Answer:
10.3 g of oxygen are formed when 26.4 g of potassium chlorate is heated
Explanation:
This is the balanced equation:
2KClO₃(s) → 2KCl(s) + 3O₂(g)
Ratio beteween the salt and oxygen is 2:3
Molar mass of KClO₃ = 122.55 g/m
Let's find out the moles of salt
Mass / Molar mass
26.4 g /122.55 g/m = 0.215 moles
So, this is the final rule of three:
If 2 moles of KClO₃ make 3 moles of oxygen
0.215 moles of KClO₃ make (0.215 .3) /2 = 0.323 moles of O₂ are produced
Molar mass O₂ = 32 g/m
Moles . molar mass = mass
0.323 m . 32g/m = 10.3 g
Answer:
The majority of the molecules move from higher to lower concentration, although there will be some that move from low to high. The overall (or net) movement is thus from high to low concentration.
hope this helps!<3
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.
The correct answer is C. The water is the solvent because the green pellets dissolved in it.
Explanation:
In solutions, the are two substances involved, the solvent and the solute. The solvent is usually a liquid substance; additionally, the solvent dissolves another substance, which is known as the solute. For example, if you dissolve a spoon of salt in a glass of water, the solute is the salt which is the substance dissolved and the solvent is the water because the solute is dissolved in it. According to this, in the case presented the water is the solvent because the green pellets which are the solute dissolve in it.
Melting point- the temperature at which a substance has changed from a solid to a liquid
freezing- the temperature at which a substance chanes from liquid to a solid
boiling point- the temperature at which a substance changes from a liquid to a gas phase