Ppm = mass of solute mg / mass of solvent kg
0.008 * 1000 = 8.0 mg ( solute )
1000 / 1000 = 1.0 kg (solvent )
ppm = 8 / 1
= 8.0 ppm
hope this helps!
Answer:
Without trees formerly forested areas would become drier and more prone to extreme droughts. When and if rain did come flooding would be disastrous and massive erosion would impact oceans smothering coral reefs and other marine habitats
Explanation:
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Answer:
carbon
Explanation:
The properties of carbon forms the basis for all life. Explanation: The atoms of carbon are found in the molecules of all living species.
Answer:
(a) -0.00017 M/s;
(b) 0.00034 M/s
Explanation:
(a) Rate of a reaction is defined as change in molarity in a unit time, that is:

Given the following reaction:

We may write the rate expression in terms of reactants firstly. Since reactants are decreasing in molarity, we're adding a negative sign. Similarly, if we wish to look at the overall reaction rate, we need to divide by stoichiometric coefficients:
![r = -\frac{\Delta [N_2O_5]}{2 \Delta t}](https://tex.z-dn.net/?f=r%20%3D%20-%5Cfrac%7B%5CDelta%20%5BN_2O_5%5D%7D%7B2%20%5CDelta%20t%7D)
Reaction rate is also equal to the rate of formation of products divided by their coefficients:
![r = \frac{\Delta [NO_2]}{4 \Delta t} = \frac{\Delta [O_2]}{\Delta t}](https://tex.z-dn.net/?f=r%20%3D%20%5Cfrac%7B%5CDelta%20%5BNO_2%5D%7D%7B4%20%5CDelta%20t%7D%20%3D%20%5Cfrac%7B%5CDelta%20%5BO_2%5D%7D%7B%5CDelta%20t%7D)
Let's find the rate of disappearance of the reactant firstly. This would be found dividing the change in molarity by the change in time:

(b) Using the relationship derived previously, we know that:
![-\frac{\Delta [N_2O_5]}{2 \Delta t} = \frac{\Delta [NO_2]}{4 \Delta t}](https://tex.z-dn.net/?f=-%5Cfrac%7B%5CDelta%20%5BN_2O_5%5D%7D%7B2%20%5CDelta%20t%7D%20%3D%20%5Cfrac%7B%5CDelta%20%5BNO_2%5D%7D%7B4%20%5CDelta%20t%7D)
Rate of appearance of nitrogen dioxide is given by:
![r_{NO_2} = \frac{\Delta [NO_2]}{\Delta t}](https://tex.z-dn.net/?f=r_%7BNO_2%7D%20%3D%20%5Cfrac%7B%5CDelta%20%5BNO_2%5D%7D%7B%5CDelta%20t%7D)
Which is obtained from the equation:
![-\frac{\Delta [N_2O_5]}{2 \Delta t} = \frac{\Delta [NO_2]}{4 \Delta t}](https://tex.z-dn.net/?f=-%5Cfrac%7B%5CDelta%20%5BN_2O_5%5D%7D%7B2%20%5CDelta%20t%7D%20%3D%20%5Cfrac%7B%5CDelta%20%5BNO_2%5D%7D%7B4%20%5CDelta%20t%7D)
If we multiply both sides by 4, that is:
![-\frac{4 \Delta [N_2O_5]}{2 \Delta t} = \frac{\Delta [NO_2]}{\Delta t}](https://tex.z-dn.net/?f=-%5Cfrac%7B4%20%5CDelta%20%5BN_2O_5%5D%7D%7B2%20%5CDelta%20t%7D%20%3D%20%5Cfrac%7B%5CDelta%20%5BNO_2%5D%7D%7B%5CDelta%20t%7D)
This yields:
[tex]r_{NO_2} = \frac{\Delta [NO_2]}{\Delta t} = -2\frac{\Delta [N_2O_5]}{ \Delta t} = -2\cdot (-0.00017 M/s) = 0.00034 M/s[tex]