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victus00 [196]
2 years ago
13

What phase does chloride occur in?

Chemistry
1 answer:
Bumek [7]2 years ago
8 0

Answer:

solid

Explanation:

chlorine is liquid

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28 ml of 0.10 m hcl is added to 60 ml of 0.10 m sr(oh)2. determine the concentration of oh− in the resulting solution.
Blababa [14]
Here is an acid-base reaction. Hydrochloric acid (HCl) reacts with strontium hydroxide [ Sr(OH)2 ]

Ions H+ and OH- neutralize each other. If the amounts are not equal, one of them will be in excess. 
Follow the steps as

1. Find moles of ions: mole= Molarity * Volume (in liter) ; n= M * V OR millimole = Molarity * Volume (in ml) ;
2. Write the equation
3. Find out excess ion
4. Use final volume (V acid + V base ) to calculate concentration of excess ion.

n HCI = 28 ml * 0.10 M = 0.28 mmol, releases 0.28 mmol H+ ions
n Sr(OH)2= 60 ml * 0.10 M= 0.60 mmol, releases 2* 0.60=1.20 mmol OH- ions
since Sr(OH)2⇒ Sr2+ + 2OH-

Neutralization reaction is OH- + H+ ---> H2O. The ratio is 1:1. That means 1 mmol hydroxide ions will neutralize 1 mmol hydrogen ions. Since OH- ions are greater in amount, they will be in excess

n(OH-) - n(H+)= 1.20 - 0.28 = 0.92 mmol OH- ions UNREACTED.

Total volume= V acid + V base= 28 ml + 60 ml = 98 ml

Molarity of OH- ions= mole / Vtotal = 0.92/98= 0.009 M

The answer is 0.009 M.


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For the following reaction, C6H12O6 and O2 are the
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3 years ago
A sample of flammable liquid is placed into an enclosed cylinder which is then fitted with a movable piston. Initially the cylin
polet [3.4K]

Answer:

12.09 L

Explanation:

Step 1: Convert 826.1 mmHg to atm

We will use the conversion factor 760 mmHg = 1 atm.

826.1 mmHg × 1 atm/760 mmHg = 1.087 atm

Step 2: Convert 427.8 J to L.atm

We will use the conversion factor 101.3 J = 1 L.atm.

427.8 J × 1 L.atm/101.3 J = 4.223 L.atm

Step 3: Calculate the change in the volume

Assuming the work done (w) is 4.223 L.atm against a pressure (P) of 1.087 atm, the change in the volume is:

w = P × ΔV

ΔV = w/P

ΔV = 4.223 L.atm/1.087 atm = 3.885 L

Step 4: Calculate the final volume

V₂ = V₁ + ΔV

V₂ = 8.20 L + 3.885 L = 12.09 L

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