Answer:
pCH₄ = 105.1 - 0.42 = 104.68 torr
pI₂ = 7.96 -0.42 = 7.54 torr
pCH₃I = 0.42 torr
pHI = 0.42 torr
Explanation:
Kp is the equilibrium constant for the partial pressure of the gases in the reaction, and it is calculated for a general equation:
aA(g) + bB(g) ⇄ cC(g) + dD(g)
, where p is the partial pressure in the equilibrium. By the reaction given:
CH₄(g) + I₂(g) ⇄ CH₃I(g) + HI(g)
105.1 torr 7.96 torr 0 0 <em> initial partial pressure</em>
-x -x +x +x <em> react</em>
105.1-x 7.96-x x x <em>equilibrium</em>
Then:


x² = 0.1891 - 0.0255x -2.26x10⁻⁴x²
0.9997x² + 0.0255x - 0.1891 = 0
Using Bhaskara's rule:
Δ = (0.0255)² - 4x(0.9997)x(-0.1891)
Δ = 0.7568

Using only the positive term, x = 0.42 torr.
So,
pCH₄ = 105.1 - 0.42 = 104.68 torr
pI₂ = 7.96 -0.42 = 7.54 torr
pCH₃I = 0.42 torr
pHI = 0.42 torr
Answer:
a)If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.
b)If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.
c)If concentration of
is changed to 0.0001 M than rate will be increased by the factor of 0.01.
d) If concentration when [sucrose] and
both are changed to 0.1 M than rate will be increased by the factor of 1.
Explanation:
Sucrose +
fructose+ glucose
The rate law of the reaction is given as:
![R=k[H^+][sucrose]](https://tex.z-dn.net/?f=R%3Dk%5BH%5E%2B%5D%5Bsucrose%5D)
![[H^+]=0.01M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.01M)
[sucrose]= 1.0 M
..[1]
a)
The rate of the reaction when [Sucrose] is changed to 2.5 M = R'
..[2]
[2] ÷ [1]
![\frac{R'}{R}=\frac{[0.01 M][2.5 M]}{k[0.01M][1.0 M]}](https://tex.z-dn.net/?f=%5Cfrac%7BR%27%7D%7BR%7D%3D%5Cfrac%7B%5B0.01%20M%5D%5B2.5%20M%5D%7D%7Bk%5B0.01M%5D%5B1.0%20M%5D%7D)

If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.
b)
The rate of the reaction when [Sucrose] is changed to 0.5 M = R'
..[2]
[2] ÷ [1]
![\frac{R'}{R}=\frac{[0.01 M][0.5 M]}{k[0.01M][1.0 M]}](https://tex.z-dn.net/?f=%5Cfrac%7BR%27%7D%7BR%7D%3D%5Cfrac%7B%5B0.01%20M%5D%5B0.5%20M%5D%7D%7Bk%5B0.01M%5D%5B1.0%20M%5D%7D)

If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.
c)
The rate of the reaction when
is changed to 0.001 M = R'
..[2]
[2] ÷ [1]
![\frac{R'}{R}=\frac{[0.0001 M][1.0M]}{k[0.01M][1.0 M]}](https://tex.z-dn.net/?f=%5Cfrac%7BR%27%7D%7BR%7D%3D%5Cfrac%7B%5B0.0001%20M%5D%5B1.0M%5D%7D%7Bk%5B0.01M%5D%5B1.0%20M%5D%7D)

If concentration of
is changed to 0.0001 M than rate will be increased by the factor of 0.01.
d)
The rate of the reaction when [sucrose] and
both are changed to 0.1 M = R'
..[2]
[2] ÷ [1]
![\frac{R'}{R}=\frac{[0.1M][0.1M]}{k[0.01M][1.0 M]}](https://tex.z-dn.net/?f=%5Cfrac%7BR%27%7D%7BR%7D%3D%5Cfrac%7B%5B0.1M%5D%5B0.1M%5D%7D%7Bk%5B0.01M%5D%5B1.0%20M%5D%7D)

If concentration when [sucrose] and
both are changed to 0.1 M than rate will be increased by the factor of 1.
If the atom is neutral (meaning, not charged) the number of electron is equal to the number of protons. The mass number of an atom is the sum of the number of proton and the number of neutrons. From the given above, the mass number of gallium is 31 + 39. The answer is letter D. 70.
Answer:
Explanation:
specific heat of granite s = .79 J / g / k
let the mass of granite = m
heat lost by granite = heat gained by water
heat lost = mass x specific heat x drop in temperature
= m x .79 x (80 - 20.45)
heat gained by water
= 3000 x 4.186 x (20.45- 20)
heat lost by granite = heat gained by water
m x .79 x 59.55 = 3000 x 4.186 x .45
m = 120.12 g .
This is an incomplete question, the given sketch is shown below.
Answer : The name of given unit cell is, FCC (face-centered cubic unit cell)
Explanation :
Unit cell : It is defined as the smallest 3-dimensional portion of a complete space lattice which when repeated over the and again in different directions produces the complete space lattice.
There are three types of unit cell.
- SCC (simple-centered cubic unit cell)
- BCC (body-centered cubic unit cell)
- FCC (face-centered cubic unit cell)
In SCC, the atoms are arranged at the corners.

The number of atoms of unit cell = Z = 1
In BCC, the atoms are arranged at the corners and the body center.

The number of atoms of unit cell = Z = 2
The given unit cell is, FCC because the atoms are arranged at the corners and the center of the 6 faces.

The number of atoms of unit cell = Z = 4
Thus, the name of given unit cell is, FCC (face-centered cubic unit cell)