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3 years ago

How is the behavior of food coloring in the pool and heated water different? What caused the difference in their behavior?

2 answers:

7 0

In hot water, the food coloring diffuses through the water fast. In cold water, the food coloring diffuses through the water slowly. At higher temperatures, particles move faster

solniwko [45]3 years ago

4 0

Answer )In hot water, the food coloring diffuses (spreads out) through the water quickly. In cold water, the food coloring diffuses (spreads out) through the water slowly. The hot water causes the food coloring to diffuse faster. At higher temperatures, particles move faster.

i hope it is helpful

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A sample consisting of 1.0 mol of perfect gas molecules with CV = 20.8 J K−1 is initially at 4.25 atm and 300 K. It undergoes re

Marat540 [252]

Answer : The value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

Explanation :

First we have to calculate the value of $\gamma$ .

$\gamma=\frac{C_p}{C_v}$

As, $C_p=R+C_v$

So, $\gamma=\frac{R+C_v}{C_v}$

Given :

$C_v=20.8J/K\\\\R=8.314J/K$

$\gamma=\frac{8.314+20.8}{20.8}=1.4$

Now we have to calculate the initial volume of gas.

Formula used :

$P_1V_1=nRT_1$

where,

$P_1$ = initial pressure of gas = 4.25 atm

$V_1$ = initial volume of gas = ?

$T_1$ = initial temperature of gas = 300 K

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

$(4.25atm)\times V_1=(1.0mol)\times (0.0821L.atm/mol.K)\times (300K)$

$V_1=5.80L$

Now we have to calculate the final volume of gas by using reversible adiabatic expansion.

$P_1V_1^{\gamma}=P_2V_2^{\gamma}$

where,

$P_1$ = initial pressure of gas = 4.25 atm

$P_2$ = final pressure of gas = 2.50 atm

$V_1$ = initial volume of gas = 5.80 L

$V_2$ = final volume of gas = ?

$\gamma$ = 1.4

Now put all the given values in above formula, we get:

$(4.25atm)\times (5.80L)^{1.4}=(2.50atm)\times V_2^{1.4}$

$V_2=8.47L$

Now we have to calculate the final temperature of gas.

Formula used :

$P_2V_2=nRT_2$

where,

$P_2$ = final pressure of gas = 2.50 atm

$V_2$ = final volume of gas = 8.47 L

$T_2$ = final temperature of gas = ?

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in above formula, we get:

$(2.50atm)\times (8.47L)=(1.0mol)\times (0.0821L.atm/mol.K)\times T_2$

$T_2=257.9K\approx 258K$

Now we have to calculate the work done.

$w=nC_v(T_2-T_1)$

where,

w = work done = ?

n = number of moles of gas =1.0 mol

$T_1$ = initial temperature of gas = 300 K

$T_2$ = final temperature of gas = 258 K

$C_v=20.8J/K$

Now put all the given values in above formula, we get:

$w=(1.0mol)\times (20.8J/K)\times (258-300)K$

$w=-873.6J$

Therefore, the value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

8 0

3 years ago

What happens to electrons in an ionic bond? *

dolphi86 [110]

In ionic bonding, atoms transfer electrons to each other. Ionic bonds require at least one electron donor and one electron acceptor.

3 0

3 years ago

What is the volume of 0.98 mol oxygen gas at 275 k and a pressure of 2.0 atm

Virty [35]

The volume of 0.98 mol oxygen gas at 275 k and a pressure of 2.0 atm is 11.06L.

<h3>How to calculate volume?</h3>

The volume of a given mass of gas can be calculated using the following formula:

PV = nRT

Where;

P = pressure
V = volume
R = gas law constant
T = temperature
n = number of moles

According to this question, 0.98 moles of oxygen gas at 275 k contains a pressure of 2.0 atm. The volume is calculated as follows:

2 × V = 0.98 × 0.0821 × 275

2V = 22.13

V = 11.06L

Therefore, the volume of 0.98 mol oxygen gas at 275 k and a pressure of 2.0 atm is 11.06L.

Learn more about volume at: brainly.com/question/12357202

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6 0

2 years ago

What is the molarity of a solution containing 400g cuso4 in 4.00 l of solution

Murrr4er [49]

Calculation-
400gCuSo4
---------------- ] faction 2.5 over 4 which EQUALS .625MCuSO4 in decimal form
1

8 0

3 years ago

Solve for x 5 to the power 2 x + 4 - 25 to the power x - 1 is equals to

Dafna11 [192]

Answer:

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4 0

3 years ago

Read 2 more answers