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3 years ago

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An experiment was preformed to test the effects of different types of fertilizers on the number of tomatoes produced by one type

of tomato plant. What is the experimental (independent) variable in this investigation?
A color of plant
B type of fertilizer
C number of tomatoes
D type of plant

1 answer:

alisha [4.7K]3 years ago

4 0

B Type of fertilizer

Hope it helps!

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How is a new moon the same as or different from a total lunar eclipse

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Explanation:

Both new moon and lunar eclipse occur when the moon is in the same plane the sun and the earth. However, both are different because of the relative position of the moon in its orbit – relative to earth. In a new moon, the moon is in between the earth and the sun, while in a lunar eclipse, the earth us in between the moon and the sun.

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3 years ago

0.0278, how many significant figures

My name is Ann [436]

Answer:

There are three significant figures

Explanation:

When counting sig figs you don't count the zeros unless it is between a number greater than zero. The two zeros aren't between the greater numbers so there are only 3.

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3 years ago

Describe the method to make pure, dry crystals of magnesium sulfate from a metal oxide and dilute acid.

emmasim [6.3K]

Answer:

H2SO4(aq) + MgO(s) → H2O(l) + MgSO4(aq)

Explanation:

We must recall that the oxides of metals are bases. These metal oxides can react with dilute acids to yield salt and water.

Bearing that in mind, we want to obtain magnesium sulfate from a metal oxide and dilute acid.

In this case we need magnesium oxide and dilute sulphuric acid. The reaction occurs as follows;

H2SO4(aq) + MgO(s) → H2O(l) + MgSO4(aq)

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3 years ago

Why is it important to have long term measurements when studying ecosystems

Alex Ar [27]

Because ecosystems don't change over one day, they change over a long period of time. Plus, something that happens may be because of more than just one change.

5 0

3 years ago

Consider the reaction

SOVA2 [1]

Answer :

(a) The average rate will be:

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

(b) The average rate will be:

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$5Br^-(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)$

The expression for rate of reaction :

$\text{Rate of disappearance of }Br^-=-\frac{1}{5}\frac{d[Br^-]}{dt}$

$\text{Rate of disappearance of }BrO_3^-=-\frac{d[BrO_3^-]}{dt}$

$\text{Rate of disappearance of }H^+=-\frac{1}{6}\frac{d[H^+]}{dt}$

$\text{Rate of formation of }Br_2=+\frac{1}{3}\frac{d[Br_2]}{dt}$

$\text{Rate of formation of }H_2O=+\frac{1}{3}\frac{d[H_2O]}{dt}$

Thus, the rate of reaction will be:

$\text{Rate of reaction}=-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{d[BrO_3^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}=+\frac{1}{3}\frac{d[H_2O]}{dt}$

<u>Part (a) :</u>

<u>Given:</u>

$\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s$

As,

$-\frac{1}{5}\frac{d[Br^-]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}$

and,

$\frac{d[Br_2]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}$

$\frac{d[Br_2]}{dt}=\frac{3}{5}\times 1.56\times 10^{-4}M/s$

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

<u>Part (b) :</u>

<u>Given:</u>

$\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s$

As,

$-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}$

and,

$-\frac{1}{6}\frac{d[H^+]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}$

$\frac{d[H^+]}{dt}=\frac{6}{5}\times 1.56\times 10^{-4}M/s$

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

5 0

4 years ago