<h3>
Answer:</h3>
Empirical formula is CrO
<h3>
Explanation:</h3>
<u>We are given;</u>
- Mass of sample of Chromium as 7.337 gram
- Mass of the metal oxide formed as 9.595 g
We are required to determine the empirical formula of the metal oxide.
<h3>Step 1 ; Determine the mass of oxygen used </h3>
Mass of oxygen = Mass of the metal oxide - mass of the metal
= 9.595 g - 7.337 g
= 2.258 g
<h3>Step 2: Determine the moles of chromium and oxygen</h3>
Moles of chromium metal
Molar mass of chromium = 51.996 g/mol
Moles of Chromium = 7.337 g ÷ 51.996 g/mol
= 0.141 moles
Moles of oxygen
Molar mass of oxygen = 16.0 g/mol
Moles of Oxygen = 2.258 g ÷ 16.0 g/mol
= 0.141 moles
<h3>Step 3: Determine the simplest mole number ratio of Chromium to Oxygen</h3>
Mole ratio of Chromium to Oxygen
Cr : O
0.141 mol : 0.141 mol
1 : 1
Empirical formula is the simplest whole number ratio of elements in a compound.
Thus the empirical formula of the metal oxide is CrO
2 in front of H2. Technically nothing in front of O2 otherwise put a 1 then a 2 in front of H2O
2H20 + 1O2 —— > 2H2O
Answer:
Ver explicacion
Explanation:
<u>a) 2,2,4-trimetilhexano</u>
En esta molecula tenemos una cadena lineal de seis carbonos. En los carbonos 1,2 y 4 tenemos grupos metilos.
<u>b) 3,6-dietil-2-metiloctano</u>
Para esta molecula tenemos una cadena lineal de 8 carbonos. En el carbono dos tenemos un grupo metilo y en los carbonos tres y seis tenemos dos grupos etilo.
<u>C) 3,5-dietil-2,8-dimetil-6-propilnonano</u>
Finalmente, para esta molecula tenemos una cadena de nueve carbonos. En los carbonos dos y ocho tenemos 2 grupos metil, en los carbonos tres y cinco tenemos 2 grupos etilo y en el carbono seis tenemos un grupo propilo.
Ver figura 1
Espero sea de ayuda!
Answer:
296.1g of Ag is the maximum amount of silver
Explanation:
A solution of 6.3% Ag by mass contains 6.3g of Ag per 100g of solution. Thus, you need to calculate the mass of the solution and then, the mass of Ag present in solution, thus:
Mass of solution:
Assuming a density of 1g/mL:
If the solution contains 6.3g of Ag per 100g of solution, the mass of Ag in 4700L is:
4700L × (6.3g Ag / 100g) =
<h3>296.1g of Ag is the maximum amount of silver</h3>
helium, beryllium, magnesium, calcium, strontium, barium, radium. i believe any of these are proper answers.