phosphorus is obtained primarily from ores contain...
phosphorus is obtained primarily from ores containing calcium phosphate.if a particular ore contains 57.1 % calcium phosphate, what minimum mass of the ore must be processed to obtain 1.00 kg of phosphorus?
we are asked to determine the minimum mass of the ore containing calcium phosphate (ca3(po4)2) that must be processed to obtain 1.00 kg of phosphorus (p). to solve this problem, we shall follow these steps:step 1: convert grams to moles of phosphorus using the molar mass of phosphorus.step 2: perform a mole-to-mole comparison between phosphorus(p) and calcium phosphate (ca3(po4)2), 2 mol p: 1 mol ca3(po4)2.step 3: compute for the mass of calcium phosphate (ca3(po4)2) using the obtained moles of calcium phosphate and the molar mass of calcium phosphate (ca3(po4)2). step 4: convert the mass from g to kg. take into account the percentage given in the problem to determine the number of kilograms of the ore that is required.[readmore]step 1: convert grams to moles of phosphorus using the molar mass of phosphorus. we are given:mass of phosphorus = 1.00 kgrecall:1 kg = 103 g molar mass of p (from the periodic table) = 30.97 g/molmole of phosphorus=1.00 kg p(103 g p1 kg p)(1 mol p30.97 g p)mole of phosphorus = 32.289 molstep 2: perform a mole-to-mole comparison between phosphorus(p) and calcium phosphate (ca3(po4)2), 2 mol p: 1 mol ca3(po4)2.mole of ca3(po4)2 =32.289 mol p(1 mol ca3(po4)22 mol p)mole of ca3(po4)2 = 16.145 molstep 3: compute for the mass of calcium phosphate (ca3(po4)2) using the obtained moles of calcium phosphate and the molar mass of calcium phosphate (ca3(po4)2). the molar mass of calcium phosphate (ca3(po4)2) is:ca3(po4)2          3 × 40.078 g/mol ca = 120.234 g/mol                          2 × 30.97 g/mol p = 61.94 g/mol                          8 × 16.00 g/mol o = 128 g/mol                         sum = 310.174 g/molmass of ca3(po4)2=16.145 mol ca3(po4)2(310.174 g ca3(po4)21 mol ca3(po4)2)mass of ca3(po4)2 = 5,007.759 g step 4: convert the mass from g to kg. take into account the percentage given in the problem to determine the number of kilograms of the ore that is required.recall:1 kg = 103 g percent (%) = per 100 (i.e. 57.1 kg calcium phosphate per 100 kg ore)mass of ca3(po4)2=5,007.759 g ca3(po4)2(1 kg ca3(po4)2103 g ca3(po4)2)(100 kg ore 57.1kg ca3(po4)2)mass of ore = 8.77 kgthe mass of ore, containing calcium phosphate, needed to obtain 1.00 kg of phosphorus is 8.77 kg.
hey there! as answered previously:-
crystallization is the process of slow precipitation of solid
crystals of a substance from a concentrated homogeneous solution it
is basically a separation method in which a substance is
transferred from the liquid (i e solution) phase to the solid state
the substance is thus obtained in pure crystalline form and has no
impurity it is used by chemists for the purification of solid
substances
following is an example of using crystallization to obtain pure
copper sulphate from an impure sample
1) take a small amount (5 g) of an impure sample of copper
sulphate in a china dish
2) dissolve the sample in 10 ml of water (we basically have to
use minimum amount of water in order to prepare a saturated
solution)
3) then, filter it to remove impurities
4) evaporate the filtrate on a water bath till small crystals
are formed indicating that a concentrated solution has been
obtained
5) after this, cover the solution with a filter paper and leave
it undisturbed at room temperature for the rest of the day
![](https://img-nm%20mnimgs%20com/img/study_content/lp/1/9/8/114/318/760/662/5-6-08_sonali_9_chemistry_9%202%202%202%201_gsx_sg_ss_html_3c9d32fa%20png)
it will be observed that crystals of pure copper sulphate are
obtained and the impurities are left behind in the solution hi.
ya calcium is to strengthen your bones to carry fetus and for babies bone development and iron is to improve your hb which ll lead for complication if hb is less and for babies blood flow..do not miss any tablets at any cost to have healthy pregnancy..take care osteoporosis treatments include the “basic cde’s” -- calcium(c), vitamin d (d), weight-bearing exercise (e), prevention of falls (f), and bone-friendly medicines. see more... answer: the molecular mass of an element is defined as the sum of the masses of the elements present in the molecule. molecular mass is obtained by multiplying the atomic mass of an element with the number of atoms in the molecule and then adding the masses of all the elements in the molecule. o2 is a diatomic molecule. the mass of an oxygen atom = 16 amu. ∴ mass of o2 molecule = 2 × 16 = 32 amu. mass of one molecule of oxygen → 32/ avogadro constant ( 6.023 × 1023) → 5.31 × 10-23g. a) ans. concentration = no. of moles of solute * 1000/ vol. in m;

no. of moles of solute = concentration * vol / 1000 = 1.4 * 270 / 1000 = 0.378 = 0.378 mol.



b)

molality = 1.65

molality = moles of solute / mass of solvent in kg

1.65 = moles of solute / 5.5 x 10^-5 kg

moles of solute = 9.075 x 10^-5

c)

mass of sucrose = 75 x 1.4 / 100 = 1.05g

moles = 1.05 / 342 = 3.07 x 10^-3 (i) adsorption is a surface phenomenon that causes the accumulation of molecules of a substance at the surface of a solid or liquid rather than in the bulk. in adsorption, the substance gets concentrated at the surface only. it does not penetrate through the surface to the bulk of the solid or liquid. for example, when a chalk stick is dipped into an ink solution, only its surface becomes coloured.


on the other hand, the process of absorption is a bulk phenomenon. in absorption, the absorbed substance gets uniformly distributed throughout the bulk of the solid or liquid. for example, when a sponge is dipped in water the whole sponge gets wets.


(ii)according to the schulze-hardy rule, the effectiveness of the salt causing flocculation depends on the charge on the ion of opposite sign to the charge on the sol particles. the greater the magnitude of the opposite charge, the higher the ability of a salt to coagulate the sol.  
thus, trivalent salt alcl3 is more effective in causing the coagulation of a negatively charged sol than divalent salt mgcl2.


(iii)out of sulphur sol and proteins, sulphur sol forms multimolecular colloids. sulphur sol consists of particles containing a thousand or more s8 molecules. on the other hand, proteins form macromolecular colloids. pure substances are made up of same type of particles or atoms. all the elements and compounds are pure substances because they contain the same kinds of particles. examples: hydrogen, carbon dioxide use the ka equation: 
take the acid as ha
this will dissociate : 
ha ↔ h+ + a- 
ka = [h+] [a-] /[ha] 
because dissociation is very small we can use 0.1m for [ha] 
[h+] = 2/50*0.1 = 0.0004m 
[a-] = 0.0004m 
 ka 0.0004 * 0.0004 / 0.1 
ka = 1.6*10^-5 * 0.1 
ka = 1.6*10^-4 magnesium atom loses 2 electrons to attain a stable electronic configuration and becomes a cation. example: a non-metallic atom like chlorine gains 1 electron to attain a stable electronic configuration and becomes an anion. example: cations and anions are oppositely charged particles which attract one another to form an electrovalent bond leading to the formation of an electrovalent compound here magnesium donates one electron each with two chlorine atoms resulting in the formation of magnesium chloride. According to a common rule of thumb, if something having less than 50 calories is consumed by you while fasting, then there is nothing to worry about as you will remain in the state of fast.


Coffee does not contain any calories itself, thus you can easily take black coffee while fasting if you want to take it. In fact, most of the calorie-free drinks can be consumed while fasting. However, diet drinks containing artificial sweeteners are not recommended by some dietitians.