The equilibrium constant for the reaction 2 C3H6(g) ↔ C2H4(g) + C4H8(g) Is found to fit the expression: lnK = A + B/T + C/T2, between 300 and 600 K, where A = -1.04, B = -1088 K, C = 1.51x105 K2. Calculate the standard reaction enthalpy and standard reaction entropy at 400 K.

The noble gas configurations for

Li= [He] 2s¹, Na=[Ne] 3s¹ and K =[Ar] 4s¹.

Noble gas configurations are the term for the electronic configurations of the atoms for which the electronic configuration becomes very larger i.e, for the higher atomic numbers.

For the noble gas electronic configurations, the noble gas, which is before that atom is written.

So, the Lithium can be represented as [He] 2s¹.

So, the Sodium can be represented as [Ne] 3s¹.

So, the Potassium can be represented as [Ar] 4s¹.

Hence, the noble gas configurations for Lithium= [He] 2s¹, Sodium=[Ne] 3s¹ and potassium =[Ar] 4s¹.

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Answer:

The answer is C

Answer:

Option 3

Explanation:

Heat is absorbed, so the reaction is endothermic.

Answer:

A. Along warm-water currents from warmer regions to colder regions

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Explanation:

Answer:

Ammonia is easily made in the laboratory by heating an ammonium salt, such as ammonium chloride NH4Cl with a strong alkali, such as sodium hydroxide or calcium hydroxide.

The gas may also be made by warming concentrated ammonium hydroxide.

Explanation:

2NH4Cl + Ca(OH)2 → CaCl2 + 2H2O + 2NH3(g)

Answer:

2 Fe(HCO₃)₃ + 3 CaO ---> 1 Fe₂O₃ + 3 Ca(HCO₃)₂

Explanation:

The unbalanced equation:

Fe(HCO₃)₃ + CaO ---> Fe₂O₃ + Ca(HCO₃)₂

Reactants: 1 iron, 3 hydrogen, 3 carbon, 10 oxygen, 1 calcium

Products: 2 iron, 2 hydrogen, 2 carbon, 9 oxygen, 1 calcium

As you can see, the reaction is clearly unbalanced because the number of each element on both sides is different. The best way to go about balancing an equation is by trying to first balance an element located in only one molecule on either side. For example, I started trying to balance  iron. After balancing iron, I looked to balance other elements that would have been affected by the new coefficient (i.e., hydrogen, carbon).

The balanced equation:

2 Fe(HCO₃)₃ + 3 CaO ---> 1 Fe₂O₃ + 3 Ca(HCO₃)₂

Reactants: 2 iron, 6 hydrogen, 6 carbon, 21 oxygen, 3 calcium

Products: 2 iron, 6 hydrogen, 6 carbon, 21 oxygen, 3 calcium

Now, with these coefficients, the quantity of each element on both sides is the same.

Answer:

The time required to decompose the substance is 360 seconds.

Answer:

2.52 g H₂

Explanation:

To find the mass of hydrogen gas, you need to (1) convert moles H₂O to moles H₂ (via mole-to-mole ratio from reaction coefficient) and then (2) convert moles H₂ to grams H₂ (via molar mass from periodic table values). When multiplying the given value by the mole-to-mole ratio, you need to use the coefficients of the balanced equation. An reaction is balanced once there is an equal amount of each element on both sides.

(Step 1)

The unbalanced equation:

Ca(s) + H₂O(l) ----> Ca(OH)₂(aq) + H₂(g)

Reactants: 1 calcium, 2 hydrogen, 1 oxygen

Products: 1 calcium, 4 hydrogen, 2 oxygen

The balanced equation:

1 Ca(s) + 2 H₂O(l) ----> 1 Ca(OH)₂(aq) + 1 H₂(g)

Reactants: 1 calcium, 4 hydrogen, 2 oxygen

Products: 1 calcium, 4 hydrogen, 2 oxygen

(Step 2)

Molar Mass (H₂): 2(1.008 g/mol)

