Which of the following sequences lists the relative sizes of particles in a water mixture from smallest to largest? a. Solutions, suspensions, colloids
b. Solutions, colloids, suspensions
c. Colloids, solutions, suspensions
d. Colloids, suspensions, solutions
e. Suspensions, colloids, solutions

Answers

Answer 1
Answer:  mostly A., im guessing 

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Polly is pushing a box across the floor with a force of 30 N. The force of gravity is –8 N, and the normal force is 8 N. Which value could describe the force of friction if Polly could not move the box?

Answers

The answer is -30 N

The explanation:

- to make the box move. so,the horizontal force must be greater than the frictional force.

-and when the horizontal force applied is 30 N. So, the box will remain stationary if the frictional force is equal in magnitude but opposite in direction to the horizontal force. means frictional force = -30 N

This is also being indicate that vertical force due to gravity is equivalent to the normal force when gravity force = 8 N and normal force =8 , meaning the box does not move vertically.

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the answer is -30*N.................


What is the volume 0.0023 moles of CO2

Answers

1 mole ---------------- 22.4 ( at STP )
0.0023 moles -------- ?

v = 0.0023 * 22.4 / 1

v = 0.05152 L

hope this helps!

Final answer:

In order to calculate the volume of 0.0023 moles of CO2, we use the ideal gas law and understand that one mole of an ideal gas at standard temperature and pressure occupies a volume of 22.4 liters. The volume for 0.0023 moles of CO2 is found to be approximately 0.05152 liters.

Explanation:

To solve for the volume of the 0.0023 moles of CO2, we need to use the ideal gas law, PV=nRT, where P depicts the pressure, V signifies the volume we need to find, n equals the number of moles, R denotes the universal gas constant, and T denotes the temperature. Assuming the relationship takes place at standard temperature and pressure (STP, 0 degrees Celsius, 1 atm), we can solve for V.

At STP, 1 mole of any ideal gas occupies a volume of 22.4 liters. Therefore, V = n * (Volume of 1 mole at STP), which equals 0.0023 moles * 22.4 L/mole = 0.05152 liters.

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Which one of the following statements is not true? a. Basalt magmas, in general, have higher temperatures than rhyolite magmas.
b. When magma reaches the surface, its dissolved gas content increases.
c. Melting temperatures of silicate rocks are lowered by small amounts of water.
d. Melting temperatures of silicate rocks increase with increased pressure.

Answers

Answer:

When magma reaches the surface, its dissolved gas content increases is true

Explanation:

The volcanic eruptions happen because of magma that is expelled on the earth’s surface. At the earth’s depth, all magma have gas dissolved in liquid. When the pressure has decreased the magma rises towards the earth’s surface creating a separate vapour phase.  

As pressure reduces the volume of gas will  expands and giving magma its 'explosive character'. Thus, as magma reaches the surface the dissolved gas content decreases and magma comes out of earth’s surface.

Final answer:

The untrue statement is b, as the dissolved gas content in magma decreases as it reaches the surface due to reduced pressure. Basalt magma has higher temperatures than rhyolite magma. Water reduces the melting temperature of silicate rocks, and increased pressure raises it.

Explanation:

The subject here pertains to volcanic activity and magma characteristics. The statement that is not true is: b. When magma reaches the surface, its dissolved gas content increases. In fact, as magma rises to the surface, the pressure decreases which allows the dissolved gases to escape, thus its dissolved gas content actually decreases. Statements a, c and d are true - Basalt magmas do generally have higher temperatures than rhyolite magmas. Additionally, the presence of water does indeed lower the melting temperature of silicate rocks, and increased pressure results in a higher melting temperature for these rocks.

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After 32 days, 5 milligrams of an 80-milligram sample of a radioactive isotope remains unchanged. What is the half-life of this element?(1) 8 days (3) 16 days
(2) 2 days (4) 4 days

Answers

Unstable heavy atoms will undergo radioactive decay to produce stable species. The half life time of the isotope which undergone a decay of 75 mg in 32 days is 18 days.

