Answer:
Precipitation Reactions
They contain two aqueous reactants, one aqueous product, and one solid product. In this reaction, two soluble products, Pb(NO3)2 and KI, combine to form one soluble product, KNO3, and one insoluble product, PbI2. This is a precipitation reaction, and PbI2 is the precipitate.
The type of substance that is most likely to contain a covalent bond is ONE THAT IS COMPOSED OF ONLY NON METALS.
Covalent bond is a type of chemical bond in which electron pairs are shared among the participating elements in order to achieve the octet form. Covalent bond is usually found among non metals.
Answer:
Explanation:
If the reaction is really exothermic (and it is) then the water would spatter all over the place. It would boil off if the container could hold it. It would also react according to the following reaction.
You are talking about a reaction like
2K + 2HOH = 2KOH + H2
Answer:
CH₂ ; 67.1 %
Explanation:
To determine the empirical formula we need to find what the mole ratio is in whole numbers of the atoms in the compound. To do that we will first need the atomic weights of C and H and then perform our calculation
Assume 100 grams of the compound.
# mol C = 85.7 g / 12.01 g/mol = 7.14 mol
# mol H = 14.3 g / 1.008 g/mol = 14.19 mol
The proportion is 14.9 mol H/ 7.14 mol C = 2 mol H/ 1 mol C
So the empirical formula is CH₂
For the second part we will need to first calculate the theoretical yield for the 12.03 g NaBH₄ reacted and then calculate the percent yield given the 0.295 g B₂H₆ produced.
We need to calculate the moles of NaBH₄ ( M.W = 37.83 g/mol )
1.203 g NaBH₄ / 37.83 g/mol = 0.0318 mol
Theoretical yield from balanced chemical equation:
0.0318 mol NaBH₄ x 1 mol B₂H₆ / mol NaBH₄ = 0.0159 mol B₂H₆
Theoretical mass yield B₂H₆ = 0.0159 mol x 27.66 g/ mol = 0.440 g
% yield = 0.295 g/ 0.440 g x 100 = 67.1 %
Answer: The mole ratio of sodium to sodium chloride 2:2.
Explanation:
As the given reaction equation is as follows.
Here, 2 moles of sodium reacts with 1 mole of 
and leads to the formation of 2 moles of NaCl.
This means that 2 moles of sodium gives 2 moles of NaCl on reaction with chlorine.
Hence, the ratio of moles of sodium to sodium chloride is 2:2.
Thus, we can conclude that the mole ratio of sodium to sodium chloride 2:2.
