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Alinara [238K]
3 years ago
11

Why is Cu+O=2CuO unbalanced ​

Chemistry
2 answers:
sesenic [268]3 years ago
6 0
Because the two before the cuo means there is 2 moles of each atom and in the reactants there is only one mole of each so to balance the equation u need to put 2 before the cu and before the o
stepan [7]3 years ago
3 0

Answer:

because the product of CuO contains 2 molecules so it is unbalanced

Explanation:

to balance it 2Cu+2O=4CuO now it is balanced

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What does MVLMEN stand for
Soloha48 [4]
There are a couple things that this can stand for
3 0
3 years ago
The following are electronegativity. What value should be where the M is?0.981.57M2.553.043.443.98No data
ipn [44]

Answer 2.04

Explanation

Electronegativity decreases down the group and increases across the period in the periodic table.

In the period two of the periodic table,we have the following values for electronegativities with respect to its elements.

Li...........0.98

Be.........1.57

B..........2.04

C.........2.55

N..........3.04

O...........3.44

F...........3.98

Ne........n.a

The value that should be where M is is 2.04

6 0
1 year ago
In the coal-gasification process, carbon monoxide is converted to carbon dioxide via the following reaction: CO (g) + H2O (g) ⇌
Oksana_A [137]

Answer: Equilibrium constant is 0.70.

Explanation:

Initial moles of  CO = 0.35 mole

Volume of container = 1 L

Initial concentration of CO=\frac{moles}{volume}=\frac{0.35moles}{1L}=0.35M

Initial moles of  H_2O = 0.40 mole

Volume of container = 1 L

Initial concentration of H_2O=\frac{moles}{volume}=\frac{0.40moles}{1L}=0.40M

equilibrium concentration of CO=\frac{moles}{volume}=\frac{0.18moles}{1L}=0.18M [/tex]

The given balanced equilibrium reaction is,

                            CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)

Initial conc.            0.35 M       0.40M       0     0

At eqm. conc.    (0.35-x) M   (0.40-x) M   (x) M    (x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[CO_2]\times [H_2O]}{[CO]\times [H_2O]}

K_c=\frac{x\times x}{(0.40-x)(0.35-x)}

we are given : (0.35-x)= 0.18

x = 0.17

Now put all the given values in this expression, we get :

K_c=\frac{0.17\times 0.17}{(0.40-0.17)(0.35-0.17)}

K_c=0.70

Thus the value of the equilibrium constant is 0.70.

5 0
3 years ago
I will give the first person brainliest! please helppp
Murrr4er [49]

okay so the  Answer is c

7 0
3 years ago
Based on your knowledge of factors that affect the rates of chemical reactions, predict the trend in the last column of the expe
Anika [276]

Factors that increases reaction rate such as increase in concentration or pressure will reduce reaction time whereas factors that decrease reaction rate such as inhibitors will increase reaction time.

<h3>What are the factors that affect reaction rate?</h3>

Factors that affect reaction rate are those factors which increase or decrease the rate of chemical reaction.

The factors that affect reaction rate include:

  • temperature
  • concentration/pressure
  • catalysts
  • surface area
  • nature of substance

Any factor that increases reaction rate such as increase in concentration or pressure will reduce reaction time whereas factors that decrease reaction rate such as inhibitors will increase reaction time.

Learn more about factors affecting reaction rate at: brainly.com/question/14817541

#SPJ1

6 0
2 years ago
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