Here is a link to a website that explains these observed trends.
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Book%3A_Introductory_Chemistry_(CK-12)/06%3A_The_Periodic_Table/6.15%3A_Periodic_Trends%3A_Atomic_Radius
Answer: ![2MgI_2+Mn(SO_3)_2\rightarrow 2MgSO_3+MnI_4](https://tex.z-dn.net/?f=2MgI_2%2BMn%28SO_3%29_2%5Crightarrow%202MgSO_3%2BMnI_4)
Explanation:
Double displacement reaction: it is a chemical reaction in which the reactants exchanges their ions to form new compounds as a products.
All the reaction are example of double displacement reaction beside reaction where magnesium iodide is reacting with manganese(II) sulfate to give magnesium sulfate and manganese(IV) iodide .
![2MgI_2+Mn(SO_3)_2\rightarrow 2MgSO_3+MnI_4](https://tex.z-dn.net/?f=2MgI_2%2BMn%28SO_3%29_2%5Crightarrow%202MgSO_3%2BMnI_4)
In this reaction , charge on manganese have changed from 2+ to 4+. Manganese in getting oxidized. Example of an oxidation reaction. Hence, this reaction is not an example of double displacement reaction.
Answer:
<h2>
The answer is $6.52.</h2>
Explanation:
Firstly, let's check out the given:
1. Sports Utility Vehicle's (Original SUV) avg gas mileage - 17.5 miles per gallon (mpg)
2. SUV's (Hybrid SUV) avg gas mileage - 21.5 miles per gallon (mpg)
3. Gas price - $2.59 per gallon
What is asked? How much would you save driving 237 miles in the hybrid SUV compared to the original SUV?
Next, you have to know how much gallons each car consumes upon driving 237 miles. You can do division in this step.
1. Original SUV = 237 miles ÷
= 13.54 gallons
2. Hybrid SUV = 237 miles ÷
= 11.02 gallons
Now that you know how much gallons each car consumes, you can now get the gas price for the gallons.
1. Original SUV = 13.54 gallons × $2.59 per gallon = $35.06
2. Hybrid SUV = 11.02 gallons × $2.59 per gallon = $28.54
In order to check the difference, you have to subtract.
- $35.06 - $28.54 = $6.52 (this is the amount of money that you can save driving 237 miles in the hybrid SUV compared with the original SUV)
Heavy elements such as radium, uranium, and thorium, all have alpha decaying nuclei. An Rn (radon) nuclei are created when a Ra (radium) nucleus decays, emitting an alpha particle in the process.
Discussion about radiation:
The radioactive process known as alpha decay, sometimes known as α radiation, involves the ejection of an alpha particle from the nucleus, which includes 2 neutrons and 2 protons.
- A helium atom's nucleus and an alpha particle have many similarities. Instability is seen as being present in all nuclei with an atomic number (Z) more than 82. These are frequently subject to alpha decay and are thought to be "neutron-rich." Heavy elements such as radium, uranium, thorium, etc., all have alpha decaying nuclei. An Rn (radon) nucleus is created when a Ra (radium) nucleus decays, emitting an alpha particle in the process.
- The mass number (A) and atomic number (Z) are typically lowered by two and four, respectively, during alpha decay. For instance, Ra-226 with atomic numbers 88 and 226 is converted into Rn-222 with the mass number 222 and atomic number 86 by alpha decay.
Learn more about alpha decay here:
brainly.com/question/27870937
#SPJ4
Answer The value of
and
is, 40.79 kJ and 37.7 kJ respectively.
Explanation :
Heat released at constant pressure is known as enthalpy.
The formula used for change in enthalpy of the gas is:
![\Delta Q_p=\Delta H=40.79kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20Q_p%3D%5CDelta%20H%3D40.79kJ%2Fmol)
Now we have to calculate the work done.
Formula used :
![w=-P\Delta V\\\\w=-P\times (V_2-V_1)](https://tex.z-dn.net/?f=w%3D-P%5CDelta%20V%5C%5C%5C%5Cw%3D-P%5Ctimes%20%28V_2-V_1%29)
where,
w = work done = ?
P = external pressure of the gas = 1 atm
= initial volume = ![1.88\times 10^{-5}m^3=1.88\times 10^{-5}\times 10^3L=1.88\times 10^{-2}L](https://tex.z-dn.net/?f=1.88%5Ctimes%2010%5E%7B-5%7Dm%5E3%3D1.88%5Ctimes%2010%5E%7B-5%7D%5Ctimes%2010%5E3L%3D1.88%5Ctimes%2010%5E%7B-2%7DL)
= final volume = ![3.06\times 10^{-2}m^3=3.06\times 10^{-2}\times 10^3L=3.06\times 10^{1}L](https://tex.z-dn.net/?f=3.06%5Ctimes%2010%5E%7B-2%7Dm%5E3%3D3.06%5Ctimes%2010%5E%7B-2%7D%5Ctimes%2010%5E3L%3D3.06%5Ctimes%2010%5E%7B1%7DL)
Now put all the given values in the above formula, we get:
![w=-(1atm)\times (3.06\times 10^{1}-1.88\times 10^{-2})L](https://tex.z-dn.net/?f=w%3D-%281atm%29%5Ctimes%20%283.06%5Ctimes%2010%5E%7B1%7D-1.88%5Ctimes%2010%5E%7B-2%7D%29L)
![w=-30.5812L.atm=-30.5812\times 101.3J=-3097.87556J=-3.09\times 10^3J=-3.09kJ](https://tex.z-dn.net/?f=w%3D-30.5812L.atm%3D-30.5812%5Ctimes%20101.3J%3D-3097.87556J%3D-3.09%5Ctimes%2010%5E3J%3D-3.09kJ)
Now we have to calculate the change in internal energy.
![\Delta U=q+w](https://tex.z-dn.net/?f=%5CDelta%20U%3Dq%2Bw)
![\Delta U=40.79kJ+(3.09kJ)](https://tex.z-dn.net/?f=%5CDelta%20U%3D40.79kJ%2B%283.09kJ%29)
![\Delta U=37.7kJ](https://tex.z-dn.net/?f=%5CDelta%20U%3D37.7kJ)
Thus, the value of
and
is, 40.79 kJ and 37.7 kJ respectively.