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Lynna [10]
3 years ago
14

Moritz is watching his little brother play in a mud puddle. Moritz notices that when his brother stirs it with a stick, after a

few moments the dirt settles back to the bottom of the puddle. His brother is creating a
.

Anjali knows that whole milk has more fat than skim milk. However, the solid fat doesn’t seem to separate from the liquid milk even after it has been in the fridge for a few days. This is evidence that milk is a.
Chemistry
2 answers:
PIT_PIT [208]3 years ago
8 0

<em>Moritz is watching his little brother play in a mud puddle. Moritz notices that when his brother stirs it with a stick, after a few moments the dirt settles back to the bottom of the puddle. His brother is creating a; </em>SUSPENSION  

<em>Anjali knows that whole milk has more fat than skim milk. However, the solid fat doesn’t seem to separate from the liquid milk even after it has been in the fridge for a few days. This is evidence that milk is a;</em>  COLLOID

Advocard [28]3 years ago
3 0

Answer:

The first case is an illustration of a suspension, and the second case of a milk is an example of a colloid.

Explanation:

In the field of chemistry, a heterogeneous mixture comprising the solute particles that do not get dissolve, however, get suspended all through the majority of the solvent, and floats freely around in the medium is known as a suspension.  

On the other hand, a homogenous non-crystalline component comprising ultramicroscopic particles or bigger molecules of one material getting dispersed via a second material is known as a colloid. The colloids comprise emulsions, gels, and sols.  

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MArishka [77]
If P=M*V than P=30kg*5m/s. P=150.
P=momentum
M=mass
V=Velocity

Now the last time i have done physics was last year. but i'm pretty confident in this answer. Hope this helps!
8 0
3 years ago
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A decomposition reaction, with a rate that is observed to slow down as the reaction proceeds, has a half-life that does not depe
mr_godi [17]

Answer: A plot of the natural log of the concentration of the reactant as a function of time is linear.

Explanation:

Since it was explicitly stated in the question that the half life is independent of the initial concentration of the reactant then the third option must necessarily be false. Also, the plot of the natural logarithm of the concentration of reactant against time for a first order reaction is linear. In a first order reaction, the half life is independent of the initial concentration of the reactant. Hence the answer.

3 0
3 years ago
Abnormal behavior is defined as a?
hodyreva [135]

Answer:

Abnormality is a behavioral characteristic assigned to those with conditions regarded as rare or dysfunctional. Behavior is considered abnormal when it is atypical or out of the ordinary, consists of undesirable behavior, and results in impairment in the individual's functioning

Explanation:

5 0
3 years ago
Which of these metals reacts more vigorously with water? <br><br> 1. Zinc <br> 2. Magnesium
GalinKa [24]

Answer:

magnesium  

Explanation:

because zinc does not react with water because it too forms a protective layer of insoluble.

8 0
3 years ago
I need help with 1,2,3, and 4
Schach [20]

Answer:

  • Problem 1: 1.85atm
  • Problem 2: 110mL
  • Problem 3: 290 mL
  • Problem 4: 1.14 atm

Explanation:

Problem 1

<u>1. Data</u>

<u />

a) P₁ = 3.25atm

b) V₁ = 755mL

c) P₂ = ?

d) V₂ = 1325 mL

r) T = 65ºC

<u>2. Formula</u>

Since the temeperature is constant you can use Boyle's law for idial gases:

          PV=constant\\\\P_1V_1=P_2V_2

<u>3. Solution</u>

Solve, substitute and compute:

         P_1V_1=P_2V_2\\\\P_2=P_1V_1/V_2

        P_2=3.25atm\times755mL/1325mL=1.85atm

Problem 2

<u>1. Data</u>

<u />

a) V₁ = 125 mL

b) P₁ = 548mmHg

c) P₁ = 625mmHg

d) V₂ = ?

<u>2. Formula</u>

You assume that the temperature does not change, and then can use Boyl'es law again.

          P_1V_1=P_2V_2

<u>3. Solution</u>

This time, solve for V₂:

           P_1V_1=P_2V_2\\\\V_2=P_1V_1/P_2

Substitute and compute:

        V_2=548mmHg\times 125mL/625mmHg=109.6mL

You must round to 3 significant figures:

        V_2=110mL

Problem 3

<u>1. Data</u>

<u />

a) V₁ = 285mL

b) T₁ = 25ºC

c) V₂ = ?

d) T₂ = 35ºC

<u>2. Formula</u>

At constant pressure, Charle's law states that volume and temperature are inversely related:

         V/T=constant\\\\\\\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}

The temperatures must be in absolute scale.

<u />

<u>3. Solution</u>

a) Convert the temperatures to kelvins:

  • T₁ = 25 + 273.15K = 298.15K

  • T₂ = 35 + 273.15K = 308.15K

b) Substitute in the formula, solve for V₂, and compute:

        \dfrac{V_1}{T_1`}=\dfrac{V_2}{T_2}\\\\\\\\\dfrac{285mL}{298.15K}=\dfrac{V_2}{308.15K}\\\\\\V_2=308.15K\times285mL/298.15K=294.6ml

You must round to two significant figures: 290 ml

Problem 4

<u>1. Data</u>

<u />

a) P = 865mmHg

b) Convert to atm

<u>2. Formula</u>

You must use a conversion factor.

  • 1 atm = 760 mmHg

Divide both sides by 760 mmHg

       \dfrac{1atm}{760mmHg}=\dfrac{760mmHg}{760mmHg}\\\\\\1=\dfrac{1atm}{760mmHg}

<u />

<u>3. Solution</u>

Multiply 865 mmHg by the conversion factor:

    865mmHg\times \dfrac{1atm}{760mmHg}=1.14atm\leftarrow answer

3 0
3 years ago
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