Mass of SiC = 2 g
<h3>Further explanation</h3>
Given
Reaction
SiO₂(s) + 3C(s) → SiC(s) + 2CO(g)
3.00 g of SiO₂
4.50 g of C
Required
mass of SiC
Solution
mol SiO₂ (MW=60,08 g/mol) :
= 3 g : 60.08 g/mol
= 0.0499
mol C(Ar = 12 g/mol) :
= 4.5 g : 12 g/mol
= 0.375
mol : coefficient of reactants =
SiO₂ : 0.0499/1 = 0.0499
C : 0.375/3 = 0.125
SiO₂ as a limiting reactant(smaller ratio)
Mol SiC based on mol SiO₂ = 0.0499
Mass SiC :
= mol x MW
= 0.0499 x 40,11 g/mol
= 2 g
Moles/MxL
1.709moles/2.10L =<span>0.814M</span>
Answer:
MgSO4.7H2O
Explanation:
Let the formula for the hydrated magnesium sulphate be MgSO4.xH2O
Mass of the hydrated salt (MgSO4.xH2O) = 12.845g
Mass of anhydrous salt (MgSO4) = 6.273g
Mass of water molecule(xH2O) = Mass of the hydrated salt — Mass of anhydrous salt = 12.845 — 6.273 = 6.572g
Now,we can obtain the number of mole of water molecule present in the hydrated salt as follows:
Molar Mass of hydrated salt (MgSO4.xH2O) = 24 + 32 + (16x4) + x(2 + 16) = 24 + 32 + 64 + x(18) = 120 + 18x
Mass of xH2O/ Molar Mass of MgSO4.xH2O = Mass of water / mass of hydrated salt
18x/120 + 18x = 6.572/12.845
Cross multiply to express in linear form
18x x 12.845 = 6.572(120 + 18x)
231.21x = 788.64 + 118.296x
Collect like terms
231.21x — 118.296x = 788.64
112.914x = 788.64
Divide both side by 112.914
x = 788.64 /112.914
x = 7
Therefore the formula for the hydrated salt (MgSO4.xH2O) is MgSO4.7H2O
The alpha particles that were fired at the gold foil were positively charged. ... These experiments led Rutherford to describe the atom as containing mostly empty space, with a very small, dense, positively charged nucleus at the center, which contained most of the mass of the atom, with the electrons orbiting the nucleus.
hope this helped
If an atom suffers from a collision, that causes an electron to jump from a lower to higher state, it is called collisional excitation