What do a prism, a magnifying glass, a microscope, and eyeglasses ALL have in common?

The relationship between the endoplasmic reticulum and the Golgi complex is most similar to which of these examples? A a factory

Schach [20]

Answer: Answer is A

Explanation:

7 0

3 years ago

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How do waves interact with matter ?

velikii [3]

Waves interact with matter in several ways. The interactions occur when waves pass from one medium to another. Besides bouncing back like an echo, waves may bend or spread out when they strike a new medium. These three ways that waves may interact with matter are called reflection, refraction, and diffraction.

6 0

3 years ago

If 156.06 g of propane, C3H8, is burned in excess oxygen, how many grams of water are formed? C3H8 + O2 → CO2 + H2O Select one:

Nastasia [14]

Answer:

The correct option is;

a. 255.0 g

Explanation:

The given information are;

Mass of propane, C₃H₈ in the combustion reaction = 156.06 g

The equation of the combustion reaction is C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

From the balanced chemical equation of the reaction, we have;

One mole of propane, C₃H₈ reacts with five moles oxygen gas, O₂, to form three moles of carbon dioxide, CO₂, and four moles of water, H₂O

The molar mass of propane gas = 44.1 g/mol

The number of moles, n, of propane gas = Mass of propane gas/(Molar mass of propane gas) = 156.06/44.1 = 3.54 moles

Given that one mole of propane gas produces 4 moles of water molecule (steam) H₂O, 3.54 moles of propane gas will produce 4×3.54 = 14.16 moles of (steam) H₂O

The mass of one mole of H₂O = 18.01528 g/mol

The mass of 14.16 moles of H₂O = 14.16 × 18.01528 = 255.0 g

The mass of H₂O produced = 255.0 g

3 0

3 years ago

g A solution contains 100mM NaCl, 20mM CaCl2, and 20mM urea. We would say this solution is __________ compared to a 300 mOsM sol

ICE Princess25 [194]

Answer:

A solution contains 100mM NaCl, 20mM CaCl2, and 20mM urea. We would say this solution is hypotonic compared to a 300 mOsM solution and hypotonic compared to a cell with 300 mOsM (non-penetrating solutes) interior.

Explanation:

The osmolarity is calculated from the molar concentration of the active particles in the solution. We have a solution that is composed of NaCl, CaCl₂ and urea.

When they are dissolved in water, they dissociate into particles as follows:

NaCl → Na⁺ + Cl⁻ (2 particles per compound)

CaCl₂ → Ca²⁺ + 2 Cl⁻ (3 particles per compound)

urea: not dissociation (1 particle per compound)

Then, we have to calculate the osmolarity of the solution. We multiply the molarity of each compound by the number of particles produced by the compound in water:

Osm = (100 mM NaCl x 2) + (20 mM CaCl₂ x 3) + (20 mM urea x 1) = 280 mOsm

Compared with 300 mOsm, 280 mOsm has a lower osmolarity, so it is a hypotonic solution.

To compare with a cell's osmolarity, we have to consider only the non-penetrating solutes. Urea is considered a penetrating solute for mammalian cells. So, the osmolarity of non-penetrating solutes (NaCl and CaCl₂) is calculated as:

Osm (non-penetrating solutes) = (100 mM NaCl x 2) + (20 mM CaCl₂ x 3) = 260 mOsm

Therefore, we have:

Compared to 300 mOsm solution ⇒ 280 mOsm solution is a hypotonic solution

Compared to a cell with 300 mOsm ⇒ 260 mOsm solution is hypotonic

4 0

3 years ago

1 mol super cooled liquid water transformed to solid ice at -10 oC under 1 atm pressure.

Arada [10]

Answer:

Explanation:

Given that:

number of moles of super cooled liquid water = 1

Melting enthalpy of ice = 6020 J/mol

Freezing point =0 °C = (0 + 273 K)= 273 K

The decrease in entropy of the system during freezing for 1 mol (i.e during transformation from liquid water to solid ice ) = - 6020 J/mol × 1 mol /273 K = -22.051 J/K

Entropy change during further cooling from 0 °C (273 K) to -10 °C (263 K)

$\Delta \ S = \int\limits^{T_2}_{T_1}\dfrac{nC_p(s)dT}{T}$

$\Delta \ S = {nC_p(s)In \dfrac{T_2}{T_1}$

$\Delta \ S = {(1*37.7)In \dfrac{263}{273}$

Δ S = -1.4 J/K

Total entropy change of the system = - 22.05 J/K - 1.4 J/K = - 23.45 J/K

Entropy change of universe = entropy change of the system+ entropy change of the surrounding

According to the second law of thermodynamics

Entropy change of universe >0

SO,

Entropy change of the system + entropy change in the surrounding > 0

Entropy change in the surrounding > - entropy change of the system

Entropy change in the surrounding > - (- 23.53 J/K)

Entropy change in the surrounding > 23.53 J/K

b) Make some comments on entropy changes from the obtained data.

From the data obtained; we will realize that the entropy of the system decreases as cooling takes place when water is be convert to ice , randomness of these molecules reduces and as cooling proceeds , hence, entropy reduces more as well and the liberated heat will go into the surrounding due to this entropy of the surrounding increasing.

4 0

3 years ago