Answer:
water evaportaes from heat it then turns into a gas then can go into a solid one day and repeat the cycle
Explanation:
<h3>
![\tt Kc=\dfrac{[CO_2]}{[C][O_2]}](https://tex.z-dn.net/?f=%5Ctt%20Kc%3D%5Cdfrac%7B%5BCO_2%5D%7D%7B%5BC%5D%5BO_2%5D%7D)
</h3><h3>Further explanation</h3>
Given
Reaction
C+02 = CO2
Required
The equilibrium constant
Solution
The equilibrium constant is the ratio of concentration or pressure between the product and the reactant with each reaction coefficient raised
The equilibrium constant is based on the concentration (Kc) in a reaction
pA + qB -----> mC + nD
![\large {\boxed {\bold {Kc ~ = ~ \frac {[C] ^ m [D] ^ n} {[A] ^ p [B] ^ q}}}}](https://tex.z-dn.net/?f=%5Clarge%20%7B%5Cboxed%20%7B%5Cbold%20%7BKc%20~%20%3D%20~%20%5Cfrac%20%7B%5BC%5D%20%5E%20m%20%5BD%5D%20%5E%20n%7D%20%7B%5BA%5D%20%5E%20p%20%5BB%5D%20%5E%20q%7D%7D%7D%7D)
So for the reaction :
C+O₂ ⇔ CO₂
Answer is: mass of the ore is 8.54kg.<span>
</span>ω(Ca₃(PO₄)₂ - calcium phosphate) = 58.6% ÷ 100% = 0.586.
m(P) = 1.00 kg · 1000 g/kg.
m(P) = 1000 g.
In one molecule of calcium phosphate there are two phosphorus atoms:
M(Ca₃(PO₄)₂) = 310.18 g/mol.
M(P) = 30.97 g/mol.
For one kilogram of phosphorus, we need:
M(Ca₃(PO₄)₂) : 2M(P) = m(Ca₃(PO₄)₂) : m(P).
310.18 g/mol : 61.94 g/mol = m(Ca₃(PO₄)₂) : 1000 g.
m(Ca₃(PO₄)₂) = 5007.75 g ÷ 1000 g/kg = 5.007 kg.
Mass of ore find from proportion:
m(Ca₃(PO₄)₂) : m(ore) = 56% : 100%.
m(ore) = 100% · 5.007 kg ÷ 58.6%.
m(ore) = 8.54kg.
81. There is 1 carbon, 2 chlorine and fluorine atoms in Freon 12. To draw them it forms a cross with C in the middle and Cl and F both on the opposite side.
Cl
l
F - C- F
l
Cl
82. Freon-12 and Freon-14 are called halocarbons or just halides.
