Silver is not a compound. It's a mixture
Answer:
Balanced equation have equal number of atoms of different elements in the side of reactants and products.
Answer:
Problem, Variables, Hypothesis, Research, Procedure, Data, Conclusion.
Explanation:
First, you have to know what the problem that you're talking about is. Then, you have to know the independent, dependent, and controlled variables are in order to make a hypothesis. Then do research about the topic. After that, you make the procedure your experiment, if that's what your doing. Then after the experiment, you find the data. Using that data, you make a conclusion.
Answer:
At Equilibrium
[COCl₂] = 0.226 M
[CO] = 0.054 M
[Cl₂] = 0.054 M
Explanation:
Given that;
equilibrium constant Kc = 1.29 × 10⁻² at 600k
the equilibrium concentrations of reactant and products = ?
when 0.280 moles of COCl2(g) are introduced into a 1.00 L vessel at 600 K. [COCl²]
Concentration of COCl₂ = 0.280 / 1.00 = 0.280 M
COCl₂(g) ----------> CO(g) + Cl₂(g)
0.280 0 0 ------------ Initial
-x x x
(0.280 - x) x x ----------- equilibrium
we know that; solid does not take part in equilibrium constant expression
so
KC = [CO][Cl₂] / COCl₂
we substitute
1.29 × 10⁻² = x² / (0.280 - x)
0.0129 (0.280 - x) = x²
x² = 0.003612 - 0.0129x
x² + 0.0129x - 0.003612 = 0
x = -b±√(b² - 4ac) / 2a
we substitute
x = [-(0.0129)±√((0.0129)² - 4×1×(-0.003612))] / [2 × 1 ]
x = [-0.0129 ± √( 0.00017 + 0.01445)] / 2
x = [-0.0129 ± 0.1209] / 2
Acceptable value of x =[ -0.0129 + 0.1209] / 2
x = 0.108 / 2
x = 0.054
At equilibrium
[COCl₂] = (0.280 - x) = 0.280 - 0.054 = 0.226 M
[CO] = 0.054 M
[Cl₂] = 0.054 M
Answer:
We have:
M
g
(
s
)
+
H
C
l
(
a
q
)
→
M
g
C
l
2
(
a
q
)
+
H
2
(
g
)
LHS
Mg= 1
H= 1
Cl= 1
RHS
Mg= 1
Cl=2
H=2
To make Cl and H have 2 on both sides, we place a 2 in front of the HCl
M
g
(
s
)
+
2
H
C
l
(
a
q
)
→
M
g
C
l
2
(
a
q
)
+
H
2
(
g
)
LHS
Mg=1, H=2, Cl=2
RHS
Mg=1, H=1, Cl=2
Thus the equation is balanced and the coefficient of Mg is 1, which is understood so it is not written in the final equation.
Explanation:
