1. 12.992 L
2. 2.42 moles
3. 275.52 L
4. 567.844 g
<h3>Further explanation</h3>
Given
moles and volume at STP
Required
mass, volume and moles
Solution
Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters / mol.
1. 0.58 moles ammonia :
Volume = 0.58 moles x 22.4 L = 12.992 L
2. 77.5 grams of O₂ :
Moles = 77.5 grams x (1 mol/32 grams) = 2.42
3. 12.3 mole of Bromine gas :
Volume = 12.3 mole x (22.4 L/1 mole) = 275.52 L
4. 4.8 moles iron(II)chloride :
Mass = 4.48 moles x molar mass ( 126,751 g/mol) = 567.844 g
Answer:
A simple microscope is one that uses a single lens for magnification, such as a magnifying glass while a compound microscope uses several lenses to enhance the magnification of an object. It uses a lens to enlarge an object through angular magnification alone, giving the viewer an erect enlarged virtual image.
If its wrong correct me.
Answer:
they have an equal number of positive and negative charges
The ore contains 55.4% calcium phosphate (related to the mineral apatite) so the amount of Ca3(PO4)2 is 55.4%x=1000g so x=1000/0.554= 1.805kg. Now for the % of P in this amount of calcium phosphate, use all the masses of the elements in Ca3PO4= Ca=40.078 x 3= 120.23 and (PO4)2= (30.974+64)2=189.95 (NB oxygen is 16 mass x 4 =64) so the total mass is 310.2 and we have 61.95 of P (Pmass x 2) so 61.95/3102.= 0.19 or 19% P. So of the 1.805 x 0.19= 0.34kg of phosphorus.
