Answer:
Old and inefficient mining smokestacks contaminate the soils around abandoned mine sites.
Explanation:
A smokestack, is a very tall channel commonly used in many instances to release gases produced by combustion processes directly into the air. These high towers are aimed at dispersing the gaseous pollutants over a wider area thereby minimizing their impact.
Old and inefficient smokestack do not contaminate the soil since they are very high towers that discharge gases directly into the atmosphere. Hence they are not part of the sources of soil contamination in abandoned mines.
Answer:
They have properties of both metals and nonmetals
Explanation:
- Elements in the periodic table may be divided into Metals, non-metals, and metalloids.
- Metals are the elements that react by losing electrons to form stable positively charged ions known as cations. Examples are group 1, 2, and 3 elements together with transition elements.
- Non-metals are those elements that react by gaining electrons to form stable negatively charged ions called anions. Examples include oxygen, carbon, sulfur, etc.
- Metalloids, on the other hand, are elements that have both metallic and non-metallic properties.
- Metalloids occur between metals and non-metals in the periodic table. Examples include Boron and silicon among others.
An electron is a negatively charged subatomic particle, whereas a proton is positively charged, and a neutron has no charge.
Answer:
0.295 L
Explanation:
It seems your question lacks the final concentration value. But an internet search tells me this might be the complete question:
" A chemist must dilute 47.2 mL of 150. mM aqueous sodium nitrate solution until the concentration falls to 24.0 mM. He'll do this by adding distilled water to the solution until it reaches a certain final volume. Calculate this final volume, in liters. Be sure your answer has the correct number of significant digits. "
Keep in mind that if your value is different, the answer will be different as well. However the methodology will remain the same.
To solve this problem we can<u> use the formula</u> C₁V₁=C₂V₂
Where the subscript 1 refers to the concentrated solution and the subscript 2 to the diluted one.
- 47.2 mL * 150 mM = 24.0 mM * V₂
And <u>converting into L </u>becomes:
- 295 mL *
= 0.295 L
