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IRISSAK [1]
3 years ago
5

In an experiment, what is an independent variable?

Chemistry
1 answer:
Andrews [41]3 years ago
8 0

Answer:

The independent variable (IV) is the characteristic of a psychology experiment that is manipulated or changed by researchers, not by other variables in the experiment.

For example, in an experiment about the effect of nutrients on crop growth: The independent variable is the amount of nutrients added to the crop field. The dependent variable is the biomass of the crops at harvest time.

Therefore : 3. the variable that responds to what the scientist changes in an experiment

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- How many moles of H2O are required to react to form 2.5 grams of CH ?
abruzzese [7]

Answer:

0.312 moles of H2O

Explanation:

no. of moles of ch4= mass ÷ molar mass

                               =2.5 ÷ 16.04

                               =0.156 moles of ch4

According to balanced chemical equation

CH4        :        H2O

1 mole     :        2 moles

0.156 moles :       x moles  

by cross multiplication

x=  (0.156x2) ÷ 1

 = 0.312 moles of H2O

7 0
3 years ago
A 7.27-gram sample of a compound is dissolved in 250. grams of benzene. The freezing point of this solution is 1.02°C below that
UkoKoshka [18]

Answer:

The correct answer is 146 g/mol

Explanation:

<em>Freezing point depression</em> is a colligative property related to the number of particles of solute dissolved in a solvent. It is given by:

ΔTf = Kf x m

Where ΔTf is the freezing point depression (in ºC), Kf is a constant for the solvent and m is the molality of solution. From the problem, we know the following data:

ΔTf = 1.02ºC

Kf = 5.12ºC/m

From this, we can calculate the molality:

m = ΔTf/Kf = 1.02ºC/(5.12ºC/m)= 0.199 m

The molality of a solution is defined as the moles of solute per kg of solvent. Thus, we can multiply the molality by the mass of solvent in kg (250 g= 0.25 kg) to obtain the moles of solute:

0.199 mol/kg benzene x 0.25 kg = 0.0498 moles solute

There are 0.0498 moles of solute dissolved in the solution. To calculate the molar mass of the solute, we divide the mass (7.27 g) into the moles:

molar mass = mass/mol = 7.27 g/(0.0498 mol) = 145.9 g/mol ≅ 146 g/mol

<em>Therefore, the molar mass of the compound is 146 g/mol </em>

6 0
3 years ago
What is the molarity (M) of the following solutions?
Dennis_Churaev [7]

Answer:

The molarity (M) of the following solutions are :

A. M = 0.88 M

B. M = 0.76 M

Explanation:

A. Molarity (M) of 19.2 g of Al(OH)3 dissolved in water to make 280 mL of solution.

Molar mass of Al(OH)3 = Mass of Al + 3(mass of O + mass of H)

                                      = 27 + 3(16 + 1)

                                      = 27 + 3(17) = 27 + 51

                                      = 78 g/mole

Al(OH)_3 = 78 g/mole

Given mass= 19.2 g/mole

Mole = \frac{Given\ mass}{Molar\ mass}

Mole = \frac{19.2}{78}

Moles = 0.246

Molarity = \frac{Moles\ of\ solute}{Volume\ of\ solution(L)}

Volume = 280 mL = 0.280 L

Molarity = \frac{0.246}{0.280)}

Molarity  = 0.879 M

Molarity  = 0.88 M

B .The molarity (M) of a 2.6 L solution made with 235.9 g of KBr​

Molar mass of KBr = 119 g/mole

Given mass = 235.9 g

Mole = \frac{235.9}{119}

Moles = 1.98

Volume = 2.6 L

Molarity = \frac{Moles\ of\ solute}{Volume\ of\ solution(L)}

Molarity = \frac{1.98}{2.6)}

Molarity = 0.762 M

Molarity = 0.76 M

4 0
2 years ago
Atomic weight is equal to the number of protons and neutrons in the nucleus. atomic weight is equal to the number of protons and
dlinn [17]
Atomic weight is actually calculated by the sum of protons and neutrons of that atom. It is not equal to the number. You must add them for the result
7 0
2 years ago
Coal, which is primarily carbon, can be converted to natural gas, primarily CH4, by the following exothermic reaction: C(s)+2H2(
LuckyWell [14K]

Answer:

A. The reaction will proceed forward forming more CH4

B. The reaction will proceed forward forming more CH4

C. Since the reaction is exothermic, raising the temperature will cause the reaction to proceed backward, thus forming C and H2.

D. Lowering the volume makes the gas particles to be more close together thereby enhancing their collisions leading to reaction. Therefore the reaction will proceed forward forming more CH4

E. Catalyst only reduce the activation energy so the reaction can proceed faster. The reaction will proceed forward forming.

F. The following will favour CH4 at equilibrium

i. Catalyst to the reaction mixture,

ii. Both adding more H2 to the reaction mixture and lowering the volume of the reaction mixture

iii. Adding more C to the reaction mixture.

4 0
3 years ago
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