x} x M NO2 + \frac{2x}{1+x} M NO+ \frac{x}{1+x} M O2

Putting the values as molecular weight of NO2, NO,O2

\frac{46(1-2x) +30(2x)+32x}{1+x}

34.61= \frac{46}{1+x}

x= 0.33

Kc= \frac{4x^2)x}{1-2x^2}

    putting the values in the above equation

Kc = 0.00662

     

5 0
2 years ago
Which of the following is a characteristic of a scientific theory
Andru [333]

Answer:

number one

Explanation:

hope I helped

8 0
2 years ago
Ascorbic acid, or vitamin C (C6H8O6, molar mass = 176 g/mol), is a naturally occurring organic compound with antioxidant propert
nydimaria [60]

Answer : 50.69 mg of ascorbic acid does not meet the daily requirement.

Solution : Given,

Molar mass of Ascorbic acid = 176 g/mole

Moles of Ascorbic acid = 2.88\times 10^{-4}moles

Formula used :

Moles=\frac{Mass}{\text{ Molar mass}}

or, \text{ Mass of ascorbic acid}=\text{ Moles of ascorbic acid}\times \text{ Molar mass of ascorbic acid}

Now put all the given values in this formula, we get the mass of ascorbic acid.

\text{ Mass of ascorbic acid}=(2.88\times 10^{-4}moles)\times (176g/mole)=0.050688g=50.69mg

Conversion : (1g=1000mg)

As per question, a healthy adult’s daily requirement of vitamin C is 70-90 mg. But calculate mass of vitamin C is 50.69 mg. So, 50.69 mg of ascorbic acid does not meet the daily requirement.

5 0
3 years ago
A mixture of XO2 (P = 3.00 atm) and O2 (P = 1.00 atm) is placed in a container. This elementary reaction takes place at 27 °C: 2
sukhopar [10]

Answer:

a) \triangle G^{0} = 7.31 kJ/mol

b) K_{-1} = 0.0594 m^{-1} s^{-1}

Explanation:

Equation of reaction:

                                     2 XO_{2} (g) + O_{2} (g) \rightleftharpoons 2XO_{3} (g)

Initial pressure                  3              1              0

Pressure change             2P           1P             2P

Total pressure = (3-2P) + (1-P) + (2P)

Total Pressure = 3.75 atm

(3-2P) + (1-P) + (2P) = 3.75

4 - P = 3.75

P = 4 - 3.75

P = 0.25 atm

Let us calculate the pressure of each of the components of the reaction:

Pressure of XO2 = 3 - 2P = 3 - 2(0.25)

Pressure of XO2 =2.5 atm

Pressure of O2 = 1 - P = 1 -0.25

Pressure of O2 = 0.75 atm

Pressure of XO3 = 2P = 2 * 0.25

Pressure of XO3 = 0.5 atm

From the reaction, equilibrium constant can be calculated using the formula:

K_{p} = \frac{[PXO_{3}] ^{2} }{[PXO_{2}] ^{2}[PO_{2}] }

K_{p} = \frac{0.5^2}{2.5^2 *0.75} \\K_{p} = 0.0533 = K_{eq}

Standard free energy:

\triangle G^{0} = - RT ln k_{eq} \\\triangle G^{0} = -(0.008314*300* ln0.0533)\\\triangle G^{0} = 7.31 kJ/mol

b) value of k−1 at 27 °C, i.e. 300K

K_{1} = 7.8 * 10^{-2} m^{-2} s^{-1}

K_{c} = K_{p}RT\\K_{c} = 0.0533* 0.0821 * 300\\K_{c} = 1.313 m^{-1}

K_{-1} = \frac{K_{1} }{K_{c} } \\K_{-1} = \frac{7.8 * 10^{-2}  }{1.313 }\\K_{-1} = 0.0594 m^{-1} s^{-1}

6 0
3 years ago
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