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Virty [35]
1 year ago
8

Brooke decides to model a lunar eclipse. She attaches a large poster of the Sun to her wall to represent the Sun. She then decid

es that her head will represent Earth, and the direction she is facing will indicate the direction an observer on Earth is facing. She will then hold a small ball in her hand to represent the Moon. In order to model a person watching a lunar eclipse, Brooke faces the "Sun," stretches out her hand, and faces the "Moon" in front of her. What aspect of Brooke's model should be corrected?
A. Brooke should turn so that the Sun is on her right, and hold the Moon to her left.
B. Brooke should turn so that her back is to the Sun, and hold the Moon behind her.
C. Brooke should turn so that the Sun is on her left, and hold the Moon in front of her.
D. Brooke should turn so that her back is to the Sun, and hold the Moon in front of her.
PLEASE HELP ITS A TEST :(((
Chemistry
1 answer:
koban [17]1 year ago
7 0

Answer:

its D

Explanation:

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Why it is impossible for an isolated atom to exist in the hybridized state?
kakasveta [241]

Hybridization refers to the mixing of atomic orbitals in an atom. The number of hybrid orbitals needs to be equal to the number of orbitals that have involved in prior to mixing.  

The isolated atoms cannot prevail in a hybridized state as the atom in an isolated state do not form any kind of bond with the other atom, due to which the atomic orbitals do not go through the process of hybridization.  


7 0
2 years ago
If the lead can be extracted with 92.5% efficiency, what mass of ore is required to make a lead sphere with a 3.50 cm radius? le
ElenaW [278]

The volume of sphere can be calculated using the following formula:


V=\frac{4}{3}\pi r^{3}


Here, r is radius of the sphere which is 3.5 cm. Putting the value,


V=\frac{4}{3}\pi r^{3}=\frac{4}{3}(3.14)(3.5cm)^{3}=180 cm^{3}


This is equal to the volume of lead, density of lead is 11.34 g/cm^{3} thus, mass of lead can be calculated as follows:


m=d×V


Putting the values,


m=11.34 g/cm^{3}\times 180 cm^{3}=2041.2 g


Let the mass of ore be 1 g, 68.6% of galena is obtained by mass, thus, mass of galena obtained will be 0.686 g.


Now, 86.6% of lead is obtained from this gram of galena, thus, mass of lead will be:


m=0.686×0.866=0.5940 g


Therefore, 0.5940 g of lead is obtained from 1 g of ore for 100% efficiency, thus, for 92.5% efficiency

m=\frac{92.5}{100}\times 0.5940=0.54945 g

1 g of lead obtain from\frac{1}{0.54945}=1.82 grams of ore.


Thus, 2041.2 g of lead obtain from:


2041.2\times 1.82=3.715\times 10^{3}g or 3.715 kg


Therefore, mass of ore required to make lead sphere is 3.715 kg.


6 0
2 years ago
<img src="https://tex.z-dn.net/?f=PCl_5%20%5Crightleftarrows%20PCl_3%20%2B%20Cl_2" id="TexFormula1" title="PCl_5 \rightleftarrow
Butoxors [25]

<u>Answer:</u> The total pressure of the container will be 2.00 atm

<u>Explanation:</u>

We are given:

Initial moles of phosphorus pentachloride = 1.00 atm

For the given chemical reaction:

PCl_5\rightleftharpoons PCl_3+Cl_2

By Stoichiometry of the reaction:

1 mole of PCl_5 produces 1 mole of PCl_3 and 1 mole of chlorine gas

So, 1.00 atm of PCl_5 will also produce 1.00 atm of PCl_3 and 1.00 atm of chlorine gas when the reaction goes to completion.

Total pressure of the container when the reaction goes to completion  = 1.00 + 1.00 = 2.00 atm

Hence, the total pressure of the container will be 2.00 atm

3 0
2 years ago
What is the percent yield of this reaction if 22.o g of Mgl2 is produced by the reaction of 25.0 g of Mg with 25.0 g of l2?
expeople1 [14]

% yield = 80.719

<h3>Further explanation</h3>

Given

22.0 g of Mgl₂

25.0 g of Mg

25.0 g of l₂

Required

The percent yield

Solution

Reaction

Mg + I₂⇒ MgI₂

mol Mg = 25 g : 24.305 g/mol = 1.029

mol I₂ = 25 g : 253.809 g/mol = 0.098

Limiting reactant = I₂

Excess reactant = Mg

mol MgI₂ based on I₂, so mol MgI₂ = 0.098

Mass MgI₂ (theoretical):

= mol x MW

= 0.098 x 278.114

= 27.255 g

% yield = (actual/theoretical) x 100%

% yield = (22 / 27.255) x 100%

% yield = 80.719

8 0
1 year ago
A person throws a ball up into the air, and the ball falls back toward Earth. At which point would the kinetic energy be the low
dybincka [34]
It would be d bc when it’s at its lowest point it means not that much energy and that would be its lowest point which is D
6 0
2 years ago
Read 2 more answers
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