Molar Mass (H₂): 2.016 g/mol

(Step 3)

 2.50 moles H₂O              1 mole H₂                  2.016 g
---------------------------  x  -------------------------  x  ----------------------  =  2.52 g H₂
                                        2 moles H₂O                1 mole

Answer:

nitrogen compounds in the air

Explanation:

Answer:

92.9%

Explanation:

You have been given the actual yield of the reaction. First, you need to find the theoretical yield of the reaction. To do this, you need to (1) convert grams Fe₂O₃ to moles Fe₂O₃ (via molar mass from periodic table values), then (2) convert moles Fe₂O₃ to moles Fe (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles Fe to grams Fe (via molar mass).

Once you have found the theoretical yield, you need to use the percent yield equation to calculate the final answer. This number should have 3 sig figs to match the given values.

(Step 1)

Molar Mass (Fe₂O₃): 2(55.845 g/mol) + 3(15.998 g/mol)

Molar Mass (Fe₂O₃): 159.684 g/mol

1 Fe₂O₃(s) + 3 CO(g) ---> 2 Fe(s) + 3 CO₂(g)

Molar Mass (Fe): 55.845 g/mol

50.0 g Fe₂O₃           1 mole              2 moles Fe           55.845 g
--------------------  x  ------------------  x  ---------------------  x  ----------------  = 35.0 g Fe
                               159.684 g         1 mole Fe₂O₃          1 mole

(Step 2)

                                 Actual Yield
Percent Yield  =  ---------------------------  x  100%
                             Theoretical Yield

                                32.5 g Fe
Percent Yield  =  ----------------------  x  100%  =  92.9%
                                35.0 g Fe

Answer:

pH=2.3

Explanation:

One H2SO4 molecule produce two H+ ions.

So conc. of H+ = 2(0.0025)

=0.005

Now, pH=-log (conc. of H+)

=-log(0.005)

=2.3

Answer:

40 g= 6.022×10²³

x=1.4×10¹⁵

x=40g×6.022×10²³/1.4×10¹⁵

x=17.77×10⁸

Answer:

1.20 M (3 s.f.)

Explanation:

In this question, we are trying to find the concentration of KCl after dilution. Note that the number of moles of KCl does not change.

[tex]g \div \frac{g}{mol} = g \times \frac{mol}{g} = mol[/tex]

Number of moles of KCl

= 44.7 ÷74.55

= 0.59960 mol (5 s.f.)

New volume

= 500 mL

= (500 ÷1000) L

= 0.5 L

Concentration after dilution

= 0.59960 ÷0.5

= 1.1992 mol/L

= 1.20 M (3 s.f.)

* M is the symbol for Molarity, which is also equal to mol/L

Answer:

4.458 L

Explanation:

Assuming the pressure remains constant  and assuming 24° C and 58° C

V1/T1  = V2/T2       Where T is in Kelvin

V1/T1  *  T2  = V2

4 / ( 273.15 +24)    *    ( 273.15 + 58)    =  4.458 L

Electrochemical cell is one in which energy is released by a chemical action.

An electrochemical cell is one in which energy is released by a chemical action.

Now we have the reaction;

Zn^2 (aq) + 2e- = Zn(s)

Pb²(aq) + 2e- → Pb(s)

and

Cr2O7 + 14H + 6e- 2Cr³+ 7H₂O

H₂O2 + 2H+ + 2e- →2H₂O

In the first case, the Ecell is obtained from;

(-0.13 V) - (-0.76 V)

= 0.63 V

In the second case;

1.78 V - 1.33 V

= 0.45 V

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Answer: The true volume of the gas will be lower than predicted by the ideal gas law because the volume of the actual particles is more significant under high pressure.

Answer:

A

The H+ from acid and OH- from base forms water, the other parts form the salt.

