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zhuklara [117]
3 years ago
8

what would happen to an igneous rock exposed at the earth's surface? A.it would melt B.it would weather and erode C.it would cha

nge into a metamorphic
Chemistry
2 answers:
Alex17521 [72]3 years ago
8 0
Hey there,

Answer : C, It would change into a metamorphic

Hope this helps :D

<em>~Top♥</em>
sasho [114]3 years ago
3 0
It would erode if an igneous rock was at the center of the earth it would erode hope this helps

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There is a 30g of be-11 it has a half-life of about 14 seconds how much will be left in 28 seconds
Lelu [443]

There will be 7.5 g of Be-11 remaining after 28 s.

If 14 s = 1 half-life, 28 s = 2 half-lives.

After the first half-life, ½ of the Be-11 (15 g) will disappear, and 15 g will remain.

After the second half-life, ½ of the 15 g (7.5 g) will disappear, and 7.5 g will remain.

In symbols,

<em>N</em> = <em>N</em>₀(½)^<em>n</em>

where

<em>n</em> = the number of half-lives

<em>N</em>₀ = the original amount

<em>N</em> = the amount remaining after <em>n</em> half-lives

6 0
2 years ago
0.315 to scientific notation
irga5000 [103]

Answer:

3.15 × 10^-1

or

3.15E-1

Explanation:

7 0
3 years ago
Read 2 more answers
Why doesn't glass exhibit crystalline form
ser-zykov [4K]
It doesn't have a 3 dimensional pattern
7 0
3 years ago
The Embargo Act of 1807 and the Neutrality Acts
insens350 [35]
The Embargo Act of 1807 and the Neutrality Acts of the 1930's were both attempts by the United States to avoid foreign conflicts. The correct option in regards to all the options given in the question is option "3". The Embargo Act of 1807 prohibited all kinds of exports from United states of America. This Act would treat any export as illegal. This Act was also intended to keep America from avoiding any conflicts with the warring nations. The Neutrality act was also intended towards remaining neutral towards all fighting nations.


6 0
3 years ago
A chemist prepares a solution of potassium permanganate by measuring out 26. g of potassium permanganate into a 350. mL volumetr
alekssr [168]

Answer:

0.471 mol/L

Explanation:

First, we'll begin by by calculating the number of mole of KMnO4 in 26g of KMnO4.

This is illustrated below:

Molar Mass of KMnO4 = 39 + 55 + (16x4) = 39 + 55 + 64 = 158g/mol

Mass of KMnO4 from the question = 26g

Mole of KMnO4 =?

Number of mole = Mass/Molar Mass

Mole of KMnO4 = 26/158 = 0.165mole

Now we can obtain the concentration of KMnO4 in mol/L as follow:

Volume of the solution = 350mL = 350/1000 = 0.35L

Mole of KMnO4 = 0.165mole

Conc. In mol/L = mole of solute(KMnO4)/volume of solution

Conc. In mol/L = 0.165mol/0.35

conc. in mol/L = 0.471mol/L

3 0
3 years ago
Read 2 more answers
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