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larisa86 [58]
3 years ago
14

Although carbon dioxide is the primary compound discussed with respect to global warming; of the common atmospheric components t

he global warming potential (GWP) of a carbon dioxide molecule is very low. Which statement best explains this pair of facts?
Chemistry
1 answer:
Mariana [72]3 years ago
6 0

Answer:

C.  CO2 exists in relatively high concentrations.  

Explanation:

From the available options:

A.  CO2 molecules are highly symmetrical.  

B.  CO2 has a low molar mass.  

C.  CO2 exists in relatively high concentrations.  

D.  CO2 is formed in the combustion of fossil fuels.  

<em>The focus on CO2 as far as global warming is concerned is not because It has a high global warming potential but because of its relatively high concentration in the atmosphere as a result of both natural and anthropological activities.</em>

The correct option is C.

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Will chlorine gas effuse more quickly than ammonia gas
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This statement is false due to the fact that the ammonia gas has the lower molar mass.
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3 years ago
Read 2 more answers
Where are the alkali metals and alkaline earth metals and halogens and noble gases located in the periodic table?
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The alkali metals are first column, alkali earth 2nd, halogens 2nd to last, and noble gases last. Hope it helps!
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3 years ago
Modern oil tankers weigh over a half-million tons and have lengths of up to one-fourth miles. Such massive ships require a dista
uranmaximum [27]

Answer:

Acceleration = (change in speed) / (time for the change)

Change in speed= (0 - 26 km/hr) = -26 km/hr

(-26 km/hr) x (1,000 m/km) x (1 hr / 3,600 sec) = -7.222 m/sec

Average acceleration = (-7.222 m/s) / (22 min x 60sec/min) = -0.00547 m/sec²

Average speed during the stopping maneuver =

              (1/2) (start speed + end speed) = 13 km/hr = 3.6111 m/sec

Explanation:

4 0
3 years ago
Consider the balanced equation for the following reaction:
Bad White [126]

<u>Answer:</u> The theoretical yield of the lithium chlorate is 1054.67 grams

<u>Explanation:</u>

To calculate the mass for given number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Actual moles of lithium chlorate = 9.45 moles

Molar mass of lithium chlorate = 90.4 g/mol

Putting values in above equation, we get:

9.45mol=\frac{\text{Actual yield of lithium chlorate}}{90.4g/mol}\\\\\text{Actual yield of lithium chlorate}=(9.45mol\times 90.4g/mol)=854.28g

To calculate the theoretical yield of lithium chlorate, we use the equation:

\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100

Actual yield of lithium chlorate = 854.28 g

Percentage yield of lithium chlorate = 81.0 %

Putting values in above equation, we get:

81=\frac{854.28g}{\text{Theoretical yield of lithium chlorate}}\times 100\\\\\text{Theoretical yield of lithium chlorate}=\frac{854.28\times 100}{81}=1054.67g

Hence, the theoretical yield of the lithium chlorate is 1054.67 grams

7 0
3 years ago
Calculate the standard reaction Gibbs free energy for the following cell reactions: (a) 2 Ce41(aq) 1 3 I2(aq) S 2 Ce31(aq) 1 I32
Law Incorporation [45]

<u>Answer:</u>

<u>For a:</u> The standard Gibbs free energy of the reaction is -347.4 kJ

<u>For b:</u> The standard Gibbs free energy of the reaction is 746.91 kJ

<u>Explanation:</u>

Relationship between standard Gibbs free energy and standard electrode potential follows:

\Delta G^o=-nFE^o_{cell}           ............(1)

  • <u>For a:</u>

The given chemical equation follows:

2Ce^{4+}(aq.)+3I^{-}(aq.)\rightarrow 2Ce^{3+}(aq.)+I_3^-(aq.)

<u>Oxidation half reaction:</u>   Ce^{4+}(aq.)\rightarrow Ce^{3+}(aq.)+e^-       ( × 2)

<u>Reduction half reaction:</u>   3I^_(aq.)+2e^-\rightarrow I_3^-(aq.)

We are given:

n=2\\E^o_{cell}=+1.08V\\F=96500

Putting values in equation 1, we get:

\Delta G^o=-2\times 96500\times (+1.80)=-347,400J=-347.4kJ

Hence, the standard Gibbs free energy of the reaction is -347.4 kJ

  • <u>For b:</u>

The given chemical equation follows:

6Fe^{3+}(aq.)+2Cr^{3+}+7H_2O(l)(aq.)\rightarrow 6Fe^{2+}(aq.)+Cr_2O_7^{2-}(aq.)+14H^+(aq.)

<u>Oxidation half reaction:</u>   Fe^{3+}(aq.)\rightarrow Fe^{2+}(aq.)+e^-       ( × 6)

<u>Reduction half reaction:</u>   2Cr^{2+}(aq.)+7H_2O(l)+6e^-\rightarrow Cr_2O_7^{2-}(aq.)+14H^+(aq.)

We are given:

n=6\\E^o_{cell}=-1.29V\\F=96500

Putting values in equation 1, we get:

\Delta G^o=-6\times 96500\times (-1.29)=746,910J=746.91kJ

Hence, the standard Gibbs free energy of the reaction is 746.91 kJ

7 0
3 years ago
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