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2 years ago

Use Dot and cross method to show the bonding of the atoms and state the number of atoms of each element present

Chemistry

1 answer:

goldfiish [28.3K]2 years ago

5 0

Answer:

Check the attached one..I only draw the Dot method of Magnesium and Fluorine.. since i love chemistry ..Hope this helps you...Balance you have to do it yourself..If you didn't get the answer let me know in the comments..But you have try okay..

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(01.03 MC)

vlada-n [284]

Answer:

Explanation:

What you know is that the electrical energy comes from the wind turbine. Therefore the x must identify the type of energy that the wind turbine represents.

It must represent mechanical energy because blades of the wind turbine turns to produce electrical energy.

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2 years ago

An element whose third shell contains two <br> p<br> electrons

pentagon [3]

Maybe sodium? becasue its an element and I would like some points... sorryy:((((

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3 years ago

Read 2 more answers

6. Compare and Contrast How are the properties of acids and bases similar? How are they different?

baherus [9]

$\huge\underline{\red{A}\blue{n}\pink{s}\purple{w}\orange{e}\green{r} - }$

Acids and bases both react with water and a lot of acids and bases are soluble in nature.
Both acids and bases are electrolytes which means that they’re good conductors of electricity.
Acids and bases both produce ions in water solution.
Acids release hydrogen ions (H+) whereas Bases release hydroxide ions (OH–).

4 0

2 years ago

URGENT PLEASE HELP!!

Orlov [11]

Answer:

Liters per second is 0.2 LPS

Liters per hour is 720

Explanation:

Divide the litres by the time, than multiply the time of an hour (in seconds) and than multiply the lps by time of an hour in seconds.

3 0

3 years ago

t 745 K, the reaction below has an equilibrium constant (Kc) of 5.00 × 102. H2 (g) + I2 (g) ⇌ 2 HI (g) If a mixture of 0.10 mol

spayn [35]

Answer : The concentration of HI (g) at equilibrium is, 0.643 M

Explanation :

The given chemical reaction is:

$H_2(g)+I_2(g)\rightarrow 2HI(g)$

Initial conc. 0.10 0.10 0.50

At eqm. (0.10-x) (0.10-x) (0.50+2x)

As we are given:

$K_c=5.00\times 10^2$

The expression for equilibrium constant is:

$K_c=\frac{[HI]^2}{[H_2][I_2]}$

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

$5.00\times 10^2=\frac{(0.50+2x)^2}{(0.10-x)\times (0.10-x)}$

x = 0.0713 and x = 0.134

We are neglecting value of x = 0.134 because the equilibrium concentration can not be more than initial concentration.

Thus, we are taking value of x = 0.0713

The concentration of HI (g) at equilibrium = (0.50+2x) = [0.50+2(0.0713)] = 0.643 M

Thus, the concentration of HI (g) at equilibrium is, 0.643 M

8 0

4 years ago