Answer:
Chemical name for As2S3 is Arsenic sulfide
Explanation:
<u>Answer:</u>
Nitrogen gas be a mineral only, if it is in organic forms.
<u>Explanation:</u>
Most of the forms of organic nitrogen is not be taken by plants, with the exception in the form of small organic molecules. Also plants can promptly take the nitrogen when it is in other forms like ammonia and nitrate.
The microorganisms in the soil converts the organic forms of nitrogen to mineral form when they decompose organic matters and also fresh plant residues. This type of process is called mineralisation.
<u>Answer:</u>
Steep slopes on the contour map are identified using contour lines which are closely spaced.
<u>Explanation:</u>
Contour map also known as topographic maps are used to represent the three-dimensional portion of the earth’s surface in two-dimensional space. This map is used to represent the surface of the land such as steep, slopes, valleys. The contour maps are used in geological studies to understand the configuration of the earth's surface. Terms like map scale, vertical scale, contour lines are used on the contour map. Elevations are represented using contour lines. Contour lines placed very close to each represent the steep slopes and contour lines that are spaced farther away from each other represent the gentle slopes.
Since we are only asked for the number of moles, we don't need the information of density. The concentration is expressed in terms of 0.135 M AgCl or 0.135 moles of AgCl per liter solution. The solution is as follows:
Moles AgCl = Molarity * Volume
Moles AgCl = 0.135 mol/L * 244 mL * 1 L/1000 mL
<em>Moles AgCl = 0.03294 mol </em>
Answer:

Explanation:
1. Calculate the initial moles of acid and base

2. Calculate the moles remaining after the reaction
OH⁻ + H₃O⁺ ⟶ 2H₂O
I/mol: 0.0053 0.005 00
C/mol: -0.00500 -0.005 00
E/mol: 0.0003 0
We have an excess of 0.0003 mol of base.
3. Calculate the concentration of OH⁻
Total volume = 53 mL + 25.0 mL = 78 mL = 0.078 L
![\text{[OH}^{-}] = \dfrac{\text{0.0003 mol}}{\text{0.078 L}} = \textbf{0.0038 mol/L}\\\\\text{The final concentration of OH$^{-}$ is $\large \boxed{\textbf{0.0038 mol/L}}$}](https://tex.z-dn.net/?f=%5Ctext%7B%5BOH%7D%5E%7B-%7D%5D%20%3D%20%5Cdfrac%7B%5Ctext%7B0.0003%20mol%7D%7D%7B%5Ctext%7B0.078%20L%7D%7D%20%3D%20%5Ctextbf%7B0.0038%20mol%2FL%7D%5C%5C%5C%5C%5Ctext%7BThe%20final%20concentration%20of%20OH%24%5E%7B-%7D%24%20is%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B0.0038%20mol%2FL%7D%7D%24%7D)