Answer:
2.40 M
Explanation:
The molarity of a solution tells you how many moles of solute you get per liter of solution.
Notice that the problem provides you with the volume of the solution expressed in milliliters,
mL
. Right from the start, you should remember that you must convert this volume to liters by using the conversion factor
1 L
=
10
3
mL
Now, in order to get the number of moles of solute, you must use its molar mass. Now, molar masses are listed in grams per mol,
g mol
−
1
, which means that you're going to have to convert the mass of the sample from milligrams to grams
1 g
=
10
3
mg
Sodium chloride,
NaCl
, has a molar mass of
58.44 g mol
−
1
, which means that your sample will contain
unit conversion
280.0
mg
⋅
1
g
10
3
mg
⋅
molar mass
1 mole NaCl
58.44
g
=
0.004791 moles NaCl
This means that the molarity of the solution will be
c
=
n
solute
V
solution
c
=
0.004791 moles
2.00
⋅
10
−
3
L
=
2.40 M
The answer is rounded to three sig figs, the number of sig figs you have for the volume of the solution.
Answer:
Forensic drug chemists analyze samples of unknown materials including powders, liquids and stains to determine the chemical identity or characteristics of the compounds that make up the sample. samples submitted as evidence in a drug-related case can contain one compound or a mixture of many compounds.
<h2>a)
The rate at which
is formed is 0.066 M/s</h2><h2>b)
The rate at which molecular oxygen
is reacting is 0.033 M/s</h2>
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.
Rate in terms of disappearance of
=
= 0.066 M/s
Rate in terms of disappearance of
= ![-\frac{1d[O_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1d%5BO_2%5D%7D%7Bdt%7D)
Rate in terms of appearance of
= ![\frac{1d[NO_2]}{2dt}](https://tex.z-dn.net/?f=%5Cfrac%7B1d%5BNO_2%5D%7D%7B2dt%7D)
1. The rate of formation of 
![-\frac{d[NO_2]}{2dt}=\frac{1d[NO]}{2dt}](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BNO_2%5D%7D%7B2dt%7D%3D%5Cfrac%7B1d%5BNO%5D%7D%7B2dt%7D)
![\frac{1d[NO_2]}{dt}=\frac{2}{2}\times 0.066M/s=0.066M/s](https://tex.z-dn.net/?f=%5Cfrac%7B1d%5BNO_2%5D%7D%7Bdt%7D%3D%5Cfrac%7B2%7D%7B2%7D%5Ctimes%200.066M%2Fs%3D0.066M%2Fs)
2. The rate of disappearance of 
![-\frac{1d[O_2]}{dt}=\frac{d[NO]}{2dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1d%5BO_2%5D%7D%7Bdt%7D%3D%5Cfrac%7Bd%5BNO%5D%7D%7B2dt%7D)
![-\frac{1d[O_2]}{dt}=\frac{1}{2}\times 0.066M/s=0.033M/s](https://tex.z-dn.net/?f=-%5Cfrac%7B1d%5BO_2%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%200.066M%2Fs%3D0.033M%2Fs)
Learn more about rate law
brainly.com/question/13019661
https://brainly.in/question/1297322
Answer:
There was an improvement in accuracy. There was no change in precision.
Explanation:
<em>The average mass after recalibration is closer to the mass of the standard, </em>so the recalibration improved the accuracy<em> </em>(the measurement is closer to an accepted 'true' value).
The standard deviation did not change, so the precision (or how disperse the measurements are) was not affected.
Answer:
704.6 g CO2
Explanation:
MM sucrose = 342.3 g/mol
MM CO2 = 44.01 g/mol
g CO2 = 456.7 g sucrose x (1 mol sucrose/MM sucrose) x (12 moles CO2/1 mol sucrose) x (MM CO2/1mol CO2) = 704.6 g CO2