There’s no element with symbol M
Hydrated salts are when salt crystals have water molecules bound. Anhydrous salts are when the water has been removed.
mass of water removed = hydrated salt - anhydrate salt
= 11.75 g - 9.25 g = 2.50 g
number of water moles = 2.50 g / 18 g/mol = 0.139 mol
number of cobalt (II) chloride moles = 9.25 g / 130 g/mol = 0.0712 mol
ratio of water moles to CoCl₂ moles - 0.139 mol / 0.0712 mol = 1.95
rounded off 2 moles of water for every 1 mol of CoCl₂
formula - CoCl₂.2H₂O
name - Cobalt(II) chloride dihydrate
This year course engages students in becoming skilled readers of prose written in a variety of periods, disciplines, and
rhetorical contexts and in becoming skilled writers who compose for a variety of purposes. More immediately, the course
prepares the students to perform satisfactorily on the A.P. Examination in Language and Composition given in the spring.
Both their writing and their reading should make students aware of the interactions among a writer’s purposes, audience
expectations, and subjects as well as the way generic conventions and the resources of language contribute to effectiveness
in writing. Students will learn and practice the expository, analytical, and argumentative writing that forms the basis of
academic and professional writing; they will learn to read complex texts with understanding and to write prose of
sufficient richness and complexity to communicate effectively with mature readers. Readings will be selected primarily,
but not exclusively, from American writers. Students who enroll in the class will take the AP examination.
Step one calculate the moles of each element
that is moles= % composition/molar mass
molar mass of Ca = 40g/mol, S= 32 g/mol , O= 16 g/mol
moles of Ca = 29.4 /40g/mol=0.735 moles, S= 23.5/32 =0.734 moles, O= 47.1/16= 2.94 moles
calculate the mole ratio by dividing each mole with smallest mole that is 0.734
Ca= 0.735/0.734= 1, S= 0.734/0.734 =1, O = 2.94/ 0.734= 4
therefore the emipical formula = CaSO4
Answer:
4.) 9, 1, and 4 5.) 4, 1, and 4
Explanation:
I am not quite sure about this because I cannot remember if the coefficient (the number before the elements) is applied to every element in the compound. If it is then your number of atoms are as follows: CORRECTION: you do not have to apply the coefficient to every element only the one that is after it. So when you back and fix the error your number of atoms will be as follows:
number 4
H: 9
P: 1
O: 4
number 5:
H: 4
S: 1
O: 4
you can calculate the number of atoms present in this compound by multiplying the coefficient and the subscripts of each atom.
hope this helped you :)
