CHEMISTRY HIGH SCHOOL

Which structure is found in a plant cell but not in a animal cell is ita. chloroplast
b. cell membrane or
c. ribosome

Answers

Answer 1
Answer: The one on the left is an animal cell and the one on the right is a plant cell, the answer is chloroplast
Answer 2
Answer:

Answer:

A.) Chloroplast

Explanation:

Chloroplasts are found in plant cells but not in animal cells. The chloroplast is the organelle that captures light energy and is the site where photosynthesis occurs.

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Omar wrote a hypothesis about batteries called dry cells.If more dry cells are connected end to end, a light bulb will work longer because there is more energy available.

What are the variables in his hypothesis?

Answers

Below are the choices:

The independent variable is the number of dry cells, and the dependent variable is the length of time the bulb works.
The independent variable is the length of time the bulb works, and the dependent variable is the number of dry cells.
The independent variable is the number of dry cells, and the dependent variable is the amount of energy available.
The independent variable is the amount of energy available, and the dependent variable is the number of dry cells.

I think the answer is The independent variable is the number of dry cells, and the dependent variable is the amount of energy available.

Answer:

im not sure about the answer but i did the test and "The independent variable is the number of dry cells, and the dependent variable is the amount of energy available."is wrong

describe the" fear appeal" and how it's used in public messaging. Give an example or create your own public health message using the fear appeal.

Answers

A good example is cigarretes they try to scare you into not smoking by telling you all of the downsides of smoking like lung cancer and i guess that would be fear appeal 

Find the empirical formula of a compound which contains 54.93% potassium, 38.73% boron and 6.34% hydrogen.A.
KBH
B.
KB2H4
C.
KB3H9
D.
K2B5H9

Answers

Use the percent as mass in g 
Mass g) --> g/mol 
K, B and H molar mass from periodic table

54.93 g K x (1 mol K/ 39.10 g K) = 1.405 mol K
38.73 g B x (1 mol B/ 10.81 g B) = 3.583 mol B
6.34 g H x (1 mol H/ 1.008 g H) = 6.290 mol H

Divide all three answers by the smallest value and you will get 
1.000 mol K
2.550 mol B
4.477 mol H 
now multiply these three answers by a number that will make all a whole number or (a number with a 9 as the first decimal point)
so multiply by 2
2 mol k
5 mol B
and 9 mol H
E.F. = K2B5H9 
Answer is D 
Sorry for bad explanation!!!!!!

Hello!

Find the empirical formula of a compound which contains 54.93% potassium, 38.73% boron and 6.34% hydrogen.

A.     KBH

B.     KB2H4

C.     KB3H9

D.     K2B5H9

  • We have the following data:

Potassium (K) ≈ 39 a.m.u (g/mol)

Boron (B) ≈ 11 a.m.u (g/mol)

Hydrogen (H) ≈ 1 a.m.u (g/mol)

  • We use the amount in grams (mass ratio) based on the composition of the elements, see: (in 100g solution)

K: 54.93 % = 54.93 g

B: 38.73 % = 38.73 g

H: 6.34 % = 6.34 g

  • The values ​​(in g) will be converted into quantity of substance (number of mols), dividing by molecular weight (g / mol) each of the values, we will see:

K: (54.93\:\diagup\!\!\!\!\!g)/(39\:\diagup\!\!\!\!\!g/mol) \approx 1.408\:mol

B: (38.73\:\diagup\!\!\!\!\!g)/(11\:\diagup\!\!\!\!\!g/mol) \approx 3.521\:mol

H: (6.34\:\diagup\!\!\!\!\!g)/(1\:\diagup\!\!\!\!\!g/mol) = 6.34\:mol

  • We realize that the values ​​found above are not integers, so we divide these values ​​by the smallest of them, so that the proportion does not change, let us see:

K: (1.408)/(1.408)\to\:\:\boxed{K = 1}

B: (3.521)/(1.408)\to\:\:\boxed{B \approx 2.5}

H: (6.34)/(1.408)\to\:\:\boxed{H \approx 4.5}

convert number of atomic radio into whole number

2 * (1 : 2.5 : 4.5)

= 2 : 5 : 9  ← whole number of atomic radio

K = 2

B =  5 

H = 9

  • Thus, the minimum or empirical formula found for the compound will be:

\boxed{\boxed{K_2B_5H_9}}\Longleftarrow(Empirical\:Formula)\:\:\:\:\:\:\bf\green{\checkmark}

Answer:

D.     K2B5H9

________________________

\bf\blue{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}

A solution of methanol and water has a mole fraction of water of 0.312 and a total vapor pressure of 211 torr at 39.9 ∘C. The vapor pressures of pure methan ol and pure water at this temperature are 256 torr and 55.3 torr, respectively. ls the solution ideal? If not, what can you say about the relative strengths of the solute-solvent interactions compared to the solute-solute and solvent-solvent interactions?

Answers

Answer:

Weaker

Explanation:

The strategy here is to use Raoult´s law to calculate the theoretical vapor pressure for the concentrations given and compare it with the experimental value of 211 torr.

Raoult´s law tell us that for a binary solution

P total = partial pressure A + partial pressure B = Xa PºA + Xb PºB

where Xa and Xb are the mol fractions, and  PºA and PºB are the vapor pressures of pure A and pure B, respectively

For the solution in question we have

Ptotal = 0.312 x 55.3 torr + ( 1- 0.312 ) x 256 torr     ( XA + XB = 1 )

Ptotal = 193 torr

Since experimentally, the total vapor pressure is 211 and our theoretical value is smaller ( 193 torr ), we can conclude the interactions  solute-solvent are weaker compared to the solute-solute and solvent-solvent interactions.

If the pressure, volume, and temperature of a gas are known, which can most likely be found by using the ideal gas law

Answers

the ideal gas law equation is as follows

PV = nRT

where P - pressure

V - volume

n - number of moles

R - universal gas constant

T - temperature

so if the pressure, volume and temperature are already known

we are left with n and R

since R is the universal gas constant that has a known fixed value then R too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

A.) the molar amount of gas

Hope it helps!

Consider the group 1A elements sodium (period 3), potassium (period 4), and rubidium (period 5). What would you predict about the ionization energies of these elements?

Answers

It can be predicted that the ionization energies of these elements increase as you go up a period. This means that sodium has the greatest ionization energy, while rubidium has the least. This is due to the placement or distance of the valence electrons from the nucleus of these elements. When you go up a period, this distance decreases, making it harder to overcome the attraction between the nucleus and the electrons. The harder it is to overcome this attraction, the higher the ionization energy needed.
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