Answer: The weight/weight % or percent by mass of the solute is 5.41 %.
Explanation:
Mass of the sodium sulfate,w = 9.74 g
Volume of the water = 165 mL
Density of the water = 1 g/mL
Mass of the water =
Mass of the solution, W:
Mass of solute + Mass of solvent =9.47 g + 165 g=174.47 g
The weight/weight % or percent by mass of the solute is 5.41 %.
Answer:
Energy transfers from the metal to the water and calorimeter until they are all at room temperature.
Explanation:
CHECK THE COMPLETE QUESTION BELOW;
A metal sample is heated and placed into the water in a calorimeter at room temperature. Which statement best describes how the calorimeter can be used to determine the specific heat capacity of the metal sample?
Energy transfers to the metal from the water and calorimeter until they are all at room temperature
. Energy transfers from the metal to the water and calorimeter until they are all at room temperature.
Energy transfers to the metal from the water and calorimeter until they all reach a single temperature.
Energy transfers from the metal to the water and calorimeter until they all reach a single temperature.
EXPLANATION;
Using calorimeter to determine the specific heat capacity of the metal sample can be associated to the theory of conservation of energy because heat which is a form of energy is been transfer of heat between the metal to the water and the calorimeter, this process will proceed till single temperature is attained.
The change in the amount of temperature of the water in the calorimeter is measured in order to get the difference in heat change of the calorimeter water.
CHANGE IN HEAT CAN BE CALCULATED USING THE FORMULA.
Q = cmΔT where Q is the change in heat , c is the specific heat capacity and ΔT is the change in temperature
Answer:
The answer to your question is 7160 cm
Explanation:
Data
diameter = 1 mm
length = ?
amount of gold = 1 mol
density = 17 g/cm³
Process
1.- Get the atomic mass of gold
Atomic mass = 197 g
then, 197g ------------ 1 mol
2.- Calculate the volume of this wire
density = mass/volume
volume = mass/density
volume = 197/17
volume = 5.7 cm³
3.- Calculate the length of the wire
Volume = πr²h
solve for h
h = volume /πr²
radius = 0.05 cm
substitution
h = 5.7/(3.14 x 0.05²)
h = 5.7 / 0.0025
h = 7159.2 cm ≈ 7160 cm
Answer:
The solubility of MnS will decrease on addition of KOH solution.
Explanation:
As per the equation given:
On dissolution of MnS in water it gives a basic solution as it gives hydroxide ions.
Now when the we are adding aqueous KOH solution, it will dissociate as:
Thus it will further furnish more hydroxide ion,
This will increase the concentration of hydroxide ions (present of product side), the system will try to decrease its concentration by shifting towards reactant side.
Thus the solubility of MnS will decrease on addition of KOH solution.
The balanced equation for combustion is as follows;
2CH₃OH + 3O₂ ---> 2CO₂ + 4H₂O
The stoichiometry of CH₃OH to O₂ is 2:3
the limiting reagent is the reactant that is fully consumed during the reaction. The amount of product formed is directly proportional to the amount of limiting reactant produced. The excess reagent is the reactant that is provided in excess and is not fully used up, there will be an amount of this reagent remaining after the reaction.
If methanol is the limiting reactant,
If 2 mol of methanol reacts with 3 moles of O₂
Then 24 mol of methanol reacts with - 3/2 x 24 = 36 mol of O₂ should be present
But only 15 mol of O₂ is present, therefore O₂ is the limiting reactant and methanol is in excess.
3 mol of O₂ reacts with 2 mol of CH₃OH
then 15 mol of O₂ reacts with 2/3 x 15 = 10 mol of CH₃OH
Excess reactant is methanol, 10 mol are used up therefore 24 - 10 mol = 14 mol are remaining at the end of the reaction
