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3 years ago

What is the positive effect of composed fertilizer?

Chemistry

1 answer:

Phantasy [73]3 years ago

8 0

It’s positive effect is to add nutrients to the soil i think

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Consider the reaction H2(g) + I2(g) Double headed arrow. HI(g) with an equilibrium constant of 46.3 and a reaction quotient of 5

Alika [10]

Answer:

The reaction shifts to the left.

Explanation:

Equilibrium constant (K) = 46.3

Reaction Quotient (Q) = 525

The relationship between Q and K with their implications are given as;

K = Q (No net reaction)

K > Q (Reaction shifts to the right)

K < Q (Reaction shifts to the left)

Since in this question, Q (525) > K (46.3)

The reaction shifts to the left.

5 0

2 years ago

An object is seen to either speed up, slow down, or change direction. these behaviors are caused by what ?

denpristay [2]

I believe the correct answer from the choices listed above is option D. An object is seen to either speed up, slow down, or change direction due to <span>unbalanced forces. When a system is subjected to unbalanced forces, the object would tend to move or change in motion. Hope this answers the question.</span>

7 0

3 years ago

Read 2 more answers

In addition, they are poor conductors of

MA_775_DIABLO [31]

Answer: electricity

Explanation:

6 0

3 years ago

Lewis dot structure for N3-

labwork [276]

Answer :

The steps involved in the electron dot structure of $N^{3-}$ are :

First we have to determine the total number of valence electron in $N^{3-}$ .

Number of valence electrons in N = 5

The charge on N is (-3). So, we add 3 electrons.

Total number of valence electrons = 5 + 3 = 8 electrons

The image is shown below.

4 0

3 years ago

Read 2 more answers

A ball is dropped from a height of 1.20 m and hits the floor. The ball compresses and then reforms to spring upwards from the fl

diamong [38]

Answer:

$\large \boxed{\text{98 m$\cdot$s}^{-2}}$

Explanation:

The formula for the velocity of the ball is

$v = \sqrt{2gh}$

1. Velocity at time of impact

$v = -\sqrt{2 \times 9.807 \times 1.20} = -\sqrt{23.54} = -\textbf{4.85 m/s}$

2. Velocity on rebound

The ball has enough upward velocity to reach a height of 0.86 m.

$v = \sqrt{2 \times 9.807 \times 0.86} = \sqrt{16.87} =\textbf{4.11 m/s}$

3. Acceleration

$a = \dfrac{\Delta v}{\Delta t} = \dfrac{4.11 - (-4.85)}{ 0.091} = \dfrac{8.96 }{0.091} =\textbf{98 m$\cdot$s}^{\mathbf{-2}}\\\\\text{The acceleration while the ball is in contact with the floor is $\large \boxed{\textbf{98 m$\cdot$s}^{\mathbf{-2}}}$}$

5 0

3 years ago