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NNADVOKAT [17]
3 years ago
11

An experiment was set up to test the hypothesis that all plants will grow faster in a "super soil" versus regular potting soil.

Six identical plants were tested, three in the "super soil" and three in regular potting soil. All plants received the same amount of water, light and fertilizer. Which of the following items would improve the experiment?
add more types of plants


test one plant with no fertilizer


use different soils for each plant


expose all six plants to green light
Chemistry
1 answer:
dolphi86 [110]3 years ago
5 0

Answer:

add more types of plants

Explanation:are you kidding me i’m doing this stuff and i’m in 7th grade

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I need help please ​
Alex17521 [72]

Answer: If you think about it, B. would be the most reasonable answer with the given factors.

4 0
3 years ago
Write a balanced equation for the reaction between nitric acid and lithium carbonate .
love history [14]
Li2CO3 + 2HNO3 ----> 2LiNO3 + H2O + CO2
7 0
3 years ago
Balance the following equations:
lesya [120]
<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2><h2>_____________________________________</h2>

(I) N_{2} + 3H_{2} --> 2NH_{3}

(II) P_{4}  + 5O_{2}  --> 2P_{2} O_{5}

(III) 2NaF + Br_{2}  -->2NaBr + F_{2}

(IV) 2ZnS + 3O_{2} --> 2ZnO + 2SO_2

(V) Pb(NO_3)_2 + 2NaCl --> 2NaNO_3 + PbCl_2

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7 0
3 years ago
372 mL is the volume of Aluminum, density is 2.70 g/mL what is the mass in grams
podryga [215]

Answer:

930 g

Explanation:

D = m/V     Multiply both sides by V

m = VD

<em>Data: </em>

V = 372 cm³

D = 2.70 g/cm³

<em>Calculation: </em>

m = 372 × 2.50

m = 930 g

The mass of Al is 930 g.

7 0
3 years ago
How many moles of water would form the reaction of exactly 58.3 grams of magnesium hydroxide
Marat540 [252]

Answer:

\boxed{\text{2.00 mol}}

Explanation:

We know we will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.

You don't tell us what the reaction is, but we can solve the problem so long as we balance the OH.

M_r:      58.32

          Mg(OH)₂ + … ⟶ … + 2HOH

m/g:       58.3

(a) Moles of Mg(OH)₂

\text{Moles of Mg(OH)$_{2}$} =\text{58.3 g Mg(OH)$_{2}$} \times \dfrac{\text{1 mol Mg(OH)$_{2}$}}{\text{58.32 g Mg(OH)$_{2}$}}\\\\=\text{0.9997 mol Mg(OH)$_{2}$}

(b) Moles of H₂O

The molar ratio is 2 mol H₂O = 1 mol Mg(OH)₂.

\text{Moles of H$_{2}$O}= \text{0.9995 mol Mg(OH)$_{2}$} \times \dfrac{\text{2 mol {H$_{2}$O}}}{ \text{1 mol Mg(OH)$_{2}$}}\\\\= \textbf{2.00 mol H$_{2}$O}

The reaction will form \boxed{\textbf{2.00 mol}} of water.

6 0
3 years ago
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