B. Ionic solids have higher melting points than molecular solids.
Answer: B is a gas because its volume is less than the volume of the containers.
Explanation: hope this help bye
Answer:
A wave reflection interaction (reflected by the wall)
Explanation:
In a reflection, the propagating wave is bounced off the reflecting surface because the medium of the surface prevents the propagation of the wave through it, such that wave is redirected at an angle equal to the angle of incident on the reflecting surface
Reflection takes place with the different forms of waves, such as sound wave, water waves, and light wave
The objects around us are seen with the aid of reflection of light from a light source such that the reflected light enters our eyes after being reflected on the surface of the object, and the object is seen.
Answer:
55.0 g
Explanation:
Step 1: Write the balanced equation for the production of oxygen
2 CO₂(g)⇒ 2 CO(g) + O₂(g)
Step 2: Calculate the mass of oxygen produced over a 2 hour period
The MOXIE produces 10.0 g of oxygen per hour.
2 h × 10.0 g/1 h = 20.0 g
Step 3: Calculate the moles corresponding to 20.0 g of O₂
The molar mass of O₂ is 32.00 g/mol.
20.0 g × 1 mol/32.00 g = 0.625 mol
Step 4: Calculate the number of moles of CO₂ needed to produce 0.625 moles of O₂
The molar ratio of CO₂ to O₂ is 2:1. The moles of CO₂ needed are 2/1 × 0.625 mol = 1.25 mol
Step 5: Calculate the mass corresponding to 1.25 moles of CO₂
The molar mass of CO₂ is 44.01 g/mol.
1.25 mol × 44.01 g/mol = 55.0 g
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]