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ratelena [41]
2 years ago
13

In an experiment to determine the effect of different brands of fertilizer on the rate of plant growth, what are 4 variables tha

t would need to be controlled?
Chemistry
1 answer:
kherson [118]2 years ago
5 0

The variables to control in an experiment to determine the effect of different fertilizers on the rate of plant growth include soil composition, temperature, water, and light.

<h3>What are controlled variables?</h3>

The expression controlled variables makes reference to experimental conditions that must be equal or constant between experimental groups in order to obtain better comparisons when collecting results.

In conclusion, The variables to control in an experiment to determine the effect of different fertilizers on the rate of plant growth include soil composition, temperature, water, and light.

Learn more about controlled variables here:

brainly.com/question/17328868

#SPJ1

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What is the frequency of light that is associated with green in the visible spectrum with a
Delicious77 [7]

Answer:

5.5 × 10^14 Hz

Explanation:

The wavelength can be represented using the formula;

λ = v/f

Where;

λ = wavelength (m)

v = speed of light (3 × 10^8m/s)

f = frequency (Hz)

According to this question, a light associated with the visible spectrum has a wavelength of 550 nm.

550nm = 550 × 10^-9m

f = v/λ

f = 3 × 10^8 ÷ 550 × 10^-9

f = 0.0055 × 10^ (8+9)

f = 0.0055 × 10^17

f = 5.5 × 10^14 Hz

The frequency is 5.5 × 10^14 Hz

8 0
3 years ago
a helium balloon has a volume of 2.95 liters at 25 c. the volume of the ballon decreased to 2.25 l after it is placed outside on
OleMash [197]

Answer:

The outside temperature is -45.8°C

Explanation:

When a gas keeps on constant its moles and its pressure, we can assume that volume will be increased or decreased as the T° (absolute T° in K).

V1 / T1 = V2 / T2

2.95L/298K = 2.25L / T2

(2.95L/298K ) . T2 = 2.25L

T2 = 2.25L . 298K / 2.95L

T2 = 227.2K

T°K - 273 = T°C

227.2K - 273 = -45.8°C

3 0
4 years ago
Please Please! help help! so stress
Oksanka [162]
I AM STRESSED AS WELL
8 0
3 years ago
A gaseous system undergoes a change in temperature and volume. What is the entropy change for a particle in this system if the f
deff fn [24]

Explanation:

Entropy means the amount of randomness present within the molecules of the body of a substance.

Relation between entropy and microstate is as follows.

           S = K_{b} \times ln \Omega

where,      S = entropy

             K_{b} = Boltzmann constant

             \Omega = number of microstates

This equation only holds good when the system is neither losing or gaining energy. And, in the given situation we assume that the system is neither gaining or losing energy.

Also, let us assume that \Omega = 1, and \Omega' = 0.833

Therefore, change in entropy will be calculated as follows.

     \Delta S = K_{b} \times ln \Omega' - K_{b} \times ln \Omega

                 = 1.38 \times 10^{-23} \times ln(0.833) - 1.38 \times 10^{-23} \times \times ln(1)

                 = 1.38 \times 10^{-23} \times (-0.182)

                 = -0.251 \times 10^{-23}

or,             = -2.51 \times 10^{-24}

Thus, we can conclude that the entropy change for a particle in the given system is -2.51 \times 10^{-24} J/K particle.

8 0
3 years ago
g Tibet (altitude above sea level is 29,028 ft) has an atmospheric pressure of 240. mm Hg. Calculate the boiling point of water
Marina CMI [18]

<u>Answer:</u> The boiling point of water in Tibet is 69.9°C

<u>Explanation:</u>

To calculate the boiling point of water in Tibet, we use the Clausius-Clayperon equation, which is:

\ln(\frac{P_2}{P_1})=\frac{\Delta H}{R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

P_1 = initial pressure which is the pressure at normal boiling point = 1 atm = 760 mmHg      (Conversion factor:  1 atm = 760 mmHg)

P_2 = final pressure = 240. mmHg

\Delta H_{vap} = Heat of vaporization = 40.7 kJ/mol = 40700 J/mol     (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

T_1 = initial temperature or normal boiling point of water = 100^oC=[100+273]K=373K

T_2 = final temperature = ?

Putting values in above equation, we get:

\ln(\frac{240}{760})=\frac{40700J/mol}{8.314J/mol.K}[\frac{1}{373}-\frac{1}{T_2}]\\\\-1.153=4895.36[\frac{T_2-373}{373T_2}]\\\\T_2=342.9K

Converting the temperature from kelvins to degree Celsius, by using the conversion factor:

T(K)=T(^oC)+273

342.9=T(^oC)+273\\T(^oC)=(342.9-273)=69.9^oC

Hence, the boiling point of water in Tibet is 69.9°C

3 0
3 years ago
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