Answer:
The same holds true for pH values above 7, each of which is ten times more alkaline (another way to say basic) than the next lower whole value. For example, pH 10 is ten times more alkaline than pH 9 and 100 times (10 times 10) more alkaline than pH 8.
Explanation:
1. C
2. C
3. In elastic deformation, the deformed body returns to its original shape and size after the stresses are gone. In ductile deformation, there is a permanent change in the shape and size but no fracturing occurs. In brittle deformation, the body fractures after the strength is above the limit.
4. Normal faults are faults where the hanging wall moves in a downward force based on the footwall; they are formed from tensional stresses and the stretching of the crust. Reverse faults are the opposite and the hanging wall moves in an upward force based on the footwall; they are formed by compressional stresses and the contraction of the crust. Thrust faults are low-angle reverse faults where the hanging wall moves in an upward force based on the footwall; they are formed in the same way as reverse faults. Last, Strike-slip faults are faults where the movement is parallel to the crust of the fault; they are caused by an immense shear stress.
I hope this helped! These are COMPLEX questions though! =D
Answer:
The amount/type of stain
Explanation:
You would want to ensure that the stain was the same in both samples.
212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.
Answer:
Option A.
Explanation:
Similar to Avagadro's law, there is another law termed as dilution law. As the product of volume and normality of the reactant is equal to the product of volume and normality of the product from the Avagadro's law. In dilution law, it will be as product of volume and concentration of the solute of the reactant is equal to the product of volume and concentration of solution.
So, as per the given question C1 = 5.45 M of lead nitrate and V1 has to be found. While C2 is 1.41 M of lead nitrate and V2 is 820.7 ml.
Then, 
So nearly 212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.
<h3>
Answer:</h3>
2000 atoms
<h3>
Explanation:</h3>
We are given the following;
Initial number of atoms of radium-226 as 8000 atoms
Time taken for the decay 3200 years
We are required to determine the number of atoms that will remain after 3200 years.
We need to know the half life of Radium
- Half life is the time taken by a radio active material to decay by half of its initial amount.
- Half life of Radium-226 is 1600 years
- Therefore, using the formula;
Remaining amount = Original amount × 0.5^n
where n is the number of half lives
n = 3200 years ÷ 1600 years
= 2
Therefore;
Remaining amount = 8000 atoms × 0.5^2
= 8000 × 0.25
= 2000 atoms
Thus, the number of radium-226 that will remain after 3200 years is 2000 atoms.
