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Helen [10]
3 years ago
9

A student conducts an experiment to determine how the temperature of water affects the time for sugar to dissolve. In each trial

, the
student uses a different amount of water and a different temperature of water.
What is wrong with this experimental design?
Chemistry
1 answer:
V125BC [204]3 years ago
8 0

The fact that the student used different amount of water (another independent variable) is wrong with the experimental design

WHAT ARE THE COMPONENTS OF AN EXPERIMENT?

  • An experiment aims at solving a scientific problem or answering a scientific question. An experiment should contain a variable being changed called INDEPENDENT VARIABLE and a variable being measured called DEPENDENT VARIABLE.

  • In an ideal experiment, only one independent variable should be used while every other variable should be kept constant. This is done so as not to affect the result of the experiment.

In the experiment conducted by the student in this question, two independent variables were used i.e. the different amount of water and the different temperatures. This is what is wrong about the experimental design.

  • In a nutshell, the fact that two independent variables were used by the student is what is wrong about the experimental design.

Learn more at: brainly.com/question/967776

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Now suppose, instead, that 5.678 g of a volatile solute is dissolved in 150.0 g of water. This solute also does not react with w
n200080 [17]

Answer:

59.9 g/mol is the molar mass for the solute

Explanation:

Lowering vapor pressure → ΔP = P° . Xm

P° → Vapor pressure of pure solvent

ΔP = P° - Vapor pressure of solution

Xm = Mole fraction of solute

17.54 Torr  - 17.344 Torr = 17.54 Torr . Xm

0.196 Torr / 17.54 Torr = Xm → 0.0112

These are the moles of solute / Total moles

Total moles = Moles of solute + Moles of solvent

We determine the moles of solvent → 150 g . 1mol/ 18 g = 8.33 moles

Now we can make this equation:

0.0112 = Moles of solute / Moles of solute + 8.33 mol

0.0112 Moles of solute + 0.0933 = Moles of solute

0.0933 = Moles of solute - 0.0112 Moles of solute

0.0933 = 0.9888 moles of solute → 0.0933 / 0.9888 = 0.0947 moles

Finally we can determine the molar mass (mol/g)

5.678 g / 0.0947 mol = 59.9 g/mol

4 0
3 years ago
43.2 + 50.0 + 16.0= answer in correct amount of significant figures
Molodets [167]

Answer : The correct answer is 109.

Explanation :

Significant figure : It is defined as the each digits of a number have meaning or contribute to the value of the number.

The addition of given digits is,

43.2 + 50.0 + 16.0 = 109.2

As per the question, the given digits has 3 significant figures. But the answer after addition is 109.2 has 4 significant figures.

The addition rule of significant rule is the least precise number of significant figure in any number of the problem determines the number of significant figures in the answer.

Therefore, from the addition rule, the correct answer is 109.

4 0
3 years ago
The notation for the nuclide gives information about
Sever21 [200]
I think mass and atomic number
8 0
4 years ago
You are given stocks of 4 M NaCl, 40% Glucose, and 1M Tris-HCl (pH 8.5). You need to make 400 ml of a buffer containing 0.5 M Na
OLga [1]

Answer:

50mL of 4M NaCl, 80mL of 40% glucose, 20mL of 1M Tris-HCl (pH 8.5) and 250mL of water.

Explanation:

To make 400mL containing 0.5M NaCl you need to add:

4M / 0.5M = 8 (dilution 1/8). 400mL / 8 = <em>50 mL of 4M NaCl.</em>

Glucose 8% you need to add:

40% / 8% = 5 (dilution 1/5). 400mL / 5 = <em>80 mL of 40% glucose </em>

Buffer 50mM you need to add:

1000mM / 50mM = 20 (dilution 1/20). 400mL / 20 = <em>20mL of 1M Tris-HCl (pH 8.5)</em>

<em></em>

The resting volume: 400mL - 50mL of 4M NaCl - 80mL of 40% glucose - 20mL of 1M Tris-HCl (pH 8.5) = 250 mL must be completed with water.

Thus, to make the solution you need: <em>50mL of 4M NaCl, 80mL of 40% glucose, 20mL of 1M Tris-HCl (pH 8.5) and 250mL of water.</em>

<em></em>

I hope it helps!

8 0
3 years ago
Another way of writing 0.00000731 g
Katen [24]

Answer:

nanograms - 7310

kilograms - 7.310000000000001e-9

5 0
3 years ago
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