Answer:
MnSO₄.7H₂O
Explanation:
To solve this question, we need to convert the mass of the dehydrated MnSO₄. The difference between mass of the hydrate and dehydrated compound is the mass of water. With the mass we can find the moles of water and the formula of the hydrate:
<em>Moles MnSO₄ -Molar mass: 151g/mol-:</em>
17.51g * (1mol / 151g) = 0.116 moles
<em>Moles H₂O -Molar mass: 18g/mol-:</em>
32.14g-17.51g = 14.63g * (1mol / 18g) = 0.813 moles
The ratio of moles MnSO₄: Moles H₂O represent the amount of water molecules in the hydrate:
0.813mol / 0.116mol = 7 molecules of water.
The hydrate formula is:
<h3>MnSO₄.7H₂O</h3>
Answer:
lol there's no question girly but lmk if ya need help with anything.
Explanation:
Answer:
I wanna say c
Explanation:
It makes the most sense, because non-metals don't produce an electrical current, and therefore aren't conductors.
1) Compund Ir (x) O(y)
2) Mass of iridium = mass of crucible and iridium - mass of crucible = 39.52 g - 38.26 g = 1.26 g
3) Mass of iridium oxide = mass of crucible and iridium oxide - mass of crucible = 39.73g - 38.26g = 1.47g
4) Mass of oxygen = mass of iridum oxide - mass of iridium = 1.47g - 1.26g = 0.21g
5) Convert grams to moles
moles of iridium = mass of iridium / molar mass of iridium = 1.26 g / 192.17 g/mol = 0.00656 moles
moles of oxygen = mass of oxygen / molar mass of oxygen = 0.21 g / 15.999 g/mol = 0.0131
6) Find the proportion of moles
Divide by the least of the number of moles, i.e. 0.00656
Ir: 0.00656 / 0.00656 = 1
O: 0.0131 / 0.00656 = 2
=> Empirical formula = Ir O2 (where 2 is the superscript for O)
Answer: Ir O2
Can you provide a picture
