Answer:
the reducing flame also called the carburizing flame.
Explanation:
because it gets the oxides of the unknown salts
Answer:
The Kc is 1.36 (but this is not an option, may be the options are wrong, or may be I was .. Thanks!)
Explanation:
Let's think all the situation.
2 ICl(g) ⇄ I₂(g) + Cl₂(g)
Initially 0.20 - -
Initially I have only 0.20 moles of reactant, and nothing of products. In the reaction, an x amount of compound has reacted.
React x x/2 x/2
Because the ratio is 2:1, in the reaction I have the half of moles.
So in equilibrium I will have
(0.20 - x) x/2 x/2
Notice that I have the concentration in equilibrium so:
0.20 - x = 0.060
x = 0.14
So in equilibrium I have formed 0.14/2 moles of I₂ and H₂ (0.07 moles)
Finally, we have to make, the expression for Kc and remember that must to be with concentration in M (mol/L).
As we have a volume of 2L, the values must be /2
Kc = ([I₂]/2 . [H₂]/2) / ([ICl]/2)²
Kc = (0.07/2 . 0.07/2) / (0.060/2)²
Kc = 1.225x10⁻³ / 9x10⁻⁴
Kc = 1.36
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 ![[H^+]](https://tex.z-dn.net/?f=%5BH%5E%2B%5D)
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
![R=k[0.01M][1.0 M]](https://tex.z-dn.net/?f=R%3Dk%5B0.01M%5D%5B1.0%20M%5D)
..[1]
a)
The rate of the reaction when [Sucrose] is changed to 2.5 M = R'
![R'=[0.01 M][2.5 M]](https://tex.z-dn.net/?f=R%27%3D%5B0.01%20M%5D%5B2.5%20M%5D)
..[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'
![R'=[0.01 M][0.5 M]](https://tex.z-dn.net/?f=R%27%3D%5B0.01%20M%5D%5B0.5%20M%5D)
..[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'
![R'=[0.0001 M][1.0 M]](https://tex.z-dn.net/?f=R%27%3D%5B0.0001%20M%5D%5B1.0%20M%5D)
..[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'
![R'=[0.1M][0.1M]](https://tex.z-dn.net/?f=R%27%3D%5B0.1M%5D%5B0.1M%5D)
..[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.
The student would find the water and sand, because salt dissolves in water unless it was ocean water or sea water
Answer:
The rms speed of the gas atoms after 3600 J of heat energy is added to the gas = 1150 m/s.
Explanation:
Mass of 3 moles of Helium = 3 moles × 4.00 g/mol = 12.00 g = 0.012 kg
The initial average kinetic energy of the helium atoms = (1/2)(m)(u²)
where u = initial rms speed of the gas = 850 m/s
Initial average kinetic energy of the gas = (1/2)(0.012)(850²) = 4335 J
Then, 3600 J is added to the gas,
New kinetic energy of the gas = 4335 + 3600 = 7935 J
New kinetic energy of Helium atoms = (1/2)(m)(v²)
where v = final rms speed of the gas = ?
7935 = (1/2)(0.012)(v²)
v² = (7935×2)/0.012
v² = 1,322,500
v = 1150 m/s
Hence, the rms speed of the gas atoms after 3600 J of heat energy is added to the gas = 1150 m/s.
Hope this Helps!!!
