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stepan [7]
3 years ago
6

Name one way igneous rock is different from metamorphic rock.

Chemistry
1 answer:
nasty-shy [4]3 years ago
6 0

Answer:

Metamorphic rock is classified by texture and composition. The texture of a metamorphic rock can be either foliated and appear layered or banded, or non-foliated and appear uniform in texture without banding. Foliated rocks contain many different kinds of minerals, but non-foliated rocks contain only one main mineral, which contributes to their more uniform appearance.  Igneous rocks are classified according to mode of occurrence, texture, mineralogy, chemical composition, and the geometry of the igneous body.

Explanation:

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What do potatoes, wood, and lobster shells have in common?
STatiana [176]
They are all biotic factors, meaning they were once alive or were alive. (an abiotic factor is something that has never lived.) hope this helped.
5 0
3 years ago
determine whether each of the following electron configurations is an inert gas, a halogen, an alkali metal, an alkaline earth m
mamaluj [8]

This is an incomplete question, here is a complete question.

Determine whether each of the following electron configurations is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choices.

(a) 1s^22s^22p^63s^23p^5

(b) 1s^22s^22p^63s^23p^63d^74s^2

(c) 1s^22s^22p^63s^23p^63d^{10}4s^24p^6

(d) 1s^22s^22p^63s^23p^63d^4s^1

Answer :

(a) 1s^22s^22p^63s^23p^5   → Halogen

(b) 1s^22s^22p^63s^23p^63d^74s^2    → Transition metal

(c) 1s^22s^22p^63s^23p^63d^{10}4s^24p^6   → Transition metal

(d) 1s^22s^22p^63s^23p^63d^4s^1   → Transition metal

Explanation :

Inert gas : These are the gases which lie in group 18.

Their general electronic configuration is: ns^2np^6 where n is the outermost shell.

Halogen : These are the elements which lie in group 17.

Their general electronic configuration is: ns^2np^5 where n is the outermost shell.

An alkali metal : These are the elements which lie in group 1.

Their general electronic configuration is: ns^1 where n is the outermost shell.

An alkaline earth metal : These are the elements which lie in group 2.

Their general electronic configuration is: ns^2 where n is the outermost shell.

Transition elements : They are the elements which lie between 's' and 'p' block elements. These are the elements which lie in group 3 to 12. The valence electrons of these elements enter d-orbital.

Their general electronic configuration is: (n-1)d^{1-10}ns^{0-2} where n is the outermost shell.

(a) 1s^22s^22p^63s^23p^5

The element having this electronic configuration belongs to the halogen family.

(b) 1s^22s^22p^63s^23p^63d^74s^2

The element having this electronic configuration belongs to the transition family.

(c) 1s^22s^22p^63s^23p^63d^{10}4s^24p^6

The element having this electronic configuration belongs to the transition family.

(d) 1s^22s^22p^63s^23p^63d^4s^1

The element having this electronic configuration belongs to the transition family.

4 0
3 years ago
Please help! I need an answer ASAP, I will be marking brainliest.
asambeis [7]
B because A is for radios of course and C is thermal energy and D is radioactive
6 0
3 years ago
Read 2 more answers
Percent yield refers to the relationship between the predicted amount of product
dsp73
True because it actually formed I know
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3 years ago
A student plans to use a density versus solution concentration standard curve to identify the sodium chloride concentration in a
Art [367]

Answer:

a. 0.50 g, 1.0 g, 1.5 g, 2.0 g, 2.5 g;

g. 0.1 g, 0.5 g, 1.0 g, 1.5 g, 2.0 g

Explanation:

The percent mass is defined as a ratio between the mass of a solute and mass of a solution:

\omega = \frac{m_{solute}}{m_{solution}}\cdot 100\%

Since solution only consists of a solute and solvent, express its mass as:

m_{solution} = m_{solute} + m_{solvent}

Then:

\omega = \frac{m_{solute}}{m_{solution}}\cdot 100\%=\frac{m_{solute}}{m_{solute} + m_{solvent}}

Firstly, solve for how much mass is required to prepare 3.0 %. Let's say, we have x g of the solute:

0.03 = \frac{x}{30.0 + x}\therefore x = 0.03(30.0 + x)

x = 0.90 + 0.03x

0.97x = 0.90\therefore x = 0.93 g

Similarly, solve for 6.0 %, let's say, we have x g of the solute again:

0.06 = \frac{x}{30.0 + x}\therefore x = 0.06(30.0 + x)

x = 1.80 + 0.06x

0.94x = 1.80\therefore x = 1.91 g

Hence, masses should be in a range of 0.93 g to 1.91 g.

7 0
3 years ago
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