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OLEGan [10]
3 years ago
11

Which of the following might be an indicator of global warming?

Chemistry
1 answer:
kolbaska11 [484]3 years ago
6 0

Answer: The correct answer is A. average increase in the temperatures near Earth's surface

Explanation: The reason this is correct is that global warming is it getting hotter, logically the temperatures near Earth's surface will be getting hotter as well. Hope this helps

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Consider the reaction 2N2(g) O2(g)2N2O(g) Using the standard thermodynamic data in the tables linked above, calculate Grxn for t
ratelena [41]

Answer:

\Delta G^0 _{rxn} = 207.6\ kJ/mol

ΔG ≅ 199.91 kJ

Explanation:

Consider the reaction:

2N_{2(g)} + O_{2(g)} \to 2N_2O_{(g)}

temperature = 298.15K

pressure = 22.20 mmHg

From, The standard Thermodynamic Tables; the following data were obtained

\Delta G_f^0  \ \ \ N_2O_{(g)} = 103 .8  \ kJ/mol

\Delta G_f^0  \ \ \ N_2{(g)} =0 \ kJ/mol

\Delta G_f^0  \ \ \ O_2{(g)} =0 \ kJ/mol

\Delta G^0 _{rxn} = 2 \times \Delta G_f^0  \ N_2O_{(g)} - ( 2 \times  \Delta G_f^0  \ N_2{(g)} +   \Delta G_f^0  \ O_{2(g)})

\Delta G^0 _{rxn} = 2 \times 103.8 \ kJ/mol - ( 2 \times  0 +   0)

\Delta G^0 _{rxn} = 207.6\ kJ/mol

The equilibrium constant determined from the partial pressure denoted as K_p can be expressed as :

K_p = \dfrac{(22.20)^2}{(22.20)^2 \times (22.20)}

K_p = \dfrac{1}{ (22.20)}

K_p = 0.045

\Delta G = \Delta G^0 _{rxn} + RT \ lnK

where;

R = gas constant = 8.314 × 10⁻³ kJ

\Delta G =207.6 + 8.314 \times 10 ^{-3} \times 298.15  \ ln(0.045)

\Delta G =207.6 + 2.4788191 \times \ ln(0.045)

\Delta G =207.6+ (-7.687048037)

\Delta G = 199.912952  kJ

ΔG ≅ 199.91 kJ

7 0
3 years ago
A 25.0 mL sample of 0.105 M HCl was titrated to the equivalence point with 31.5 mL of NaOH. What is the concentration of the NaO
Andrei [34K]

Answer:  The concentration of NaOH is 0.08 M.

Explanation:-

According to the neutralization law,

n_1M_1V_1=n_2M_2V_2

where,

M_1 = molarity of HCl solution = 0.105 M

V_1 = volume of HCl solution = 25 ml

M_2 = molarity of NaOH solution = ?

V_2 = volume of NaOH solution = 31.5 ml

n_1 = valency of HCl = 1

n_2 = valency of NaOH = 1

1\times 0.105M\times 25=1\times M_2\times 31.5

M_2=0.08M

Therefore, the concentration of NaOH is 0.08 M.

5 0
3 years ago
After 1911, most scientists accepted the theory that nucleus of an atom was very dense and very small and has a positive charge.
Yanka [14]
After 1911 most scientists accepted<span> the </span>theory<span> that the </span>nucleus<span> of an </span>atom<span> was </span>very dense<span> and </span>very small<span> and </span>has<span> a </span>positive charge<span>. </span>
8 0
3 years ago
A student placed 15.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then
Ede4ka [16]

<u>Answer:</u> The mass of glucose in final solution is 1.085 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}      ......(1)

Given mass of glucose = 15.5 g

Molar mass of glucose = 180.2 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

\text{Molarity of glucose solution}=\frac{15.5\times 1000}{180.2\times 100}\\\\\text{Molarity of glucose solution}=0.860M

To calculate the molarity of the diluted solution, we use the equation:

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of the concentrated glucose solution

M_2\text{ and }V_2 are the molarity and volume of diluted glucose solution

We are given:

M_1=0.860M\\V_1=35.0mL\\M_2=?M\\V_2=0.500L=500mL

Putting values in above equation, we get:

0.860\times 35.0=M_2\times 500\\\\M_2=\frac{0.860\times 35.0}{500}=0.0602M

Now, calculating the mass of glucose by using equation 1, we get:

Molarity of glucose solution = 0.0602 M

Molar mass of glucose = 180.2 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

0.0602=\frac{\text{Mass of glucose solution}\times 1000}{180.2\times 100}\\\\\text{Mass of glucose solution}=\frac{0.0602\times 180.2\times 100}{1000}=1.085g

Hence, the mass of glucose in final solution is 1.085 grams

4 0
3 years ago
Solid waste is eliminated from the body through the esophagus agree or disagree
maria [59]
I very much Disagree.  The esophagus runs from your mouth to your stomach.
7 0
3 years ago
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