Your stomach will bubble lol it might kind of tickle
Answer:
Well, it could be B or D but I would say B.
Explanation:
The structure of an organelle is usually fit to its function so it is most likely B.
Hope this helps!
Answer:
0.109 g.
Explanation:
Equation of the reaction:
Na3PO4 + 3HCl --> 3NaCl + H3PO4
Number of moles of HCl = molar concentration × volume
= 0.1 × 0.04
= 0.004 mol.
By stoichiometry, 1 mole of Na3PO4 neutralises 3 moles of HCl. Therefore, number of moles of Na3PO4 = 0.004/3
= 0.0013 mol
Mass of Na3PO4 = molar mass × number of moles
= 0.0013 × 164
= 0.219 g
Since 50% of Na3PO4 was present in the sample. Let 100 g be the total mass of the substance
= 0.219 × 50 g/100 g
= 0.109 g.
Answer:
0.453 moles
Explanation:
The balanced equation for the reaction is:
2Fe(s) + 3O2(g) ==> 2Fe2O3
From the equation, mass of O2 involved = 16 x 2 x 3 = 96g
mass of Fe2O3 involved = [(2x26) + 3 x 16] x 2
= 100g
Therefore 96g of O2 produced 100g of Fe2O3
32.2g of O2 Will produce 100x32.2/96
= 33.54g of Fe2O3
Converting it to mole using number of mole = mass/molar mass
but molar mass of Fe2O3 = 26 + (16 X 3)
= 74g/mole
Therefore number of mole of 33.54g of Fe2O3 = 33.54/74
= 0.453 moles
<u>Answer:</u> The ionic compound formed is 
(barium fluoride)
<u>Explanation:</u>
Ionic compound is formed by the complete transfer of electrons from 1 atom to another atom. The cation is formed by the loss of electrons by metals and anions are formed by gain of electrons by non metals.
Taking the metal as barium and non-metal as fluorine.
Barium is the 56th element of the periodic table having electronic configuration of ![[Xe]6s^2](https://tex.z-dn.net/?f=%5BXe%5D6s%5E2)
This element will loose 2 electrons and will form 
ion
Fluorine is the 9th element of the periodic table having electronic configuration of ![[He]2s^22p^5](https://tex.z-dn.net/?f=%5BHe%5D2s%5E22p%5E5)
This element will gain 1 electron and will form 
ion
By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.
Hence, the ionic compound formed is (barium fluoride)
