Answer:
Therefore (2.30-0.03747)= 2.2625 M of HI remains from an initial concentration of 2.30 M after 4.5 hours.
Step-by-step explanation:
second order reaction: The rate of reaction proportional to the square of the concentration of reactant.
For second order reaction
K is rate constant = 1.6×10⁻³M⁻¹hr⁻¹
a = initial concentration of reactant = 2.30 M
a-x = concentration of reactant after t h
t = time = 4.5 h
Putting the values in the above equation,
Therefore (2.30-0.03747)= 2.2625 M of HI remains from an initial concentration of 2.30 M after 4.5 hours.
Answer:
Gain: (Decrease)
Loss:
(Increase)
Step-by-step explanation:
1)
34a = 14
a = 14 ÷ 34
a = 7/17
2)
<span>-29 = -32m
-32m = -29
m = 29/32
</span>
Answer:
30
Step-by-step explanation:
y/x = 30/(x + 10)
(y + 12)/x = 54/(x + 10)
y/x + 12/x = 54/(x + 10)
30/(x + 10) + 12/x = 54/(x + 10)
30x + 12(x + 10) = 54x
30x + 12x + 120 = 54x
120 = 12x
x = 10
EG = x + x + 10
EG = 10 + 10 + 10
EG = 30
Hope this helps <span>1) </span><span>Equations with negative values for a</span><span> produce graphs that open down and equations with a positive values for a</span> produce graphs that open up.
<span>2)<span> </span></span><span>As the absolute value of a gets larger our graphs become more narrow (they shoot towards positive or negative infinity faster). This is more interesting than it might appear. If you consider the second derivative of any quadratic it will be the a</span><span> value. The second derivative represents acceleration, so the larger the a value the faster the increase of velocity and accordingly a quicker progression towards positive or negative infinity. Check this out in graphing calculator, press play to vary the value of a from -20 to 20. Notice that when the value of a approaches zero, the approximates a line, and of course when a is 0 we have the line y</span><span> = 2x</span><span> – 1.</span>
