Answer:
speed of plane in still air = 252 mph
speed of wind = 42 mph.
Explanation:
Given:
Distance travelled by the small plane = 245
Time taken to fly 245 miles = 1 hour and 10 minutes.
Time taken for return trip = 50 minutes.
To Find:
speed of the wind=?
The speed of the plane in still air=?
Solution:
We Know that
=>![\frac{245}{ \frac{70}{60}} hours](https://tex.z-dn.net/?f=%5Cfrac%7B245%7D%7B%20%5Cfrac%7B70%7D%7B60%7D%7D%20hours)
=>![\frac{245}{ \frac{7}{6}} hours](https://tex.z-dn.net/?f=%5Cfrac%7B245%7D%7B%20%5Cfrac%7B7%7D%7B6%7D%7D%20hours)
=>![245 \times { \frac{6}{7}](https://tex.z-dn.net/?f=245%20%5Ctimes%20%7B%20%5Cfrac%7B6%7D%7B7%7D)
=>210 mph against wind
on way back
=>![\frac{245}{ \frac{50}{60}} hours](https://tex.z-dn.net/?f=%5Cfrac%7B245%7D%7B%20%5Cfrac%7B50%7D%7B60%7D%7D%20hours)
=>![\frac{245}{ \frac{5}{6}} hours](https://tex.z-dn.net/?f=%5Cfrac%7B245%7D%7B%20%5Cfrac%7B5%7D%7B6%7D%7D%20hours)
=>![245 \times { \frac{6}{5}](https://tex.z-dn.net/?f=245%20%5Ctimes%20%7B%20%5Cfrac%7B6%7D%7B5%7D)
=> 294
Now
294 = plane +wind------------------------(1)
210 =plane - wind-------------------------(2)
Solving (1) and (2)
2 plane = 504
plane = 252
plane = ![\frac{504}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B504%7D%7B2%7D)
So substituting plane value in eq(2) we get,
210 =252 - wind
wind = 42 mph
Answer:
2522g
Explanation:
8Fe + S8 —> 8FeS
From the question,
8moles of Fe required 1mole of S8.
Therefore, Xmol of Fe will require 5.65mol of S8 i.e
Xmol of Fe = 8 x 5.65 = 45.2moles
Now we need to covert 45.2moles to gram. This is illustrated below below:
Molar Mass of Fe = 55.8g/mol
Number of mole of Fe = 45.2moles
Mass of Fe =?
Mass = number of mole x molar Mass
Mass of Fe = 45.2 x 55.8
Mass of Fe = 2522g
Therefore, 2522g of Fe is needed for the reaction.
Answer:
Answer 9: Oxygen is atomic number 8 on the periodic table, which means it has 8 protons!
Explanation:
Oxygen's 8 electrons are negatively charged, and they orbit the atomic nucleus and balance the positive charge of the 8 protons. The positive charge of 1 proton exactly cancels the negative charge of 1 electron.
As far as I know the Answer would be <span>True</span>
Answer: -
First Ionization energy IE 1 for element X = 801
Here X is told to be in the third period.
So principal quantum number n = 3 for X.
For 1st ionization energy the expression is
IE1 = 13.6 x Z ^2 / n^2
Where Z =atomic number.
Thus Z =( n^2 x IE 1 / 13.6)^(1/2)
Z = ( 3^2 x 801 / 13.6 )^ (1/2)
= 23
Number of electrons = Z = 23
Nearest noble gas = Argon
Argon atomic number = 18
Number of extra electrons = 23 – 18 = 5
a) Electronic Configuration= [Ar] 3d34s2
We know that more the value of atomic radii, lower the force of attraction on the electrons by the nucleus and thus lower the first ionization energy.
So more the first ionization energy, less is the atomic radius.
X has more IE1 than Y.
b) So the atomic radius of X is lesser than that of Y.
c) After the first ionization, the atom is no longer electrically neutral. There is an extra proton in the atom. Due to this the remaining electrons are more strongly pulled inside than before ionization. Hence after ionization, the radii of Y decreases.