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Mashutka [201]
3 years ago
8

Be sure to answer all parts. One gallon of gasoline in an automobiles engine produces on average 9.50 kg of carbon dioxide, whic

h is a greenhouse gas; that is, it promotes the warming of Earth's atmosphere. Calculate the annual production of carbon dioxide in kilograms if there are exactly 40.0 million cars in the United States and each car covers a distance of 5790 mi at a consumption rate of 24.1 miles per gallon. Enter your answer in scientific notation. × 10 kg
Chemistry
1 answer:
Sever21 [200]3 years ago
6 0

Answer:

The annual production of carbon dioxide is 9.12\times 10^{10} kg.

Explanation:

Distance covered by each car = 5790 miles

Rate of consumption of gasoline =24. mile/gal

For every 24.1 mile 1 gallon of gasoline is used

Gasoline used by a single car by travelling 5790 miles =

\frac{1}{24.1}\times 5790 mile=240.24 gal

Number of cars in the United states = 40.0 million = 4\times 10^7

Total gallons of gasoline consumed by 40 million cars = 4\times 10^7\times 240.24 gal=9.60\times 10^9 gal

1 gallon of gasoline produces = 9.50 kg of CO_2

Then 9.60\times 10^9 gal of gasoline will produce:

9.60\times 10^9\times 9.50 kg =9.12\times 10^{10} kg of CO_2

The annual production of carbon dioxide is 9.12\times 10^{10} kg.

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W + BgCz2 --&gt; WCz + Bg<br> Balance this equation
yanalaym [24]

Answer: W + BgCz2 arrow WCz + Bg

2 W + BgCz2 arrow  2 WCz + Bg

Explanation:

Cz has 2 so you balcne the other side of WCz.

Since you Balcanes the Cz you changed the W and you Balcanes the other W on the left side.

3 0
3 years ago
Write the formula of the conjugate acid of HCO₂⁻.
Alex

Taking into account the Brønsted-Lowry acid-base theory, the conjugate acid of HCO₂⁻ is H₂CO₂.

<h3>Brønsted-Lowry acid-base </h3>

The Brønsted-Lowry acid-base theory (or the Brønsted-Lowry theory) identifies acids and bases based on whether the species accepts or donates protons or H⁺.

According to this theory, acids are proton donors while bases are proton acceptors. That is, an acid is a species that donates an H⁺ proton while a base is a chemical species that accepts an H⁺ proton from the acid.

So, reactions between acids and bases are H⁺ proton transfer reactions.

<h3>Conjugate base and conjugate acid</h3>

Then, a conjugate base is an ion or molecule resulting from the acid that loses the proton, while a conjugate acid is an ion or molecule resulting from the base that gains the proton:

acid + base ⇄ conjugate base + conjugate acid

<h3>Conjugate acid of HCO₂⁻</h3>

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Learn more about the Brønsted-Lowry acid-base theory:

<u>brainly.com/question/12916250?referrer=searchResults</u>

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7 0
3 years ago
File:
givi [52]

Ummm I just need to answer questions sorry!!!

8 0
4 years ago
⚠️HELP DUE IN 10 MINS WILL GIVE BRAINLIEST!!! THANKS⚠️
LenKa [72]

Answer:

hahahh noob

Explanation:

3 0
3 years ago
Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.210 M HClO(aq) with 0.210 M KOH(aq).
Degger [83]
a) before addition of any KOH : 

when we use the Ka equation & Ka = 4 x 10^-8 : 

Ka = [H+]^2 / [ HCIO]

by substitution:

4 x 10^-8 = [H+]^2 / 0.21

[H+]^2 = (4 x 10^-8) * 0.21

           = 8.4 x 10^-9

[H+] = √(8.4 x 10^-9)

       = 9.2 x 10^-5 M

when PH = -㏒[H+]

   PH = -㏒(9.2 x 10^-5)

        = 4  

b)After addition of 25 mL of KOH: this produces a buffer solution 

So, we will use Henderson-Hasselbalch equation to get PH:

PH = Pka +㏒[Salt]/[acid]


first, we have to get moles of HCIO= molarity * volume

                                                           =0.21M * 0.05L

                                                           = 0.0105 moles

then, moles of KOH = molarity * volume 

                                  = 0.21 * 0.025

                                  =0.00525 moles 

∴moles HCIO remaining = 0.0105 - 0.00525 = 0.00525

and when the total volume is = 0.05 L + 0.025 L =  0.075 L

So the molarity of HCIO = moles HCIO remaining / total volume

                                        = 0.00525 / 0.075

                                        =0.07 M

and molarity of KCIO = moles KCIO / total volume

                                    = 0.00525 / 0.075

                                    = 0.07 M

and when Ka = 4 x 10^-8 

∴Pka =-㏒Ka

         = -㏒(4 x 10^-8)

         = 7.4 

by substitution in H-H equation:

PH = 7.4 + ㏒(0.07/0.07)

∴PH = 7.4 

c) after addition of 35 mL of KOH:

we will use the H-H equation again as we have a buffer solution:

PH = Pka + ㏒[salt/acid]

first, we have to get moles HCIO = molarity * volume 

                                                        = 0.21 M * 0.05L

                                                        = 0.0105 moles

then moles KOH = molarity * volume
                            =  0.22 M* 0.035 L 

                            =0.0077 moles 

∴ moles of HCIO remaining = 0.0105 - 0.0077=  8 x 10^-5

when the total volume = 0.05L + 0.035L = 0.085 L

∴ the molarity of HCIO = moles HCIO remaining / total volume 

                                      = 8 x 10^-5 / 0.085

                                      = 9.4 x 10^-4 M

and the molarity of KCIO = moles KCIO / total volume

                                          = 0.0077M / 0.085L

                                          = 0.09 M

by substitution:

PH = 7.4 + ㏒( 0.09 /9.4 x 10^-4)

∴PH = 8.38

D)After addition of 50 mL:

from the above solutions, we can see that 0.0105 mol HCIO reacting with 0.0105 mol KOH to produce 0.0105 mol KCIO which dissolve in 0.1 L (0.5L+0.5L) of the solution.

the molarity of KCIO = moles KCIO / total volume

                                   = 0.0105mol / 0.1 L

                                   = 0.105 M

when Ka = KW / Kb

∴Kb = 1 x 10^-14 / 4 x 10^-8

       = 2.5 x 10^-7

by using Kb expression:

Kb = [CIO-] [OH-] / [KCIO]

when [CIO-] =[OH-] so we can substitute by [OH-] instead of [CIO-]

Kb = [OH-]^2 / [KCIO] 

2.5 x 10^-7 = [OH-]^2 /0.105

∴[OH-] = 0.00016 M

POH = -㏒[OH-]

∴POH = -㏒0.00016

           = 3.8
∴PH = 14- POH

        =14 - 3.8

PH = 10.2

e) after addition 60 mL of KOH:

when KOH neutralized all the HCIO so, to get the molarity of KOH solution

M1*V1= M2*V2

 when M1 is the molarity of KOH solution

V1 is the total volume = 0.05 + 0.06 = 0.11 L

M2 = 0.21 M 

V2 is the excess volume added  of KOH = 0.01L

so by substitution:

M1 * 0.11L = 0.21*0.01L

∴M1 =0.02 M

∴[KOH] = [OH-] = 0.02 M

∴POH = -㏒[OH-]

           = -㏒0.02 

           = 1.7

∴PH = 14- POH

       = 14- 1.7 

      = 12.3 
8 0
3 years ago
Read 2 more answers
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