Answer:
c. chemistry
Explanation:
it uses science and math which are based on facts.
pH stands for
the power of hydrogen. pH that ranges form 1-6.9 are acid substances. At pH 1 -
2.9, the substance is highly acidic which will have the color from red to red - orange. At pH 3 - 4, the substance is mildly
acidic and the color range is from red - orange to orange. At pH 4.1 – 6.9, the substance is weakly acidic and the color range is from orange to yellow. At pH 7, it is
neutral and it is green in color. At pH 7.1-14, it is basic. At pH 7.1 - 10.9, the substance is weakly
basic the color range is from green to blue. At pH 11 - 13, the substance is mildly basic and the color range is from blue to purple. At pH 13.1 – 14, the
substance is highly basic and the color range is from purple to light purple.
Given the balanced equation:
( Reaction type : double replacement)
CaF2 + H2SO4 → CaSO4 + 2HFI
We can determine the number of grams prepared from the quantity of 75.0 H2SO4, and 63.0g of CaF2 by converting these grams to moles per substance.
This can be done by evaluating the atomic mass of each element of the substance, and totaling it to find the molecular mass.
For H2SO4 or hydrogen sulfate it's molecular mass is the sum of the quantity of atomic mass per element. H×2 + S×1 + O×4 = ≈1.01×2 + ≈32.06×1 + ≈16×4 = 2.02 + 32.06 + 64 = 98.08 u (Dalton's or Da) or g / mol.
For CaF2 or calcium fluoride, it's molecular mass adds 1 atomic mass of calcium and 2 atomic masses of fluoride due to the number of atoms.
Ca×1 + F×2 = ≈40.07×1 + ≈19×2 = 40.08 + 38 = 78.07 u (Da or Dalton's) or g / mol.
Answer:
_5_ AsO2−(aq) + 3 Mn2+(aq) + _2_ H2O(l) → _5_ As(s) + _3_ MnO4−(aq) + _4_ H+(aq)
Explanation:
Step 1:
The unbalanced equation:
AsO2−(aq) + 3 Mn2+(aq) + H2O(l) → As(s) + MnO4−(aq) + H+(aq)
Step 2:
Balancing the equation.
AsO2−(aq) + 3Mn2+(aq) + H2O(l) → As(s) + MnO4−(aq) + H+(aq)
The above equation can be balanced as follow:
There are 3 atoms of Mn on the left side of the equation and 1 atom on the right side. It can be balance by putting 3 in front of MnO4− as shown below:
AsO2−(aq) + 3Mn2+(aq) + H2O(l) → As(s) + 3MnO4−(aq) + H+(aq)
There are 12 atoms of O on the right side and a total of 3 atoms on the left side. It can be balance by putting 5 in front of AsO2− and 2 in front of H2O as shown below:
5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → As(s) + 3MnO4−(aq) + H+(aq)
There are 4 atoms of H on the left side and 1 atom on the right side. It can be balance by putting 4 in front of H+ as shown below:
5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → As(s) + 3MnO4−(aq) + 4H+(aq)
There are 5 atoms of As on the left side and 1 atom on the right side. It can be balance by putting 5 in front of As as shown below:
5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → 5As(s) + 3MnO4−(aq) + 4H+(aq)
Now the equation is balanced
Answer:
The volume of sodium hydroxide at the equivalence point is:
- <u>14.9 mL of sodium hydroxide</u>.
Explanation:
<u>The equivalence point occurs when, in this case, the HCl is completely neutralized with the solution of NaOH, how you can see this doesn't occur in the last point but occurs in the nineteenth point, where the pH is no more acid (below to 7) but is 11 approximately</u>, then you must see in the X-axis from this point and you can see the volume is almost 15, by this reason I calculate the valor of 14.9 milliliters.
