Answer:
Ea= -175.45J
A= 3.5×10^14
k=3.64 ×10^14 s^2.
Explanation:
From
ln k= -(Ea/R) (1/T) + ln A
This is similar to the equation of a straight line:
y= mx + c
Where m= -(Ea/R)
c= ln A
y= ln k
a)
Therefore
21.10 3 104= -(Ea/8.314)
Ea=-( 21.10 3 104×8.314)
Ea= -175.45J
b) ln A= 33.5
A= e^33.5
A= 3.5×10^14
c)
k= Ae^-Ea/RT
k= 3.5×10^14 × e^ -(-175.45/8.314×531)
k = 3.64 ×10^14 s^2.
CH4+(x)O2=CO2 +(Y)H2O
C=1 +H=4 +O=? = C=1 +O=2+? +H=?
H=4>>Y=2
C=1 +H=4 +O=? = C=1 +O=(2+2) +H=4
C=1 +H=4 +O=4 = C=1 +O=4 +H=4
O=4>>X=2
CH4+(2)O2 =CO2 +(2)H2O
Answer:
The answer to your question is SrCrO₄ + H₂O
Explanation:
Data
H₂CrO₄ + Sr(OH)₂ ⇒
We can notice that this is a Redox reaction or neutralization reaction because the reactants are an acid (H₂CrO₄) and a base (Sr(OH)₂). These reactions are also called double displacement reactions.
In these kind of reactions the products are always a binary or ternary salt and water.
Then, for this reaction,
H₂CrO₄ + Sr(OH)₂ ⇒ SrCrO₄ + H₂O
45 m. If each student needs 750 mm of tubing, the teacher should order 45 m of tubing.
a) Find the <em>length in millimetres</em>
Length = 60 students x (750 mm tubing/1 student) = 45 000 mm tubing
b) Convert <em>millimetres to metres
</em>
Length = 45 000 mm tubing x (1 m tubing/1000 mm tubing) = 45 m tubing
