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PIT_PIT [208]
3 years ago
13

What data should be plotted to show that experimental concentration data fits a first-order reaction? A) 1/[reactant] vs. time B

) [reactant) vs. time C) In[reactant] vs. time D) In(k) vs. 1/1
Chemistry
1 answer:
natita [175]3 years ago
7 0

Answer:

C) In[reactant] vs. time

Explanation:

For a first order reaction the integrated rate law equation is:

A = A_{0}e^{-kt}

where A(0) = initial concentration of the reactant

A = concentration after time 't'

k = rate constant

Taking ln on both sides gives:

ln[A] = ln[A]_{0}-kt

Therefore a plot of ln[A] vs t should give a straight line with a slope = -k

Hence, ln[reactant] vs time should be plotted for a first order reaction.

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3 0
2 years ago
_A1Br3+_K=_KBr+_A1<br> fill in the correct coefficients.
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3 0
3 years ago
C6H12O6(s) + 6O2(g) --&gt; 6H2O(g) + 6CO2(g)
vredina [299]
One
Balance the Equation.
This has been done for you or it is given. Anyway this step is finished, but it must always be done.

Two
Find the molar mass of C6H12O6
6C = 6 * 12 = 72
12H = 12*1 = 12
CO = 6 * 16 = 96
Mol Mass = 180 grams / mol

Three
Find the mols of C6H12O6
n = ???
Molar Mass = 180 grams / mol
given mass = 13.2 grams.

n = given mass / molar mass
n = 13.2 / 180
n = 0.07333333 mols.

Four
Find the mols CO2
1 mol C6H12O6 will produce 6 mols CO2
0.0733333 mols will produce x

1/0.073333 = 6/x Cross multiply
x = 0.073333 * 6
x = 0.4399 moles.

Five
Find the volume given the conditions for temperature and pressure are STP conditions.
PV = nRT
R = 0.082057 
n = 0.43999
P = 1 atmosphere
T = 0 + 273 = 273
V = ???
1 * V = 0.43999 * 082057 * 273
V = 10.2 L
Answer: B

Note if you give out Brainly awards I'd sure appreciate one. This was a lot of typing


8 0
3 years ago
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