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:
A
Explanation:
The answer to this question is A. Both ripening and spoiling are chemical reactions.
Spoiling is a chemical reaction because spoiled food has bad smell and taste and it changes colour too.
Ripening of fruits is a chemical change. For example the colour could change as well as the texture.
Answer: we learned this not to long ago i think its a,c
Explanation:
Answer:
6.564×10¹⁶ fg.
Explanation:
The following data were obtained from the question:
Mass of beaker = 76.9 g
Mass of beaker + salt = 142.54 g
Mass of salt in fg =?
Next, we shall determine the mass of the salt in grams (g). This can be obtained as follow:
Mass of beaker = 76.9 g
Mass of beaker + salt = 142.54 g
Mass of salt =?
Mass of salt = (Mass of beaker + salt) – (Mass of beaker)
Mass of salt = 142.54 – 76.9
Mass of salt = 65.64 g
Finally, we shall convert 65.64 g to femtograms (fg) as illustrated below:
Recall:
1 g = 1×10¹⁵ fg
Therefore,
65.64 g = 65.64 g × 1×10¹⁵ fg / 1g
65.64 g = 6.564×10¹⁶ fg
Therefore, the mass of the salt is 6.564×10¹⁶ fg.
Respuesta:
2400 mL
Explicación:
Paso 1: Información dada
- Volumen de solución: 3 L (3000 mL)
- Concentración de naranja: 20 % v/v
Paso 2: Calcular el volumen de naranja
La concentración de naranja es de 20 % v/v, es decir, cada 100 mL de solución hay 20 mL de naranja.
3000 mL Sol × 20 mL Naranja/100 mL Solución = 600 mL Naranja
Paso 3: Calcular el volumn de agua
El volumen de soluciónes igual a la suma de los volúmenes de naranja y agua.
VSolución = VNaranja + VAgua
VAgua = VSolución - VNaranja
VAgua = 3000 mL - 600 mL = 2400 mL
