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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Explanation:
mass H2O2 = 55 mL(1.407 g/mL) = 80.85 g
molar mass H2O2 = 2(1.01 g/mol) + 2(16.00 g/mol) = 34.02 g/mol
moles H2O2 = 80.85 g/34.02 g/mol = 2.377 moles H2O2
For each mole of H2O2 you obtain 0.5 mole of O2 (see the equation).
moles O2 = 2.377 moles H2O2 (1 mole O2)/(2 moles H2O2) = 1.188 moles O2
Now, you need the temperature. If you are at STP (273 K, and 1.00 atm) then 1 mole of an ideal gas at STP has a volume of 22.4 L. Without temperature you are not really able to continue. I will assume you are at STP.
Volume O2 = 1.188 moles O2(22.4 L/mole) = 0.0530 L of O2.
which is 53 mL.
Answer:
76,6 kg
Explanation:
A kg it's equal to 1x10^3 grams
A Gigagrams it's equal to 1x10^9 grams
Knowing this, a kg it's equal to 1x10^6 gigagrams
![7,66*10^{-5}[gigagram]*\frac{1*10^6 [kg]}{1 [gigagram]}= 76.6 [kg]](https://tex.z-dn.net/?f=7%2C66%2A10%5E%7B-5%7D%5Bgigagram%5D%2A%5Cfrac%7B1%2A10%5E6%20%5Bkg%5D%7D%7B1%20%5Bgigagram%5D%7D%3D%2076.6%20%5Bkg%5D)
the energy required to raise one kilogram of the substance one degree
Answer:
23.8
Explanation:
Formula
weight % = weight of solute/ weight of solution x 100
weight of solution = weight of salt + weight of water
weight of solution = 1.62 lb + 5.20 lb = 6.82 lb
weight % = 1.62 / 6,82 x 100
weight % = 0.238 x 100
weight % = 23.8
