Explanations:- As per the significant figures rule, In multiplication and division, we go with least number of sig figs.
4.03 has three sig figs where as 0.0000035 has two sig figs only, The zeros in this number are not sig figs as they are just holding the place values. As the least number of sig figs here is two, the answer needs to be reported with two sig figs only.

Half-life<span> is the time required for a quantity to reduce to half its initial value. </span><span>If four half-lives have elapsed for calcium-45, then it would be 4x162.7 = 650.8 days have passed. Hope this answers the question. Have a nice day. Feel free to ask more questions.</span>
To determine the Keq, we need the chemical reaction in the system. In this case it would be:
CO + 2H2 = CH3OH
The Keq is the ration of the amount of the product and the reactant. We use the ICE table for this. We do as follows:
CO H2 CH3OH
I .42 .42 0
C -0.13 -2(0.13) 0.13
-----------------------------------------------
E = .29 0.16 0.13
Therefore,
Keq = [CH3OH] / [CO2] [H2]^2 = 0.13 / 0.29 (0.16^2)
Keq = 17.51
Answer is: <span>unbalanced electronegativity of the hydrogens and oxygens as they share electrons.
Oxygen has greater electronegativity than hydrogen, because of that oxygen is partially negative and hydrogen is partially positive.
</span>Electronegativity<span> is a </span>chemical property<span> that describes the tendency of an </span>atom<span> to attract a shared pair of </span>electrons<span> towards itself.</span>
Explanation:
It is known that equation for steady state concentration is as follows.

where, Q = flow rate
k = rate constant
V = volume
C = concentration of the entering air
Formula for volume of the box is as follows.
V =
= 
= 
Now, expression to determine the discharge is as follows.
Q = Av
= 
= 0.4 
And, m (loading) = 10kg/s,
k = 0.20/hr
as 1
L (if u want kg/L as concentration)
Now, calculate the concentration present inside as follows.

= 25
Now, we will calculate the concentration present outwards as follows.
,
and, t =
= 25000 s or 6.94 hr
Hence,
= 10.47 
Thus, we can conclude that the the steady-state concentration if the air is assumed to be completely mixed is
and
.