This is called
Nuclear fusion
Example is two hydrogen atoms fuse to form a helium atom
Where some of the atoms of hydrogen turn in to energy
So the answer is A
Answer:
p3=0.36atm (partial pressure of NOCl)
Explanation:
2 NO(g) + Cl2(g) ⇌ 2 NOCl(g) Kp = 51
lets assume the partial pressure of NO,Cl2 , and NOCl at eequilibrium are P1 , P2,and P3 respectively
![Kp=\frac{[NOCl]^{2} }{[NO]^{2} [Cl_2] }](https://tex.z-dn.net/?f=Kp%3D%5Cfrac%7B%5BNOCl%5D%5E%7B2%7D%20%7D%7B%5BNO%5D%5E%7B2%7D%20%5BCl_2%5D%20%7D)
![Kp=\frac{[p3]^{2} }{[p1]^{2} [p2] }](https://tex.z-dn.net/?f=Kp%3D%5Cfrac%7B%5Bp3%5D%5E%7B2%7D%20%7D%7B%5Bp1%5D%5E%7B2%7D%20%5Bp2%5D%20%7D)
p1=0.125atm;
p2=0.165atm;
p3=?
Kp=51;
On solving;
p3=0.36atm (partial pressure of NOCl)
Answer:
0.73L
Explanation:
The following data were obtained from the question :
V1 = 0.65 L
P1 = 3.4 atm
T1 = 19°C = 19 + 273 = 292K
V2 =?
P2 = 3.2 atm
T2 = 36°C = 36 + 273 = 309K
The bubble's volume near the top can be obtain as follows:
P1V1 /T1 = P2V2 /T2
3.4 x 0.65/292 = 3.2 x V2 /309
Cross multiply to express in linear form as shown below:
292 x 3.2 x V2 = 3.4 x 0.65 x 309
Divide both side by 292 x 3.2
V2 = (3.4 x 0.65 x 309) /(292 x 3.2)
V2 = 0.73L
Therefore, the bubble's volume near the top is 0.73L
Answer:
solubility of X in water at 17.0
is 0.11 g/mL.
Explanation:
Yes, the solubility of X in water at 17.0
can be calculated using the information given.
Let's assume solubility of X in water at 17.0
is y g/mL
The geochemist ultimately got 3.96 g of crystals of X after evaporating the diluted solution made by diluting the 36.0 mL of stock solution.
So, solubility of X in 1 mL of water = y g
Hence, solubility of X in 36.0 mL of water = 36y g
So, 36y = 3.96
or, y =
= 0.11
Hence solubility of X in water at 17.0
is 0.11 g/mL.
The ionization energy for a hydrogen atom in the n = 2 state is 328 kJ·mol⁻¹.
The <em>first ionization energy</em> of hydrogen is 1312.0 kJ·mol⁻¹.
Thus, H atoms in the <em>n</em> = 1 state have an energy of -1312.0 kJ·mol⁻¹ and an energy of 0 when <em>n</em> = ∞.
According to Bohr, Eₙ = k/<em>n</em>².
If <em>n</em> = 1, E₁= k/1² = k = -1312.0 kJ·mol⁻¹.
If <em>n</em> = 2, E₂ = k/2² = k/4 = (-1312.0 kJ·mol⁻¹)/4 = -328 kJ·mol⁻¹
∴ The ionization energy from <em>n</em> = 2 is 328 kJ·mol⁻¹
.