Answer: 1.00×10^26=1000×10^23
1000×10^23/6.022×10^23=166.05×10^23 moles of water.
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Explanation:
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:
![r = \frac{\Delta c}{\Delta t}](https://tex.z-dn.net/?f=r%20%3D%20%5Cfrac%7B%5CDelta%20c%7D%7B%5CDelta%20t%7D)
Given the following reaction:
![2 N_2O_5 (g)\rightleftharpoons 4 NO_2 (g) + O_2 (g)](https://tex.z-dn.net/?f=2%20N_2O_5%20%28g%29%5Crightleftharpoons%204%20NO_2%20%28g%29%20%2B%20O_2%20%28g%29)
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:
![r_{N_2O_5} = \frac{0.066 M - 0.100 M}{200.00 s - 0.00 s} = -0.00017 M/s](https://tex.z-dn.net/?f=r_%7BN_2O_5%7D%20%3D%20%5Cfrac%7B0.066%20M%20-%200.100%20M%7D%7B200.00%20s%20-%200.00%20s%7D%20%3D%20-0.00017%20M%2Fs)
(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]
Methane is the primary component of natural gas and is used to produce heat and electricity around the world.
This is false. The amount of electrons in lithium (Li) is not 7, it is 3. There are 3 electrons in lithium.