Answer : The radii of the two ions Cl⁻ ion and Na⁺ ion is, 181 and 102 pm respectively.
Explanation :
As we are given that the Na⁺ radius is 56.4% of the Cl⁻ radius.
Let us assume that the radius of Cl⁻ be, (x) pm
So, the radius of Na⁺ = ![x\times \frac{56.4}{100}=(0.564x)pm](https://tex.z-dn.net/?f=x%5Ctimes%20%5Cfrac%7B56.4%7D%7B100%7D%3D%280.564x%29pm)
In the crystal structure of NaCl, 2 Cl⁻ ions present at the corner and 1 Na⁺ ion present at the edge of lattice.
Thus, the edge length is equal to the sum of 2 radius of Cl⁻ ion and 2 radius of Na⁺ ion.
Given:
Distance between Na⁺ nuclei = 566 pm
Thus, the relation will be:
![2\times \text{Radius of }Cl^-+2\times \text{Radius of }Na^+=\text{Distance between }Na^+\text{ nuclei}](https://tex.z-dn.net/?f=2%5Ctimes%20%5Ctext%7BRadius%20of%20%7DCl%5E-%2B2%5Ctimes%20%5Ctext%7BRadius%20of%20%7DNa%5E%2B%3D%5Ctext%7BDistance%20between%20%7DNa%5E%2B%5Ctext%7B%20nuclei%7D)
![2\times x+2\times 0.564x=566](https://tex.z-dn.net/?f=2%5Ctimes%20x%2B2%5Ctimes%200.564x%3D566)
![2x+1.128x=566](https://tex.z-dn.net/?f=2x%2B1.128x%3D566)
![3.128x=566](https://tex.z-dn.net/?f=3.128x%3D566)
![x=180.9\approx 181pm](https://tex.z-dn.net/?f=x%3D180.9%5Capprox%20181pm)
The radius of Cl⁻ ion = (x) pm = 181 pm
The radius of Na⁺ ion = (0.564x) pm = (0.564 × 181) pm =102.084 pm ≈ 102 pm
Thus, the radii of the two ions Cl⁻ ion and Na⁺ ion is, 181 and 102 pm respectively.
The formula for force is F=ma. Because weight is a measure of force, then we can substitute the weight of the meteor, 3204 N, for F in the the equation for force. We also know that the acceleration of gravity on Earth is 9.8 m/s^2. To find the mass, simply divide both sides of the equation by the value of acceleration to get
![m = \frac{f}{a} = \frac{3204}{9.8} = 326.9](https://tex.z-dn.net/?f=m%20%3D%20%20%20%5Cfrac%7Bf%7D%7Ba%7D%20%20%3D%20%20%5Cfrac%7B3204%7D%7B9.8%7D%20%20%3D%20326.9)
Therefore, the value of the mass of the meteor is 326.9 kg.
Answer:
At -13
, the gas would occupy 1.30L at 210.0 kPa.
Explanation:
Let's assume the gas behaves ideally.
As amount of gas remains constant in both state therefore in accordance with combined gas law for an ideal gas-
![\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}](https://tex.z-dn.net/?f=%5Cfrac%7BP_%7B1%7DV_%7B1%7D%7D%7BT_%7B1%7D%7D%3D%5Cfrac%7BP_%7B2%7DV_%7B2%7D%7D%7BT_%7B2%7D%7D)
where
and
are initial and final pressure respectively.
and
are initial and final volume respectively.
and
are initial and final temperature in kelvin scale respectively.
Here
,
,
,
and
Hence ![T_{2}=\frac{P_{2}V_{2}T_{1}}{P_{1}V_{1}}](https://tex.z-dn.net/?f=T_%7B2%7D%3D%5Cfrac%7BP_%7B2%7DV_%7B2%7DT_%7B1%7D%7D%7BP_%7B1%7DV_%7B1%7D%7D)
![\Rightarrow T_{2}=\frac{(210.0kPa)\times (1.30L)\times (250K)}{(150.0kPa)\times (1.75L)}](https://tex.z-dn.net/?f=%5CRightarrow%20T_%7B2%7D%3D%5Cfrac%7B%28210.0kPa%29%5Ctimes%20%281.30L%29%5Ctimes%20%28250K%29%7D%7B%28150.0kPa%29%5Ctimes%20%281.75L%29%7D)
![\Rightarrow T_{2}=260K](https://tex.z-dn.net/?f=%5CRightarrow%20T_%7B2%7D%3D260K)
![\Rightarrow T_{2}=(260-273)^{0}\textrm{C}=-13^{0}\textrm{C}](https://tex.z-dn.net/?f=%5CRightarrow%20T_%7B2%7D%3D%28260-273%29%5E%7B0%7D%5Ctextrm%7BC%7D%3D-13%5E%7B0%7D%5Ctextrm%7BC%7D)
So at -13
, the gas would occupy 1.30L at 210.0 kPa.
Love is the hormones inside you giving u a sense of emotion that comes from your brain
Answer:
hunting for other animals or when they are really hungry
Explanation: