I think the amount would be a 0.4998 mol
I did moles=mass(g)/A,r
=12.5/24.3 to get that
The empirical formula of the following compounds 0.903 g of phosphorus combined with 6.99 g of bromine.
<h3>What is empirical formula?</h3>
The simplest whole number ratio of atoms in a compound is the empirical formula of a chemical compound in chemistry. Sulfur monoxide's empirical formula, SO, and disulfur dioxide's empirical formula, S2O2, are two straightforward examples of this idea. As a result, both the sulfur and oxygen compounds sulfur monoxide and disulfur dioxide have the same empirical formula.
<h3>
How to find the empirical formula?</h3>
Convert the given masses of phosphorus and bromine into moles by multiplying the reciprocal of their molar masses. The molar masses of phosphorus and bromine are 30.97 and 79.90 g/mol, respectively.
Moles phosphorus = 0.903 g phosphorus = 0.0293 mol
Moles bromine 6.99 g bromine=0.0875 mol
The preliminary formula for compound is P0.0293Bro.0875. Divide all the subscripts by the subscript with the smallest value which is 0.0293. The empirical formula is P1.00Br2.99 ≈ P₁Br3 or PBr3
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Answer:
B. To change from a liquid state to a solid state is called Freezing
Aniline can be produced from nitro-benzene by reduction. When nitro-benzene reacts with tin (Sn) + hydrochloride acid (HCl) then aniline is produced. The Sn + HCl forms free hydrogen ions which converts the nitro group of the benzene ring to amine group, which is aniline. In place of Sn + HCl. One can use palladium (Pd) hydrogen mixture in presence of ethanol (EtOH). There produces a side product in this reaction which is cyclohexyl amine. The reaction can be shown as.
Answer:
Explanation:
How many mols do you have?
1 mol = 6.02 * 10^23 atoms
x mol = 6.25 * 10 ^32 atoms
1/x = 6.02*10^23 / 6.25 * 10^32 Cross multiply
6.02 * 10^23 * x = 1 * 6.25 * 10^32 Divide by 6.02 * 10^23
x = 6.25 * 10*32/ 6.02 ^10^23
x = 1.038 * 10^9 mols which is quite large.
Find the number of grams. (Use the value for copper on your periodic table. I will just use an approximate number.)\
1 mol of copper = 63 grams.
1.038 * 10^9 mols of copper = x
1/1.038 * 10^9 = 63/x Cross multiply
x = 1.038 * 10^9 * 63
x = 6.54 * 10^10 grams of copper.