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Rina8888 [55]
3 years ago
8

What charge will an ion of phosphorus form?

Chemistry
2 answers:
Fed [463]3 years ago
8 0

Answer:

A charge of -3

Explanation:

If you look on a periodic table you will find that phosphorus is 3 elements away from the nearest noble gas argon. This means that phosphorus would prefer a charge of -3!

Ludmilka [50]3 years ago
3 0

Answer:typically it will be -3

Explanation:

It prefers to take up 3 electrons and thereby become phosphide with a -3 charge. since it is a nonmetal it will take up electrons making it negative as an ion. In order to fill up its last shell, to become stable, to form an octet it will need 3 added electrons which will give it that -3 charge. An atom of phophorus will be neutral, An ion of phsphorus will have a -3 charge.

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1. calculate the the reaction of gas, F 2 (g) with H 2 O(l) water form and O 2 (e).
Sever21 [200]

hope this helped.:)

Explanation:

Yes, the number of moles of oxygen gas produced by your reaction under those conditions for pressure and temperature will be 0.0025.

Hydrogen peroxide,  

H

2

O

2

, decomposes to give water and oxygen gas according to the balanced chemical equation

2

H

2

O

2

(

a

q

)

→

2

H

2

O

(

l

)

+

O

2

(

g

)

You've collected 0.061 L of oxygen gas at 295.15 K and 1 atm, so you've got all the data you need to calculate the number of moles of oxygen gas produced by using the ideal gas law equation

P

V

=

n

R

T

⇒

n

=

P

V

R

T

n

O

2

=

1

atm

⋅

0.061

L

0.082

L

⋅

atm

mol

⋅

K

=

0.0025 moles

So, if this was your first question, then yes, your reaction produced 0.0025 moles of oxygen gas.

I find the second part of your question to be a little confusing. You were given the density of the hydrogen peroxide solution, so are you supposed to use that to determine the theoretical number of moles of oxygen for this reaction?

I'm not sure what  

100%

H

2

O

2

=

1.02 g/mL  

means, do you have a certain volume of hydrogen peroxide solution?

SIDE NOTE According to the additional information posted by Heather, it turns out that the initial hydrogen peroxide solution had a volume of 5 mL.

Even with the volume of the initial solution, you'd need its percent concentration to try and determine exactly how many moles you had present before the reaction.

Once you know how many moles of hydrogen peroxide you had, assume that all of the react and use the  

2

:

1

mole ratio that exists between  

H

2

O

2

and  

O

2

to get the number of moles of oxygen your reaction could have produced.

7 0
3 years ago
Calcula el pH de las siguientes sustancias formadas durante una erupción volcánica:
Virty [35]

<em>Calculate the pH of the following substances formed during a volcanic eruption: </em>

<em>• Acid rain if the [H +] is 1.9 x 10-5 </em>

<em>• Sulfurous acid if [H +] = 0.10 </em>

<em>• Nitric acid if [H +] = 0.11</em>

<em />

<h3>Further explanation  </h3>

pH is the degree of acidity of a solution that depends on the concentration of H⁺ ions. The greater the value the more acidic the solution and the smaller the pH.  

pH = - log [H⁺]  

  • pH acid rain

\tt pH=-log[1.9\times 10^{-5}]\\\\pH=5-log1.9\\\\pH=4.72

  • pH Sulfurous acid

\tt pH=-log[10^{-1}]\\\\pH=1

  • pH Nitric acid

\tt pH=-log[11\times 10^{-2}]\\\\pH=2-log~11=0.959

8 0
3 years ago
How many moles are in 187.54 grams of magnesium chlorate?
Svetlanka [38]

Hey there!

Magnesium chlorate: Mg(ClO₃)₂

Find molar mass.

Mg: 1 x 24.305 = 24.305

Cl: 2 x 35.453 = 70.906

O: 6 x 16 = 96

------------------------------------

                      191.211 g/mol

We have 187.54 grams.

187.54 ÷ 191.211 = 0.9808

There are 0.9808 moles in 187.54 grams of magnesium chlorate.

Hope this helps!

3 0
4 years ago
Diatomic iodine [I2] decomposes at high temperature to form I atoms according to the reaction I2(g)⇌2I(g), Kc = 0.011 at 1200∘C
umka2103 [35]

Answer:

The concentration of I at equilibrium = 3.3166×10⁻² M

Explanation:

For the equilibrium reaction,

I₂ (g) ⇄ 2I (g)

The expression for Kc for the reaction is:

K_c=\frac {\left[I_{Equilibrium} \right]^2}{\left[I_2_{Equilibrium} \right]}

Given:

\left[I_2_{Equilibrium} \right] = 0.10 M

Kc = 0.011

Applying in the above formula to find the equilibrium concentration of I as:

0.011=\frac {\left[I_{Equilibrium} \right]^2}{0.10}

So,

\left[I_{Equilibrium} \right]^2=0.011\times 0.10

\left[I_{Equilibrium} \right]^2=0.0011

\left[I_{Equilibrium} \right]=3.3166\times 10^{-2}\ M

<u>Thus, The concentration of I at equilibrium = 3.3166×10⁻² M</u>

3 0
4 years ago
PLZ HELP Question 14 of 25 What is the name for a representation of the physical world?
JulijaS [17]

Answer:

Model

Explanation:

A model of anything is something you make to represent it in it's physical world form

8 0
3 years ago
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