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grin007 [14]
3 years ago
12

How did the pH level and the water components level change after adding water to the battery acid?

Chemistry
1 answer:
ololo11 [35]3 years ago
7 0
The pH level changed to 1.34 and the combination of the both made the mixture less acidic
the pH is 1.84, mixture had less acid and there is alot more water molecules in the mixture
the pH level is 2.13 ,, again less acidic and the water molecules has increased to 3,28 x10(25)
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What is the percent of iron (Fe) in 325g of FeCl3
saveliy_v [14]

Answer:

Element Symbol Mass Percent

Chlorine Cl 65.571%

Iron Fe 34.429%

Explanation:

8 0
3 years ago
Calculate the area of a 3.0 inch by 5.0 inch index card in square millimeters (mm). (You can look up the formula for the area of
meriva

Answer:

The area of the given rectangular index card = <u>9677.4 mm²</u>    

Explanation:

Area is defined as the space occupied by a two dimensional shape or object. The SI unit of area is square metre (m²).

<u>The area of a rectangle</u> (A) =  length (l) × width (w)

Given dimensions of the rectangle: Length (l) = 5.0 inch, Width (w) = 3.0 inch

Since, 1 inch = 25.4 millimetres (mm)

Therefore, l = 5 × 25.4 = 127 mm, and w = 3 × 25.4 = 76.2 mm

Therefore, <u>the area of the given rectangular index card</u> = A= l × w = 127 mm × 76.2 mm = <u>9677.4 mm²</u>

5 0
3 years ago
A sample of argon initially has a volume of 5.0 L and the pressure is 2 atm. If the final temperature is 30° C, the final volume
jenyasd209 [6]
Volume of Argon V1 = 5.0 L  
Pressure of Argon P1 = 2 atm 
Final temperature T2 = 30 C = 30 + 273 = 303 K 
Volume at final temperature V2= 6 L 
Pressure at final temperature P2 = 8 atm  
We know that (P1 x V1) / T1 = (P2 x V2) / T2  
(2 x 5)/ T1 = (8 x 6)/ 303 => T1 = (10 x 303) / 48 
Initial Temperature T1 = 3030 / 48 = 63.12 
Initial Temperature = -209. 8 C
4 0
3 years ago
Changes in the composition of the atmosphere have caused gradual changes in earth's _______ throughout history, causing changes
bearhunter [10]

climate

Changes in the composition of the atmosphere have caused gradual changes in earth's <u>climate</u> throughout history, causing changes in plant and animal life that contributed to mass extinctions.

The following are some of the reasons:

  • UV light
  • climate
  • pollutants
  • hydrofluorocarbons

heat

  • The surface of the Earth warms up as sunlight strikes it.
  • Surface-emitted infrared light is absorbed in the atmosphere and transformed into heat.
  • The temperature close to the surface rises as a result of this heat being trapped in the atmosphere.
<h3>UV light:</h3>
  • indirect impacts of climate change on UV radiation from the surface.
  • By changing the concentrations of ozone, UV-absorbing tropospheric gases, aerosols, and clouds in the atmosphere, climate change may have indirectly affected UV radiation levels in the past.
  • These influences are probably going to persist in the future.
<h3>climate:</h3>
  • People are at risk from food and water shortages, greater flooding, high heat, an increase in disease, and economic loss due to climate change.
  • Conflict and human migration are potential outcomes.
  • Climate change is the top hazard to world health in the twenty-first century, according to the World Health Organization (WHO).
<h3>pollutants:</h3>
  • these are also resulting in the increase of temperature of the Earth and is also damaging ozone layer.

To learn more about the changes in earth visit:

brainly.com/question/13434833?

#SPJ4

6 0
1 year ago
Carbonyl fluoride, COF2, is an important intermediate used in the production of fluorine-containing compounds. For instance, it
lions [1.4K]

<u>Answer:</u> The equilibrium concentration of COF_2 is 0.332 M

<u>Explanation:</u>

We are given:

Initial concentration of COF_2 = 2.00 M

The given chemical equation follows:

                2COF_2(g)\rightleftharpoons CO_2(g)+CF_4(g)

<u>Initial:</u>          2.00

<u>At eqllm:</u>     2.00-2x          x      x

The expression of K_c for above equation follows:

K_c=\frac{[CO_2][CF_4]}{[COF_2]^2}

We are given:

K_c=6.30

Putting values in above expression, we get:

6.30=\frac{x\times x}{(2.00-2x)^2}\\\\x=0.834,1.25

Neglecting the value of x = 1.25 because equilibrium concentration of the reactant will becomes negative, which is not possible

So, equilibrium concentration of COF_2=(2.00-2x)=[2.00-(2\times 0.834)]=0.332M

Hence, the equilibrium concentration of COF_2 is 0.332 M

4 0
3 years ago
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