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maks197457 [2]
2 years ago
7

Billy decides to test the factors that affect plant growth. He uses sunflowers, daisies, and bean plants. He gives some of the p

lants fertilizer and other plants only water. He plants the sunflowers in a garden outside but plants the daisies and bean plants in small containers inside his house. He measures the height of the plants each day using a ruler. What is wrong with Billy’s experiment?
There are too many responsive variables.
There are too many manipulated variables.
There is nothing wrong with the experiment.
There are too many controls.
Chemistry
1 answer:
Furkat [3]2 years ago
7 0

Answer:

I think B is the answer .

Explanation:

Well cause because all each plant are given only one nutrient to survive on

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A resting adult requires about 240 ml of pure oxygen/mind and breathes about 12 times every minute. if inhaled air contains 20 p
Dmitry_Shevchenko [17]

12 times breathe give 240 ml of pure O_{2}. Each breathe gives 20 ml of O_{2}.

Let us consider, volume of air per breathe= x ml.

Pure O_{2} from inhaled air= \frac{20}{100}x ml and Pure O_{2} from exhaled air= \frac{16}{100}x ml.

Pure O_{2} from inhaled and exhaled air= 20 ml

So, \frac{20}{100}x + \frac{16}{100}x = 20

Therefore, x = 55.5 ml

So, volume of air per breath= 55.5 ml.


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3 years ago
In the figure, which two lines appear to be parallel?
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I don’t understand. Is there a picture? Where are the options?
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3 years ago
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Calculate ΔGrxn for this equation, rounding your answer to the nearest whole number.
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Answer:

-958 spontaneuos

Explanation:

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Apart from boiling point and flammability, what other quality does the size of the molecules in hydrocarbons affect?​
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Viscosity

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2 years ago
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What is the standard enthalpy of a reaction for which the equilibrium constant is (a) doubled, (b) halved when the temperature i
Alexxandr [17]

Answer:

a) 48KJ

b) -48KJ

Explanation:

Given that;

ln(K2/K1) = ΔH°/R(1/T2 - 1/T1)

K2= equilibrium constant at T2

K1 = equilibrum constant at T1

R = gas constant

T1 = initial temperature

T2 = final temperature

When we double the equilibrium constant K1; K2 = 2K1

T1 = 310 K

T2 = 310 + 15 = 325 K

ln(2K1/K1) =- ΔH°/R(1/T2 - 1/T1)

ln2 = -ΔH°/8.314(1/325 - 1/310)

0.693 = -ΔH°/8.314(3.08 * 10^-3 - 3.2 * 10^-3)

0.693 = -ΔH°/8.314 (-0.00012)

0.693 = 0.00012ΔH°/8.314

0.693 * 8.314 = 0.00012ΔH°

ΔH° = 0.693 * 8.314/0.00012

ΔH° = 48KJ

b) K2 =K1/2

ln(K1/2/K1) =- ΔH°/R(1/T2 - 1/T1)

ln (1/2) = -ΔH°/8.314 (1/325 - 1/310)

-0.693 = -ΔH°/8.314  (-0.00012)

-0.693 = 0.00012ΔH°/8.314

-0.693 * 8.314 = 0.00012ΔH°

ΔH°= -0.693 * 8.314/0.00012

ΔH°= -48KJ

6 0
3 years ago
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