Answer:

Explanation:
In this case, we can start with the <u>formula of Platinum (II) Chloride</u>. The cation is the atom at the left of the name (in this case
) and the anion is the atom at the right of the name (in this case
). With this in mind, the <u>formula would be</u>
.
Now, if we used <u>metallic copper</u> we have to put in the reaction only the <u>copper atom symbol</u>
. So, we have as reagents:

The question now is: <u>What would be the products?</u> To answer this, we have to remember <u>"single displacement reactions"</u>. With a general reaction:

With this in mind, the reaction would be:

I hope it helps!
Answer:
B
Explanation:
B, H2O + Na The elements toward the bottom left corner of the periodic table are the metals that are the most active in the sense of being the most reactive. Lithium, sodium, and potassium all react with water,
Oil is sucked up through wide floating heads and pumped into storage tanks. Although suction skimmers are generally very efficient, one disadvantage is that they are vulnerable to becoming clogged by debris and ice and require constant skilled observation.
In Chemistry, to better determine the position of a certain electron, quantum numbers are used. The four quantum numbers are n, l, m and s. In the given above of n= 4, the principal quantum number is 4 and this represents the overall relative energy in the orbital. This means that we are to find the maximum number of electrons in fourth main energy level and the answer is 32.
Answer:
The correct answer is 0.10.
Explanation:
Based on the given question, in a buffer solution of 1 liter, the molarity of acetic acid is 1.420 M and the molarity of sodium acetate is 0.67. The pKa value of acetic acid given is 4.74, now the pH of buffer is,
pH of buffer = pKa + log ([CH3COONa]/[CH3COOH])
= 4.74 + log (0.67/1.420)
= 4.74 + (-0.326)
= 4.41
Now 0.10100 mol of HCl is added, the HCl reacts with sodium acetate to give,
CH3COONa + HCl = CH3COOH + NaCl
Now the concentration of CH3COONa becomes = 0.67-0.101 = 0.57 M, and the new concentration of CH3COOH becomes = 1.420 + 0.101 = 1.52 M
Now the new pH will be,
= pKa + log (0.57/1.52)
= 4.74 + (-0.426)
= 4.31
The pH change is 4.41-4.31 = 0.10