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Anvisha [2.4K]
3 years ago
5

1. The West African and______

Chemistry
1 answer:
slega [8]3 years ago
5 0
A. India-southeastern Asia monsoons
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I need D and E please I know the rest
Westkost [7]

Answer:

K2 +Br ->2KBr

K + I ->KI

actually I don't know the e option but I had tried can u pls balance it urself

6 0
3 years ago
Read 2 more answers
How is the modern periodic table arranged?
oksano4ka [1.4K]
By increasing atomic number
5 0
3 years ago
A 2.0 mol ideal gas system is maintained at a constant volume of 4.0 L. If 100 J of heat is added, what is the work done on the
LuckyWell [14K]

Answer:

Work done on the system is zero , hence no work is done since the process is <u>isochoric.</u> There is no work done if the volume remains unchanged. (Though the temperature rises, work is only accomplished when the volume of the gas changes.)

Explanation:

ISOCHORIC PROCESS - An isochoric process, also known as a constant-volume process, isovolumetric process, or isometric process, is a thermodynamic process in which the volume of the closed system undergoing the process remains constant through the process. The heating or cooling of the contents of a sealed, inelastic container is an example of an isochoric process. The thermodynamic process is the addition or removal of heat, the closed system is established by the isolation of the contents of the container, and the constant-volume condition is imposed by the container's inability to deform. It should be a quasi-static isochoric process in this case.

<u>Hence , the work done in the system is zero.</u>

7 0
3 years ago
Consider the reaction below in a closed flask. At 200 o C, the equilibrium constant (Kp) is 2.40 × 103 . 2 NO (g)  N2 (g) + O2
olga55 [171]

Explanation:

Since, the given reaction is as follows.

       2NO(g) \rightleftharpoons N_{2}(g) + O_{2}(g)

Initial:    36.1 atm                 0          0

Change:    2x                      x           x

Equilibrium: (36.1 - 2x)       x            x

Now, expression for K_{p} of this reaction is as follows.

            K_{p} = \frac{[N_{2}][O_{2}]}{[NO]^{2}}

As the initial pressure of NO is 36.1 atm. Hence, partial pressure of O_{2} at equilibrium will be calculated as follows.

              K_{p} = \frac{[N_{2}][O_{2}]}{[NO]^{2}}

        2.40 \times 10^{3} = \frac{x \times x}{(36.1 - 2x)^{2}}

                 x = 18.1 atm

Thus, we can conclude that partial pressure of O_{2} at equilibrium is 18.1 atm.

5 0
3 years ago
Read 2 more answers
Who can help me with a few chemistry balancing equtions?
hoa [83]
<span>Who can help me with a few chemistry balancing equations? 
Me. Just show me the equations :)</span>
6 0
3 years ago
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