121 g [tex]AlI_{3}[/tex] aluminum iodide, from the complete reaction of 113 grams I₂ according to the following balanced chemical equation:

2Al(s) + 3I₂(s) → 2[tex]AlI_{3}[/tex]

As per the given balanced equation,

3 moles of [tex]I_{2}[/tex] are used to give 2 moles of [tex]AlI_{3}[/tex].

Atomic weight of iodine is 127 gm/ mole

So, molecular weight of iodine molecule = 254 gm/ mole

There for , we can say that 1 mole o iodine molecule contain 254 g

So ,Given 113 gm iodine contains how many moles?

  Moles of iodine = 113/254 = 0.44 moles

from the equation we can say that 3 moles of iodine used to give 2 moles of aluminum iodide.

So, 0.44 moles of iodine gives how many moles of aluminum iodide ?

   Moles of aluminum iodide = 0.44 × 2 ÷  3

                                             = 0.2965 moles

Molecular weight of one mole of aluminum iodide = atomic weight of Al + 3( atomic weight of iodine )

      = 408 gm

∴ 1 mole of aluminum iodide = 408 gm

so, 0.2965 moles of aluminum iodide = ?

            = 0.2964 × 408

            = 121 gm  Aluminum iodide formed.

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The value of pka for [tex]NH_3[/tex] in an aqueous solution is 9.2.

Kb denotes the base dissociation constant.

pKa + pKb =14  at 25 degree celcius.

pKa + 4.8 =14

pKa = 9.2

pKb is the negative base-10 logarithm of the base dissociation constant (Kb) of a solution.

It is used to determine the strength of a base or alkaline solution.

The value of pka for [tex]NH_3[/tex] in an aqueous solution is 9.2.

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Answer:

The answer is B an open system allows energy and mass to move in and out of it

Hope it helps

Answer:

=6×10^7metre

Explanation:

By using formula,

F=GMm/d^2

Answer:

The Red Circles: Liquid, Blue and Yellow Circles: Solid, and Purple Circles: Gas

Explanation:

Should be it based on the picture and given information.

Answer:

Verdadero!
no se en que mas quieres que te ayude? o cual realmente quieres que sea la respuesta?

The production of ammonia will be favored at high pressure and low temperature.

The term exothermic reaction has to do with a reaction in which the forward reaction is favored at lower temperatures.

Now looking at the reaction coordinate and the equation of the reaction, we know that the production of ammonia will be favored at high pressure and low temperature.

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There is no delay because the effects of gravity are constantly acting on bodies of water. Hence, option D is correct.

Gravitation or just gravity is the force of attraction between any two bodies.

The Moon exerts over twice the gravitational pull of the Sun on Earth's tides because of its proximity to Earth. -The lunar tidal bulges are about twice the size of the solar tidal bulges. -The Sun's contribution to the tides is less than that of the Moon.

When the sun and moon are perfectly unaligned (they form a 90-degree angle relative to the earth), there are still tides because the moon's gravitational gradient is stronger than the sun's. The sun's gravitational gradient never completely cancels out the moon's.

Hence, option D is correct.

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In an endothermic reaction, Keq increases as the temperature increases and vice versa.

In an exothermic reaction, Keq decreases as the temperature increases and vice versa.

Therefore, the temperature is the only factor that affects the equilibrium constant of a reaction.

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The electroplating process of the coin results in the coating of the coin with copper metal.

Electroplating is the process of using coating the surface of a metal with another one electrolytic cell.

The electroplating of the coin is done using copper as the anode and the coin as the cathode. The electrolyte is copper sulfate solution.

Oxidation occurs at the anode with the copper metal being converted to copper (ii) ions.

Reduction occurred at the cathode as the copper (ii) ions in the solution accept electrons to become copper metal and the coin being coated with copper.

The half-reactions are shown below:

At the anode:

[tex]Cu(s) \rightarrow Cu^{2+}(aq) + 2e^{-}[/tex]

At the cathode:

[tex]Cu^{2+}(aq) + 2e{-} \rightarrow Cu(s)[/tex]

Therefore, the electroplating process of the coin results in the coating of the coin with copper metal.

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