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satela [25.4K]
3 years ago
13

a student takes three beakers and places 250 ml of water in each. the student then adds a cube of sugar to all three beakers. th

e first beaker is stirred, the second beaker is heated, and the third is left alone. which of the following statements is true? the first beaker dissolves faster than the other two. the second beaker dissolves faster than the other two. the third beaker does not dissolve as fast as one and two. there is not enough information to determine the answer.
Chemistry
2 answers:
Olegator [25]3 years ago
6 0
The third beaker doesnt dissolve as fast as one and two

as it has nothing changed to it

when you stir it gives the sugar more kinetic energy to dissolve
when you heat up the solution it also gives the sugar more kinetic energy

hope that helps 
rjkz [21]3 years ago
6 0

Answer : The correct statement is, the third beaker does not dissolve as fast as one and two.

Explanation :

According to the question, we know that the dissolving process is faster in case of heating as compared to the stirring. And the dissolving process also depends on the how much amount of heat are provided to the substance or the how much faster will you stirring mixture.

The second beaker dissolves faster than the other two, this statement is true but we do not know the how much heat are given to the beaker.

From the given statements we conclude that the correct statement is, the third beaker does not dissolve as fast as first and second beakers because in the third beaker there is no heat or mechanical energy are provided for mixing of the substance.

Hence, the correct statement is, the third beaker does not dissolve as fast as one and two.

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Isotopes of the same element have the same number of ________ and a different number of _____.
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<span>C.) Protons, neutrons. Hope it helps :)</span>
5 0
3 years ago
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A sample of N2 gas in a flask is heated from 27 Celcius to 150 Celcius. If the original gas is @ pressure of 1520 torr, what is
Romashka [77]

Answer:

\large \boxed{\text{B.) 2.8 atm}}

Explanation:

The volume and amount are constant, so we can use Gay-Lussac’s Law:

At constant volume, the pressure exerted by a gas is directly proportional to its temperature.

\dfrac{p_{1}}{T_{1}} = \dfrac{p_{2}}{T_{2}}

Data:

p₁ = 1520 Torr; T₁ =   27 °C

p₂ = ?;               T₂ = 150 °C

Calculations:

(a) Convert the temperatures to kelvins

T₁ = (  27 + 273.15) K = 300.15 K

T₂ = (150 + 273.15) K = 423.15 K

(b) Calculate the new pressure

\begin{array}{rcl}\dfrac{1520}{300.15} & = & \dfrac{p_{2}}{423.15}\\\\5.064 & = & \dfrac{p_{2}}{423.15}\\\\5.064\times423.15&=&p_{2}\\p_{2} & = & \text{2143 Torr}\end{array}\\

(c) Convert the pressure to atmospheres

p = \text{2143 Torr} \times \dfrac{\text{1 atm}}{\text{760 Torr}} = \textbf{2.8 atm}\\\\\text{The new pressure reading will be $\large \boxed{\textbf{2.8 atm}}$}

7 0
3 years ago
A 0.1 mm sample of human blood has approximately 6000 red blood cells. An adult typically has 5.0 L of blood. How many red blood
olga55 [171]

 3.0 × 10¹¹ RBC's    (or)      3E11 RBC's


Solution:

Step 1: Convert mm³ into L;


As,


                                            1 mm³  =  1.0 × 10⁻⁶ Liters


So,


                                         0.1 mm³  =  X  Liters


Solving for X,


                       X  =  (0.1 mm³ × 1.0 × 10⁻⁶ Liters) ÷ 1 mm³


                       X  =  1.0 × 10⁻⁷ Liters


Step 2: Calculate No. of RBC's in 5 Liter Blood:


As given


                        1.0 × 10⁻⁷ Liters Blood contains  =  6000 RBC's


So,


                         5.0 Liters of Blood will contain  =  X  RBC's


Solving for X,


                      X  =  (5.0 Liters × 6000 RBC's) ÷ 1.0 × 10⁻⁷ Liters


                      X  =  3.0 × 10¹¹ RBC's


Or,


                     X  =  3E11 RBC's



5 0
3 years ago
Which of the following best describes the formation of plasma?
Sav [38]
I say, D is the answer

5 0
3 years ago
A student weighed an empty graduated cylinder. It weighed 35.86 g. She then carefully added water to the graduated cylinder unti
Firdavs [7]

Question:

A student weighed an empty graduated cylinder. It weighed 35.86 g. She then carefully added water to the graduated cylinder until it reached the 7.5 mL mark. When she weighed the graduated cylinder again, this time with the 7.5 mL of water in it, it weighed 43.18 g. What was this student's experimental density of water?

Answer:

0.976 g/mL

Explanation:

Weight of empty cylinder = 35.86g

Volume of water = 7.5mL

Weight of cylinder + water = 43.18g

Experimental density = ?

Density of water = Mass of water / volume of water

Mass of water = (Weight of cylinder + water) - Weight of empty cylinder

Mass of water = 43.18 - 35.86 = 7.32g

Density = 7.32 / 7.5 = 0.976 g/mL

5 0
2 years ago
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