The correct answer is - deflation.
The process of deflation can be caused by the winds. It is an erosive process in which the main role has the wind that is carrying lot of sediment in the shape of very small particles with it.
Through this process, the winds manage to erode large areas, especially in the drier places where the vegetation is very sparsely distributed. By this type of erosion, the winds manage to make lot of hollows that can range significantly in size. The hollows made by the deflation can be anywhere from few cm deep and several meters long, up to several km long and 50-60 meters of depth.
This is the process that is responsible for the creation of most of the oasis in the largest desert in the world, Sahara, some even being lowered enough to be under the sea level.
Answer is: acid-base indicator or pH indicators.
Acid-base indicators are usually weak acids or bases and they are chemical<span> detectors for hydrogen or hydronium cations.</span>
Example for acid-base indicator is phenolphthalein (molecular formula C₂₀H₁₄O₄). Phenolphthalein is <span>colorless in </span>acidic<span> solutions and pink in </span>basic<span> solutions.
Another example is m</span><span>ethyl orange. It is red colour in acidic solutions and yellow colour in basic solutions.</span>
Depending upon the clumping reaction with anti A , anti B and anti Rh antibodies the blood types are determined.
Explanation:
Agglutination (clumping) will occur when blood that contains the particular antigen is mixed with the particular antibody.
A+ have Agglutination with Anti-A ,Anti-Rh and No agglutination with Anti-B.
A- have Agglutination with Anti-A and No agglutination with Anti-B and Anti-Rh.
B+ have Agglutination with Anti-B Anti-Rh and No agglutination with Anti-A.
B- have Agglutination with Anti-B and No agglutination with Anti-B and Anti-Rh.
Rh+ have Agglutination with Anti-A and Anti-Rh and No agglutination with Anti-B.
Rh- have No Agglutination with Anti-A and Anti-B and Anti-Rh.
Answer:
pH = 12.15
Explanation:
To determine the pH of the HCl and KOH mixture, we need to know that the reaction is a neutralization type.
HCl + KOH → H₂O + KCl
We need to determine the moles of each compound
M = mmol / V (mL) → 30 mL . 0.10 M = 3 mmoles of HCl
M = mmol / V (mL) → 40 mL . 0.10 M = 4 mmoles of KOH
The base is in excess, so the HCl will completely react and we would produce the same mmoles of KCl
HCl + KOH → H₂O + KCl
3 m 4 m -
1 m 3 m
As the KCl is a neutral salt, it does not have any effect on the pH, so the pH will be affected, by the strong base.
1 mmol of KOH has 1 mmol of OH⁻, so the [OH⁻] will be 1 mmol / Tot volume
[OH⁻] 1 mmol / 70 mL = 0.014285 M
- log [OH⁻] = 1.85 → pH = 14 - pOH → 14 - 1.85 = 12.15
Answer:
C) LiOH + HCl → LiCl + H₂O
General Formulas and Concepts:
<u>Chemistry - Reactions</u>
- Synthesis Reactions: A + B → AB
- Decomposition Reactions: AB → A + B
- Single-Replacement Reactions: A + BC → AB + C
- Double-Replacement Reactions: AB + CD → AD + BC
Explanation:
<u>Step 1: Define</u>
RxN A: 2Na + 2H₂O → 2NaOH + H₂
RxN B: CaCO₃ → CaO + CO₂
RxN C: LiOH + HCl → LiCl + H₂O
RxN D: CH₄ + 2O₂ → CO₂ + 2H₂O
<u>Step 2: Identify</u>
RxN A: Single Replacement Reaction
RxN B: Decomposition Reaction
RxN C: Double Replacement Reaction
RxN D: Combustion Reaction
