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3 years ago

13

How do you identify the element shown in a Bohr model?

1 answer:

krek1111 [17]3 years ago

8 0

Answer: There are three main factors that characterize the Bohr model. First, in the Bohr Model, the orbits have a set size and energy. There’s a correlation between energy and size: the smaller the orbit, the less energy it has. Electrons can move from one orbit to another by emitting or absorbing radiation.

Explanation: Hope this helped !

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Fill in the blanks to balance equation __P+__O2 ---> __P4 O10

Nana76 [90]

Answer: 4P + 3O2 → P4O6

Explanation: plz give brainliest

3 0

3 years ago

In the coal-gasification process, carbon monoxide is converted to carbon dioxide via the following reaction: CO (g) H2O (g) CO2

jolli1 [7]

Answer: $K_{eq}$ at the temperature of the experiment is 0.56.

Explanation:

Moles of $CO$ = 0.35 mole

Moles of $H_2O$ = 0.40 mole

Volume of solution = 1.00 L

Initial concentration of $CO$ = $\frac{0.35mol}{1.00L}=0.35M$

Initial concentration of $H_2O$ = $\frac{0.40mol}{1.00L}=0.40M$

Equilibrium concentration of $CO$ = $\frac{0.19mol}{1.00L}=0.19M$

The given balanced equilibrium reaction is,

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

Initial conc. 0.35 M 0.40 M 0 M 0M

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

Given: (0.35-x) = 0.19

x= 0.16 M

The expression for equilibrium constant for this reaction will be,

$K_{eq}=\frac{[CO_2]\times [H_2]}{[CO]\times [H_2O]}$

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

$K_{eq}=\frac{0.16\times 0.16}{(0.35-0.16)\times (0.40-0.16)}$

$K_{eq}=\frac{0.16\times 0.16}{(0.19)\times (0.24)}=0.56$

Thus $K_{eq}$ at the temperature of the experiment is 0.56.

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3 years ago

Where would you find the asthenosphere?

Harman [31]

Answer:

b. upper mantle

Explanation:

low velocity zone of the upper mantle

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There are (how many?) significant figures in: 0.0050110400

DanielleElmas [232]

I want to say 8 but i'm not sure. I just did a lesson on this and my notes say it could be 8 but then again I'm not sure.

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3 years ago

The name of the compound with the formula ZnBr2 is

Brilliant_brown [7]

Zinc-bromide is its name. Since you have a metal and a non metal you don’t need to use the prefix “di” for the Br.

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3 years ago

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