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tiny-mole [99]
3 years ago
9

Which answer gives both a positive impact and a negative impact associated with the effects of nitrogen- and phosphorus-enhanced

fertilizers?
A- increased algal blooms and damage to drinking water

B- increased plant growth and damage to drinking water

C- increase in denitrifying bacteria and increase in plant growth

D- increase in a limiting resource and increase in denitrifying bacteria
Chemistry
2 answers:
dalvyx [7]3 years ago
9 0

Answer:

The answer is B

Explanation:

sergiy2304 [10]3 years ago
5 0

Answer:

Which answer gives both a positive impact and a negative impact associated with the effects of nitrogen- and phosphorus-enhanced fertilizers?

1. increased algal blooms and damage to drinking water

<em><u>Right Answer 2. increased plant growth and damage to drinking water </u></em>

3. increase in denitrifying bacteria and increase in plant growth

4. increase in a limiting resource and increase in denitrifying bacteria

Explanation:

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Discuss how resistance affects current in a circuit.
Mila [183]
<span>resistance impedes flow of electricity to shorten the voltage of a circuit </span>
8 0
3 years ago
Besides the concentration of the acid and its conjugate base, what else does one need to calculate the ph of a weak acid
Marta_Voda [28]
K_{a}, the acid disassociation constant. 
3 0
4 years ago
You need a 35% alcohol solution. on hand, you have a 405 ml of a 30% alcohol mixture. you also have 80% alcohol mixture. how muc
Bumek [7]

You must add 45 mL of the 80 % alcohol to the 30 % alcohol to get a 35 % solution.

You can use a modified dilution formula to calculate the volume of 80 % alcohol

V1×C1 + V2×C2 = V3×C3

Let the volume of 80 % mixture 1 = <em>x</em> mL. Then the volume of the final 35 % mixture 3 = (405 + <em>x</em> ) mL

(<em>x</em> mL×80 % alc) + (405 mL×30 % alc) = (405 + <em>x</em>)mL × 35 % alc

80x + 12 150 = 14 175 + 35 x

45x = 2025

x = 2025/45 = 45

4 0
3 years ago
Which sample contains the greater number of atoms, a sample of Li that contains 6.8 x 10^22 atoms or a 1.97 mole sample of Na?
IrinaVladis [17]

Answer:

                     1.97 moles of Na contains greater number of atoms than 6.8 × 10²² atoms of Li.

Explanation:

                    To solve this problem we will first calculate the number of atoms contained by 1.97 moles of Na and then will compare it with 6.8 × 10²² atoms of Li.

                    As we know that 1 mole of any substance contains exactly 6.022 × 10²³ particles which is also called as Avogadro's Number. So in order to calculate the number of atoms contained by 1.97 moles of Na, we will use following relation,

          Moles  =  Number of Atoms ÷ 6.022 × 10²³ atoms.mol⁻¹

Solving for Number of Atoms,

          Number of Atoms  =  Moles × 6.022 × 10²³ Atoms.mol⁻¹

Putting values,

          Number of Atoms  =  1.07 mol × 6.022 × 10²³ Atoms.mol⁻¹

          Number of Atoms  = 1.18 × 10²⁴ Atoms of Na

Hence,

          1.97 moles of Na contains greater number of atoms than 6.8 × 10²² atoms of Li.

4 0
3 years ago
A 3.452 g sample containing an unknown amount of a Ce(IV) salt is dissolved in 250.0-mL of 1 M H2SO4. A 25.00 mL aliquot is anal
SOVA2 [1]

Answer:

1,812 wt%

Explanation:

The reactions for this titration are:

2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻

I₃⁻ + 2S₂O₃⁻ → 3I⁻ + S₄O₆²⁻

The moles in the end point of S₂O₃⁻ are:

0,01302L×0,03428M Na₂S₂O₃ = 4,463x10⁻⁴ moles of S₂O₃⁻. As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:

4,463x10⁻⁴ moles of S₂O₃⁻×\frac{1molI_{3}^-}{2molS_{2}O_{3}^-} = 2,2315x10⁻⁴ moles of I₃⁻

As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:

2,2315x10⁻⁴ moles of I₃⁻×\frac{2molCe^{4+}}{1molI_{3}^-} = 4,463x10⁻⁴ moles of Ce(IV). These moles are:

4,463x10⁻⁴ moles of Ce(IV)×\frac{140,116g}{1mol} = <em>0,0625 g of Ce(IV)</em>

As the sample has a 3,452g, the weight percent is:

0,0625g of Ce(IV) / 3,452g × 100 = <em>1,812 wt%</em>

I hope it helps!

5 0
3 years ago
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