What is half life time?

The half life time of a radioactive sample is the time taken to reduce it to half of the initial amount by decay.

The heavy unstable material have very short half life and they will easily undergoes radioactive decay by emitting certain radiation.

Radioactive decay is a firs order reaction and have the equation to find the radioactive constant as follows:

\lambda = (1)/(t) log([Ni])/([Nt])

Where, t is the time of decay and Ni and Nt be the initial and final amount respectively.

It is given that 5 mg is remaining out of 80 mg after 32 days. Thus the radioactive constant is calculated as follows:

\lambda = (1)/(32 days )  log (80)/(5) \n \n              =   0.0376. days ^(-1)

Now the half life time of the decay is calculated as below:

t(1/2) = 0.693 /decay constant

      = 0.693/0.0376

      = 18 days

Therefore, the half life time of the isotope which undergone a decay of 75 mg in 32 days is 18 days.

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m=m_0 * ((1)/(2))^(t)/(t_(1/2))
m - the mass that remains unchanged, m₀ - the inital mass, t - the time of decay, t1⁄2 - the half-life

t=32 \ days \n m=5 \ mg \n m_0 = 80 \ mg \n \n 5 = 80 * ((1)/(2))^(32)/(t_(1/2)) \ \ \ \ \ \ \ |/ 80 \n (5)/(80)= ((1)/(2))^(32)/(t_(1/2)) \n (1)/(16)=((1)/(2))^(32)/(t_(1/2)) \n ((1)/(2))^4=((1)/(2))^(32)/(t_(1/2)) \n 4=(32)/(t_(1/2)) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |* t_(1/2) \n 4t_(1/2)=32 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |/ 4 \n t_(1/2)=8

The half-life is (1) 8 days.

What are energy sublevels? (A)multiple energy levels of any magnitude within an energy level
(B)multiple orbitals of specific magnitude within an energy level
(C)only those energy levels that overlap others
(D)electrons whose energy has dropped to a lower state

Answers

Answer:

Option B

Explanation:

First, we need to know what is an energy level.

An energy level corresponds to the row of the periodic table of element. So, if you see the attached picture, you can see that we have 7 rows where the elements are distributed, so, we have 7 energy levels for all those elements.

Knowing that, the sub levels are the caps where the electrons of the atoms are carried. These sublevels or caps, are called orbitals, these can be of several types

s orbytal: can hold 2 electrons

p orbytal: can hold up to 6 electrons.

d orbytal: can hold up to 10 electrons

f orbytal: can hold up to 14 electrons

g orbytal: can hold up to 18 electrons.

Depending on the row (or energy level) and the atom, we can know how many electrons can carry an element, in which period or row is, and the sub levels. For example, the Chlorine, with an atomic number of 17, can carry up to 7 electrons in it's outer level and it's on the third row (two energy levels). This can be known with it's electronic configuration:

[Cl] = 1s^2 2s^2 2p^6 3s^2 3p^5

The last energy level is 3, so it's the third period, and the electrons of those sub level are 2 and 5, 7 electrons.

Hope this can help better

The answer is B: multiple orbitals of specific magnitude within an energy level

Which layer of the Basement Membrane of epithelial tissues is produced and secreted by cells of the underlying connective tissue? (a) Reticular Lamina (b) Lamellar Lamina (c) Papillary Lamina (d) Basal Lamina

Answers

(a) Reticular Lamina

The layer of the basement membrane of epithelial tissues that is produced and secreted by cells of the underlying connective tissue is (a) Reticular Lamina.

The basement membrane is composed of two layers: the basal lamina, which is produced by the epithelial cells themselves, and the reticular lamina, which is produced and secreted by cells of the underlying connective tissue. These two layers together provide structural support and help anchor the epithelial tissue to the underlying connective tissue.