Answer:
it forms :
1. Gold ( Au )
2. Zinc nitrate ( Zn(NO3)2 )
Explanation:
When a chunk of zinc is added to a solution of gold (III) nitrate to extract the gold. The reaction forms Gold and Zinc nitrate .
it's a single displacement reaction,
here's the balanced equation for above reaction :
3 Zn + 2 Au(NO3)3 =》3 Zn(NO3)2 + 2 Au
Answer:
b. Beta emission, beta emission
Explanation:
A factor to consider when deciding whether a particular nuclide will undergo this or that type of radioactive decay is to consider its neutron:proton ratio (N/P).
Now let us look at the N/P ratio of each atom;
For B-13, there are 8 neutrons and five protons N/P ratio = 8/5 = 1.6
For Au-188 there are 109 neutrons and 79 protons N/P ratio = 109/79=1.4
For B-13, the N/P ratio lies beyond the belt of stability hence it undergoes beta emission to decrease its N/P ratio.
For Au-188, its N/P ratio also lies above the belt of stability which is 1:1 hence it also undergoes beta emission in order to attain a lower N/P ratio.
When naming an ionic compound, write the name of the cation, which is the metal first. Then, write the name of the anion, which is the nonmetal. However, you remove the last 2-3 letters and replace suffixes.
1. RbF --> Rubidium Fluoride
Change fluorine to fluoride
2. CuO --> Copper (II) Oxide
Change oxygen to oxide. Oxide has a charge of -2. Since no subscripts are written, it means they have the same opposite charge. So, we use Copper (II).
<span>3. (NH</span>₄<span>)</span>₂<span>C</span>₂<span>O</span>₄ ---> Ammonium Oxalate
NH₄ is ammonia, but we change it to ammonium for polyatomic ions.
The final step in a typical titration, that is here an acid base one would be to finally find the concentration of your unknown substance whether that be the acid or the base. The other steps are used before this to come to the correct calculation and conclusion.
Answer:
120g
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
Sn + 2HF —> SnF2 + H2
Next, we shall determine the number of mole of HF needed to react with 3 moles of Sn.
From the balanced equation above, 1 mole of Sn reacted with 2 moles of HF.
Therefore, 3 moles of Sn will react with = 3 x 2 = 6 moles of HF.
Finally, we shall convert 6moles of HF to grams
This is illustrated below:
Number of mole of HF = 6moles
Molar Mass of HF = 1 + 19 = 20g/mol
Mass of HF =..?
Mass = number of mole x molar Mass
Mass of HF = 6 x 20
Mass of HF = 120g
Therefore, 120g of HF is needed to react with 3 moles of Sn
