Hey there!
For this we can use the combined gas law:
We are only working with pressure and temperature so we can remove volume.

P₁ = 2 atm
T₁ = 27 C
P₂ = 2.2 atm
Plug these values in:

Solve for T₂.




Convert this to kelvin and get 302.85 K, which is closest to B. 330 K.
Hope this helps!
Answer:
The answer to your question is below
Explanation:
Data
pOH = 5.9
a) [OH⁻] = ?
pOH measures the [OH⁻]
Formula
pOH = -log[OH⁻]
-Substitution
5.9 = -log[OH⁻]
[OH⁻] = antilog (-5.9)
-Result
[OH⁻] = 1.26 x 10⁻⁶ M
b) [H₃O⁺]
pH + pOH = 14
-Solve for pH
pH = 14 - pOH
-Substitution
pH = 14 - 5.9
-Result
pH = 8.1
-Calculate [H₃O⁺]
pH = -log[H₃O⁺]
-Substitution
8.1 = -log[H₃O⁺]
[H₃O⁺] = antilog(-8.1)
-Result
[H₃O⁺] = 7.9 x 10⁻⁹ M
c) This solution is alkaline because the pH is higher than 7.
Reduction takes place at the cathode and oxidation occurs at the anode.
Anode
Triprotic acid is a class of Arrhenius acids that are capable of donating three protons per molecule when dissociating in aqueous solutions. So the chemical reaction as described in the question, at the third equivalence point, can be show as: H3R + 3NaOH ⇒ Na3R + 3H2O, where R is the counter ion of the triprotic acid. Therefore, the ratio between the reacted acid and base at the third equivalence point is 1:3.
The moles of NaOH is 0.106M*0.0352L = 0.003731 mole. So the moles of H3R is 0.003731mole/3=0.001244mole.
The molar mass of the acid can be calculated: 0.307g/0.001244mole=247 g/mol.
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.