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Lady bird [3.3K]
3 years ago
15

Luke and Sian want to plant a vegetable garden in their yard. A soil testing kit measures the soil pH at 5.0, but the lettuce th

ey want to plant does best at a pH of 6.5. Should they add an acid or a base to the soil to make it the optimum pH for growing lettuce?
Chemistry
1 answer:
dexar [7]3 years ago
4 0

Answer:

So, Luke and Sian has to increase the pH of the soil by adding base to it.

Explanation:

The pH is defined as the negative logarithm of the hydrogen ion concentration in their aqueous solution.

pH=-\log[H^+]

  • With increase in hydrogen ion concentration the pH value decreases.
  • With decrease in hydrogen ion concentration the pH value increases.

The pH of the soil after testing it on a kit comes out be 5.0, but they both need pH of the soil to 6.5.

Comparison of pH of soil:

  =  5.0 < 6.5

= High hydrogen ion concentration > High hydrogen ion concentration

So, Luke and Sian has to increase the pH of the soil by adding base .Doing so will decrease the hydrogen ion concentration in the soil (where as addition of acid lower the pH of soil).

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konstantin123 [22]
I think the correct answer is the first option. It has nonpolar bonds and a symmetrical structure. The structure of a BF3 molecule shows a symmetrical trigonal geometry. The net dipole moment of the molecule is zero therefore it is polar.
6 0
3 years ago
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How many double bonds does CCL2H2 have?
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None. Both chlorines and both hydrogens are single-bonded to the central carbon atom; the molecule is comprised of four single bonds and no double bonds.

Hope this helps!
4 0
2 years ago
At an elevated temperature, Kp=4.2 x 10^-9 for the reaction 2HBr (g)---&gt; +H2(g) + Br2 (g). If the initial partial pressures o
Damm [24]

Answer : The partial pressure of H_2 at equilibrium is, 1.0 × 10⁻⁶

Explanation :

The partial pressure of HBr = 1.0\times 10^{-2}atm

The partial pressure of H_2 = 2.0\times 10^{-4}atm

The partial pressure of Br_2 = 2.0\times 10^{-4}atm

K_p=4.2\times 10^{-9}

The balanced equilibrium reaction is,

                                2HBr(g)\rightleftharpoons H_2(g)+Br_2(g)

Initial pressure    1.0×10⁻²       2.0×10⁻⁴      2.0×10⁻⁴

At eqm.            (1.0×10⁻²-2p)   (2.0×10⁻⁴+p)  (2.0×10⁻⁴+p)

The expression of equilibrium constant K_p for the reaction will be:

K_p=\frac{(p_{H_2})(p_{Br_2})}{(p_{HBr})^2}

Now put all the values in this expression, we get :

4.2\times 10^{-9}=\frac{(2.0\times 10^{-4}+p)(2.0\times 10^{-4}+p)}{(1.0\times 10^{-2}-2p)^2}

p=-1.99\times 10^{-4}

The partial pressure of H_2 at equilibrium = (2.0×10⁻⁴+(-1.99×10⁻⁴) )= 1.0 × 10⁻⁶

Therefore, the partial pressure of H_2 at equilibrium is, 1.0 × 10⁻⁶

4 0
3 years ago
In the simulation, open the Custom mode. The beaker will be filled to the 0.50 L mark with a neutral solution. Set the pH to 2.8
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3 years ago
Calculate the equilibrium constant of the reaction below if the pressures are 1.0atm, 2.0 atm, and 1.0 atm respectively. PCl3 +
Makovka662 [10]

Answer:

K = 0.5

Explanation:

Based on the reaction:

PCl₃ + Cl₂ ⇄ PCl₅

The equilibrium constant, K, is defined as:

K = P PCl₅ / P PCl₃ * P Cl₂

<em>Where P represent the pressure at the equilibrium for each one of the gases involved in the equilibrium.</em>

<em />

As:

P PCl₅ = 1.0atm

P PCl₃ = 1.0atm

P Cl₂ = 2.0atm

K = 1.0atm / 1.0atm * 2.0atm

<h3>K = 0.5</h3>
7 0
2 years ago
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