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vodka [1.7K]
3 years ago
9

The continued use of fertilizer can cause change in the pH of the growing medium. Which of the following materials will make the

pH more alkaline?
A. Ammonium sulfate
B. Calcium nitrate
C. Potassium chloride
D. Nitrate of soda
Chemistry
1 answer:
mash [69]3 years ago
6 0

Answer:

B. Calcium nitrate

Explanation:

Calcium nitrate is one of the materials that will increase the pH of the soil and make it more alkaline in nature.

Calcium nitrate is helps to increase the alkalinity of soils.

  • For a plant to take up nitrate, it must discharge or lose a hydroxyl ion.
  • A hydrogen ion is used to take up the nitrate.
  • This releases more hydroxyl into the soil than usual.
  • Ammonium sulfate reduces soil pH.
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Answer:

1.373 wt% Ca(OH)₂

Explanation:

Sample mix = 15.0g

Ca(OH)₂(aq) + 2HCl(aq) => CaCl₂(aq) + 2H₂O(l)

moles HCl = 0.2000g / 36 g·mol⁻¹ 0.0056 mol

moles Ca(OH)₂ = 1/2(moles HCl) = 1/2(0.0056 mol) = 0.0028 mol

mass Ca(OH)₂ = 0.0028 mol ( 74 g/mol ) = 0.206 g

mass % Ca(OH)₂ = (0.206/15.0)100% = 1.373 wt%

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In a coffee-cup calorimeter, 150.0 mL of 0.50 M HCl is added to 50.0 mL of 1.00 M NaOH to make 200.0 g solution at an initial te
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Answer:

51.54°C the final temperature of the calorimeter contents.

Explanation:

HCl+NaOH\rightarrow H_2O+NaCl,\Delta H=-56 kJ/mol

moles=Molarity\times Volume (L)

Molarity of HCl= 0.50 M

Volume of HCl= 150.0 mL = 0.150 L

Moles of HCl= n

n=0.50 M\times 0.150 L=0.075 mol

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Volume of NaOH= 50.0 mL = 0.050 L

Moles of NaOH= n'

n'=1.00 M\times 0.050 L=0.050 mol

Since moles of NaOH are less than than moles of HCl. so energy release will be for neutralization of 0.050 moles of naOH by 0.050 moles of HCl.

n = 0.050

-56 kJ/mol=-\frac{Q}{n}

Q=56 kJ/mol\times 0.050 kJ/mol=2.8 kJ=2800 J

(1 kJ= 1000 J)

The energy change released during the reaction = 2800 J

Volume of solution = 150.0 mL + 50.0 mL = 200.0 mL

Density of the solution (water) = 1.00g/mL

Mass of the solution , m= 200 mL × 1.00 g/mL = 200 g

Now , calculate the final temperature by the solution from :

q=mc\times (T_{final}-T_{initial})

where,

q = heat gained = 2800 J

c = specific heat of solution = 4.184 J/^oC

T_{final} = final temperature = ?

T_{initial} = initial temperature = 48.2^oC

Now put all the given values in the above formula, we get:

2800 J=200.0 g\times 4.184 J/^oC\times (T_{final}-48.2)^oC

T_{final}= 51.54^oC

51.54°C the final temperature of the calorimeter contents.

7 0
3 years ago
In a constant‑pressure calorimeter, 70.0 mL of 0.350 M Ba(OH)2 was added to 70.0 mL of 0.700 M HCl. The reaction caused the temp
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<u>Answer:</u> The amount of heat absorbed by the solution is 2.795 kJ

<u>Explanation:</u>

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Density of water = 1 g/mL

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Putting values in above equation, we get:

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To calculate the heat absorbed, we use the equation:

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where,

q = heat absorbed

m = mass of water = 140 g

c = heat capacity of water = 4.186 J/g°C

\Delta T = change in temperature = T_2-T_1=(28.74-23.97)^oC=4.77^oC

Putting values in above equation, we get:

q=140g\times 4.186J/g^oC\times 4.77^oC=2795.4J=2.795kJ

Hence, the amount of heat absorbed by the solution is 2.795 kJ

3 0
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