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kumpel [21]
3 years ago
5

What is the difference between a simile and a metaphor?

Chemistry
2 answers:
Gennadij [26K]3 years ago
4 0
A simile is a comparison using "as" or "like" and a metaphor is a comparison without using "like" or "as"
Anastasy [175]3 years ago
3 0
A metaphor makes a direct comparison be tween to things while a simile show similarities using " like or as" makes
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Please help I don’t have a lot of time left
Harrizon [31]

Answer:

So sorry if I was wrong but I think it's B. Because from the source states,

https://socratic.org/questions/which-group-on-the-periodic-table-is-the-least-reactive-why

"The least reactive elements are those who have a full outermost valence shell ie they have 8 electrons in the outer shell so elements such as helium, neon, radon or the transition elements."

5 0
3 years ago
Read 2 more answers
The radioactive atom 61/27 co emits a beta particle. write an equation showing the decay
pickupchik [31]

Answer:

\rm _{27}^{60}\text{Co} \longrightarrow \,  _{-1}^{0}\text{e} + \, _{28}^{60}\text{Ni}

Explanation:

The unbalanced nuclear equation is

\rm _{27}^{60}\text{Co} \longrightarrow \,  _{-1}^{0}\text{e} + \, ?

Let's write the question mark as a nuclear symbol.

\rm _{27}^{60}\text{Co} \longrightarrow \,  _{-1}^{0}\text{e} + \,  _{Z}^{A}\text{X}

The main point to remember in balancing nuclear equations is that the sums of the superscripts and the subscripts must be the same on each side of the equation.  

Then

60 = 0 + A, so A = 60 - 0 = 60, and

27 = -1 + Z, so Z  = 27 + 1 = 28

Your nuclear equation becomes

\rm _{27}^{60}\text{Co} \longrightarrow \,  _{-1}^{0}\text{e} + \, _{28}^{60}\text{X}

Element 28 is nickel, so the balanced nuclear equation is

\rm _{27}^{60}\text{Co} \longrightarrow \,  _{-1}^{0}\text{e} + \, _{28}^{60}\text{Ni}

7 0
3 years ago
There are two naturally occurring isotopes of copper. 63cu has a mass of 62.9296 amu. 65cu has a mass of 64.9278 amu. determine
SSSSS [86.1K]
1) You need to use the atomic mass of copper.


You can find it in a periodic table. It is 63.546 amu.


2) The atomic mass is the weigthed mass of the different isotopes.


This is, the atomic mass of one element is the atomic mass of each isotope times its corresponding abundance:


=> atomic mass of the element = abundance isotope 1 * atomic mass isotope 1 + abundance isotope 2 *  atomic mass isotope 2 + ....+abundance isotope n * atomic mass isotope n.


3) The statement tells there are two isotopes so the abundance of one is x and the abundance of the other is 1 - x


=> 63.546 amu = x * 62.9296 amu + (1-x)*64.9278


=> 63.546 = 62.9296x + 64.9278 - 64.9278x


=> 64.9278x - 62.9296 = 64.9278 - 63.546


=> 1.9982x = 1.3818


=> x = 1.3818 / 1.9982 = 0.6915 = 69.15%


=> 1 - x = 1 - 0.6915 = 0.3085 = 30.85%


Answer:


Cu-63 69.15%;


Cu-65 : 30.85%
3 0
3 years ago
Examine each of the following pie charts. Which one shows the correct makeup of Earth's atmosphere?
I am Lyosha [343]

Answer:

Answer A

Explanation:

atmosphere is made up of approx 78% N and 21% O2

5 0
2 years ago
Calculate the work, w, gained or lost by the system when a gas expands from 15 L to 45 L against a constant external pressure of
marta [7]

Answer:

Work done by the system = 4545 J

Explanation:

The expression for the calculation of work done is shown below as:

w=-P\times \Delta V

Where, P is the pressure

\Delta V is the change in volume

From the question,  

\Delta V = 45 - 15 L = 30 L

P = 1.5 atm

w=-1.5\times 30\ atmL

Also, 1 atmL = 101 J

So,  

w=-1.0\times0.75\times 101\ J=-4545\ J (negative sign implies work is done by the system)

<u>Work done by the system = 4545 J</u>

6 0
3 years ago
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