Answer:
C.
Explanation:
The electronic configuration of N (7 electrons): 1s² 2s² 2p³.
The orbital 1s is filled with two electrons and their spinning direction is opposite and also electrons of 2s.
3p contains (3 electrons) should fill the 3 orbitals firstly. Every orbital contains 1 electron and be in the same spin direction.
So, the right choice is c.
A is wrong because 2 electrons of 3p are paired in the first orbital before filling every orbital.
B is wrong because the 2 electrons of 1s and 2s are in the same direction and also 2 electrons of 3p are paired in the first orbital before filling every orbital.
D is also wrong the 2 electrons of 1s and 2s are in the same direction and the electron in the second orbital of 3p are in opposite direction of the other 2 electrons.
Answer:
Explanation:
moler mass of Cu is 63.546 g/mol. Since 63.546 g of copper has 6.022 x 10 power(23) atoms (Avogadro's number). = 9.5 x 10(power)21 atoms of copper.
Explanation:
a) The amount of heat released by coffee will be absorbed by aluminium spoon.
Thus, ![heat_{absorbed}=heat_{released}](https://tex.z-dn.net/?f=heat_%7Babsorbed%7D%3Dheat_%7Breleased%7D)
To calculate the amount of heat released or absorbed, we use the equation:
![Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})](https://tex.z-dn.net/?f=Q%3Dm%5Ctimes%20c%5Ctimes%20%5CDelta%20T%3Dm%5Ctimes%20c%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29)
Also,
..........(1)
where,
q = heat absorbed or released
= mass of aluminium = 45 g
= mass of coffee = 180 g
= final temperature = ?
= temperature of aluminium = ![24^oC](https://tex.z-dn.net/?f=24%5EoC)
= temperature of coffee = ![85^oC](https://tex.z-dn.net/?f=85%5EoC)
= specific heat of aluminium = ![0.80J/g^oC](https://tex.z-dn.net/?f=0.80J%2Fg%5EoC)
= specific heat of coffee= ![4.186 J/g^oC](https://tex.z-dn.net/?f=4.186%20J%2Fg%5EoC)
Putting all the values in equation 1, we get:
![45 g\times 0.80J/g^oC\times (T_{final}-24^oC)=-[180 g\times 4.186J/g^oC\times (T_{final}-83^oC)]](https://tex.z-dn.net/?f=45%20g%5Ctimes%200.80J%2Fg%5EoC%5Ctimes%20%28T_%7Bfinal%7D-24%5EoC%29%3D-%5B180%20g%5Ctimes%204.186J%2Fg%5EoC%5Ctimes%20%28T_%7Bfinal%7D-83%5EoC%29%5D)
![T_{final}=80.30^oC](https://tex.z-dn.net/?f=T_%7Bfinal%7D%3D80.30%5EoC)
80.30 °C is the final temperature.
b) Energy flows from higher temperature to lower temperature.Whenever two bodies with different energies and temperature come in contact. And the resulting temperature of both bodies will less then the body with high temperature and will be more then the body with lower temperature.
So, is our final temperature of both aluminium and coffee that is 80°C less than initial temperature of coffee and more than the initial temperature of the aluminum.
Answer: The percent yield is, 93.4%
Explanation:
First we have to calculate the moles of Na.
![\text{Moles of Na}=\frac{\text{Mass of Na}}{\text{Molar mass of Na}}=\frac{41.9g}{23g/mole}=1.82moles](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20Na%7D%3D%5Cfrac%7B%5Ctext%7BMass%20of%20Na%7D%7D%7B%5Ctext%7BMolar%20mass%20of%20Na%7D%7D%3D%5Cfrac%7B41.9g%7D%7B23g%2Fmole%7D%3D1.82moles)
Now we have to calculate the moles of ![Br_2](https://tex.z-dn.net/?f=Br_2)
![{\text{Moles of}Br_2} = \frac{\text{Mass of }Br_2 }{\text{Molar mass of} Br_2} =\frac{30.3g}{160g/mole}=0.189moles](https://tex.z-dn.net/?f=%7B%5Ctext%7BMoles%20of%7DBr_2%7D%20%3D%20%5Cfrac%7B%5Ctext%7BMass%20of%20%7DBr_2%20%7D%7B%5Ctext%7BMolar%20mass%20of%7D%20Br_2%7D%20%3D%5Cfrac%7B30.3g%7D%7B160g%2Fmole%7D%3D0.189moles)
![{\text{Moles of } NaBr} = \frac{\text{Mass of } NaBr }{\text{Molar mass of } NaBr} =\frac{36.4g}{103g/mole}=0.353moles](https://tex.z-dn.net/?f=%7B%5Ctext%7BMoles%20of%20%7D%20NaBr%7D%20%3D%20%5Cfrac%7B%5Ctext%7BMass%20of%20%7D%20NaBr%20%7D%7B%5Ctext%7BMolar%20mass%20of%20%7D%20NaBr%7D%20%3D%5Cfrac%7B36.4g%7D%7B103g%2Fmole%7D%3D0.353moles)
The balanced chemical reaction is,
![2Na(s)+Br_2(g)\rightarrow 2NaBr](https://tex.z-dn.net/?f=2Na%28s%29%2BBr_2%28g%29%5Crightarrow%202NaBr)
As, 1 mole of bromine react with = 2 moles of Sodium
So, 0.189 moles of bromine react with =
moles of Sodium
Thus bromine is the limiting reagent as it limits the formation of product and Na is the excess reagent.
As, 1 mole of bromine give = 2 moles of Sodium bromide
So, 0.189 moles of bromine give =
moles of Sodium bromide
Now we have to calculate the percent yield of reaction
![\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100=\frac{0.353 mol}{0.378}\times 100=93.4\%](https://tex.z-dn.net/?f=%5C%25%5Ctext%7B%20yield%7D%3D%5Cfrac%7B%5Ctext%7BActual%20yield%7D%7D%7B%5Ctext%7BTheoretical%20yield%7D%7D%5Ctimes%20100%3D%5Cfrac%7B0.353%20mol%7D%7B0.378%7D%5Ctimes%20100%3D93.4%5C%25)
Therefore, the percent yield is, 93.4%
Answer:
CO is considered as a product.
Explanation:
A general chemical equation for a combination reaction follows:
To write a chemical equation, we must follow some of the rules:
The reactants must be written on the left side of the direction arrow.
A '+' sign is written between the reactants, when more than one reactants are present.
An arrow is added after all the reactants are written in the direction where reaction is taking place. Here, the reaction is taking place in forward direction.
The products must be written on the right side of the direction arrow.
A '+' sign is written between the products, when more than one products are present.
For the given chemical equation:
are the reactants in the reaction and are the products in the reaction.
Hence, CO is considered as a product.