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Reptile [31]
3 years ago
7

Write a balanced equation for the following:

Chemistry
1 answer:
Ket [755]3 years ago
7 0

(ANS1)— P4 + 5O2 ---> 2P2O5

(ANS2)— C3H8 + 5O2---> 3CO2 + 4H20

(ANS3)— Ca2Si + 4Cl2 ---> 2CaCl2 + SiCl4

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D. Calculate the variations in the potions of the star due to movement of Earth in its orbit

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The use of nuclear power in place of coal, oil, and natural gas greatly reduces emissions of carbon dioxide and other greenhouse
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False!  It makes emissions greater
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Please help me I can describe a real-world situation in which opposite quantities combine to make zero.
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A hydrogen atom has 0 charge because its two constituents are oppositely charged; and understand p + q as the number located a distance |q| from p, in the positive or negative direction depending on whether q is positive or negative.
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Calculate the concentrations of all species present in a 0.26 M solution of ethylammonium chloride (C2H5NH3Cl).
Alina [70]

Answer:

0.00000223

Explanation:

pKa for C2H5NH3+ = 10.7

pKw = 14.0

pKa + pKb = pKw

10.7 + pKb = 14.0

pKb = 14.0 - 10.7

pKb = 3.30

C2H5NH3Cl is a salt of ethylamine and HCl so it will dissolve in water to produce  C2H5NH3^+ + Cl^-

The base hydrolysis reaction:  C2H5NH3^+(aq) + H2O(l) <=> C2H5NH2(aq) + H3O^+(aq)

This reaction is described by Kb.

Kb = [C2H5NH2][H3O^+]/[C2H5NH3^+]

Let [C2H5NH2] = [H3O^+] = x,

so [C2H5NH3^+] = 0.26 - x

Kb = x^2/(0.26 - x) = 2.00 x 10^-11  

Let's solve for x. In this equation,  It is possible to solve without the use quadratic equation. So we can assume that 0.26 - x  is approximately equal to 0.26.  We won't know until we do the calculation.

We get:  x^2 + 2.00 x 10^-11x - 4.99 x 10^-12 = 0

With the use of a quadratic calculator.

x = 2.23 x 10^-6 M = [C2H5NH2] = [H3O^+]

0.26 - x  is just 0.26 M in this problem because 2.23 x 10^-6 M is insignificant.

[C2H5NH3^+] = 0.26 M = [Cl^-]

NOTE:

pH = -log [H3O^+] = -log(2.23 x 10^-6) = 5.65

Ka is the acid dissociation constant

Kb is the base dissociation constant

5 0
3 years ago
If Kc = 4.0×10−2 for PCl3(g)+Cl2(g)⇌PCl5(g) at 520 K , what is the value of Kp for this reaction at this temperature?
Flauer [41]

Here we have to get the K_{p} of the reaction at 520 K temperature.

The K_{p} of the reaction is 1.705 atm

We know the relation between K_{p} and K_{c} is K_{p}=K_{c}(RT)^{N}, where  K_{p} = The equilibrium constant of the reaction in terms of partial pressure, K_{c}  = The equilibrium constant of the reaction in terms of concentration and N = number of moles of gaseous products - Number of moles of gaseous reactants.

Now in this reaction, PCl₃ + Cl₂ ⇄ PCl₅

Thus number of moles of gaseous product is 1, and number of moles of gaseous reactants are 2. Thus N = |1 - 2| = 1 mole

The given value of  K_{c} is 4.0×10⁻²

The molar gas constant, R = 0.082 L. Atm. mol⁻¹. K⁻¹ and temperature, T = 520 K.

On plugging the values in the equation we get,

K_{p} = 4.0 X 10^{-2}(0.082X520)^{1}

Or, K_{p} = 1.705 atm

Thus, the K_{p} of the reaction is 1.705 atm

7 0
3 years ago
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