Consider the following chemical equation. how many moles of seоg can be produced from 35
g of oxygen and excess selenium?
2se +302 - 2so3

0.73 mole

Explanation:

Data Given

mass of O₂= 35 g

moles of SeО₃ = ?

Reaction Given:

                    2Se + 3O₂ > 2SeO₃

Solution:

First find the mass of SeO₃ from the reaction that it produced from how much mass of oxygen (O₂).

Look at the balanced reaction

                     2Se  +    3O₂   >  2SeO₃

                                  3 mol               2 mol

So 3 mole of O₂ Produce 2 moles of SeO₃

Now

convert the moles into mass for which we have to know molar mass of SeO₃ and O₂

Molar mass of SeO₃

Molar mass of SeO₃ = 79 + 3(16) = 79 + 48

Molar mass of SeO₃ = 127 g/mol

mass of SeO₃

                    mass in grams = no. of moles x molar mass

                    mass of SeO₃ = 2 mol x 127 g/mol

                    mass of SeO₃ = 254 g

Molar mass of O₂ = 2(16)

Molar mass of O₂= 32 g/mol

mass of O₂

                  mass in grams = no. of moles x molar mass

                   mass of O₂ = 3 mol x 32 g/mol

                    mass of O₂ = 96 g

So,

96 g of O₂ Produce 254 g of SeO₃, then how many grams of SeO₃ will produced from 35.0 grams of O2

Apply unity Formula

                     96 g of O₂ ≅ 254 g of SeO₃

                     35 g of O₂ ≅ X g of SeO₃

By cross multiplication

                     X g of SeO₃ = 254 g x 35 g / 96 g

                     X g of SeO₃ = 92.6 g

92.6 g of SeO₃ will produce from 35 g of O₂

Now,

we will convert mass into moles

Formula used

                  no. of mole = mass in g / molar mass              

Put values in above equation

                  no. of mole = 92.6 / 127 g/mol

                  no. of mole SeO₃ = 0.73 mol

SeO₃ = 0.73 mole

kuchisake

kuchisake-onna (口裂け女, "slit mouth woman ") is a malevolent figure appearing in japanese ghost stories, possibly dating back to the edo period. she takes the form of a masked woman who approaches people and asks them if they believe she is pretty.

explanation:

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%