Answer:
The correct option (c) "is held together by ionic bonds".
Step-by-step explanation:
Table salt is NaCl. It is a combination of Sodium and Chlorine.
There is 1 valance electron in sodium. Its electronic configuration is 2,8,1
The electronic configuration of Chlorine is 2,8,7. It means that it needs 1 electron so that its octet gets completed.
Na transfers one electron to Cl. It becomes Na⁺. Cl gains one electron From Na. It becomes Cl⁻.
As a resuslt, Na⁺ and Cl⁻ forms. Both Na and Cl are ions. They held together by ionic bonds.
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24+8 = 30
30 - 12 = 18
18-9 = 8
There are 8 cards left in the stack
Answer:
Initial Amount=1
Decay Factor=0.99987
Decay Rate=0.013%
Step-by-step explanation:
If P = 0.99987ᵃ
I. Initial amount
The initial amount is when a=0
P = 0.99987⁰ = 1
The Pressure at Sea Level=1
II. Decay factor
This is like the decay constant. From the model, P = 0.99987ᵃ, the decay constant/factor is 0.99987.
III. Decay Rate = 1-0.99987 = 0.00013
Expressed in Percentage = 0.013%
Answer: The required solution is
Step-by-step explanation: We are given to solve the following differential equation :
Let us consider that
be an auxiliary solution of equation (i).
Then, we have
Substituting these values in equation (i), we get
![m^2e^{mt}+10me^{mt}+25e^{mt}=0\\\\\Rightarrow (m^2+10y+25)e^{mt}=0\\\\\Rightarrow m^2+10m+25=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow m^2+2\times m\times5+5^2=0\\\\\Rightarrow (m+5)^2=0\\\\\Rightarrow m=-5,-5.](https://tex.z-dn.net/?f=m%5E2e%5E%7Bmt%7D%2B10me%5E%7Bmt%7D%2B25e%5E%7Bmt%7D%3D0%5C%5C%5C%5C%5CRightarrow%20%28m%5E2%2B10y%2B25%29e%5E%7Bmt%7D%3D0%5C%5C%5C%5C%5CRightarrow%20m%5E2%2B10m%2B25%3D0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~%5B%5Ctextup%7Bsince%20%7De%5E%7Bmt%7D%5Cneq0%5D%5C%5C%5C%5C%5CRightarrow%20m%5E2%2B2%5Ctimes%20m%5Ctimes5%2B5%5E2%3D0%5C%5C%5C%5C%5CRightarrow%20%28m%2B5%29%5E2%3D0%5C%5C%5C%5C%5CRightarrow%20m%3D-5%2C-5.)
So, the general solution of the given equation is
Differentiating with respect to t, we get
According to the given conditions, we have
and
Thus, the required solution is
