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Elza [17]
3 years ago
7

If 3.0 liters of oxygen gas react with excess carbon monoxide at STP, how many liters of carbon dioxide can be produced under th

e same conditions?
2 CO (g) + O2 (g) ---> 2 CO2 (g)

a. 1.5 L
b. 3.0 L
c. 4.5 L
d. 6.0 L,
Chemistry
1 answer:
AlladinOne [14]3 years ago
6 0
At STP, the volume of a gas represents the number of particles.That said, from the chemical reaction one mole of oxygen reacts with two moles of co to produce the product, CO2At STP, 3 moles of Oxygen will produce 6 moles of CO2. Hence It follows that at standard temperature and pressure, 6.0 L of CO2 will be produced. Option D.
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What group is the element barium(Ba) In?<br>A. 1<br>B. 6<br>C. 56<br>D. 4​
LenaWriter [7]
Ummmmmmmmmmm 56 yes 56
7 0
2 years ago
Calculate [H3O+] and [OH−] for each of the following solutions at 25 ∘C given the pH. pH= 8.74, pH= 11.38, pH= 2.81
Gnom [1K]

Answer:

Explanation:

Given parameters;

pH  = 8.74

pH = 11.38

pH = 2.81

Unknown:

concentration of hydrogen ion and hydroxyl ion for each solution = ?

Solution

The pH of any solution is a convenient scale for measuring the hydrogen ion concentration of any solution.

It is graduated from 1 to 14

      pH = -log[H₃O⁺]

      pOH = -log[OH⁻]

 pH + pOH = 14

Now let us solve;

   pH = 8.74

             since  pH = -log[H₃O⁺]

                           8.74 =  -log[H₃O⁺]

                           [H₃O⁺] = 10⁻^{8.74}

                             [H₃O⁺]  = 1.82 x 10⁻⁹mol dm³

       pH + pOH = 14

                 pOH = 14 - 8.74

                  pOH = 5.26

                  pOH = -log[OH⁻]

                     5.26  = -log[OH⁻]

                     [OH⁻] = 10^{-5.26}

                      [OH⁻] = 5.5 x 10⁻⁶mol dm³

2.  pH = 11.38

             since  pH = -log[H₃O⁺]

                           11.38 =  -log[H₃O⁺]

                           [H₃O⁺] = 10⁻^{11.38}

                             [H₃O⁺]  = 4.17 x 10⁻¹² mol dm³

           pH + pOH = 14

                 pOH = 14 - 11.38

                  pOH = 2.62

                  pOH = -log[OH⁻]

                     2.62  = -log[OH⁻]

                     [OH⁻] = 10^{-2.62}

                      [OH⁻] =2.4 x 10⁻³mol dm³

3. pH = 2.81

             since  pH = -log[H₃O⁺]

                           2.81 =  -log[H₃O⁺]

                           [H₃O⁺] = 10⁻^{2.81}

                             [H₃O⁺]  = 1.55 x 10⁻³ mol dm³

           pH + pOH = 14

                 pOH = 14 - 2.81

                  pOH = 11.19

                  pOH = -log[OH⁻]

                     11.19  = -log[OH⁻]

                     [OH⁻] = 10^{-11.19}

                      [OH⁻] =6.46 x 10⁻¹²mol dm³

5 0
3 years ago
What is the difference in concentration between a pH of 7 and 12?
Ratling [72]

Answer:

The pH of a solution is simply a measure of the concentration of hydrogen ions,  

H

+

, which you'll often see referred to as hydronium cations,  

H

3

O

+

.

More specifically, the pH of the solution is calculated using the negative log base  

10

of the concentration of the hydronium cations.

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

pH

=

−

log

(

[

H

3

O

+

]

)

a

a

∣

∣

−−−−−−−−−−−−−−−−−−−−−−−−  

Now, we use the negative log base  

10

because the concentration of hydronium cations is usually significantly smaller than  

1

.

As you know, every increase in the value of a log function corresponds to one order of magnitude.

Explanation:

4 0
3 years ago
Read 2 more answers
How many electrons are shared in the Lewis structure
Diano4ka-milaya [45]
Hey there,

Answer:

4 valence electrons.

Hope this helps :D

<em>~Top☺</em>
3 0
3 years ago
Consider the chemical equations shown here.
vichka [17]

P₄0₆_{s} + 20₂_{g}   ⇒    P₄0₁₀_{s}

Explanation:

The overall equation for the reaction that produces  P₄0₁₀ is :

P₄0₆_{s} + 20₂_{g}   ⇒    P₄0₁₀_{s}

Now let us derive this equation:

Given equations:

   P₄_{s} + 30₂_{g}  ⇒ P₄0₆_{s}  equation 1;

   P₄_{s} + 50₂_{g} ⇒  P₄0₁₀_{s}  equation 2;

To get the overall combined equation, the equation 1 must be reversed and added to equation 2:

            P₄0₆_{s} ⇒ P₄_{s} + 30₂_{g}   equation 3

                      +

            equation 2:

 P₄_{s} + 50₂_{g}  +    P₄0₆_{s}  ⇒  P₄0₁₀_{s}  +  P₄_{s} + 30₂_{g}  

cancelling specie that appears on both sides and removing excess oxygen gas on the reactant side gives;

   

                  P₄0₆_{s} + 20₂_{g}   ⇒    P₄0₁₀_{s}

learn more:

Net equation brainly.com/question/2947744

#learnwithBrainly

5 0
3 years ago
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