KE = mv²/2
m=2*KE/v²
v=50 m/s
KE=500J
m=2*KE/v² =2*500/50²=1000/2500= 0.4 kg
Answer: is considered as the limiting reagent for this reaction.
Explanation:
Limiting reagent is the reagent which limits the formation of the product.
Excess reagent is the reagent which is present in excess in a chemical reaction.
For the combustion of acetylene, the reaction follows:
By Stoichiometry,
5 moles of oxygen gas reacts with 2 moles of acetylene.
So, 81 moles of oxygen gas will react with = = 32.4 moles of acetylene.
As, the required amount of acetylene is less than the given amount. So, it is considered as an excess reagent and oxygen gas is the limiting reagent.
Answer:
Oxygen is the limiting reactant of this reaction.
Explanation:
2 moles of acetylene reacts with 5 moles of oxygen.
Then 37 moles of acetylene will react with :
of oxygen gas.
But only 81 moles of oxygen gas is present.
Then 81.0 moles of oxygen gas will react with:
of acetylene.
And still acetylene will remain after.
So, from the above calculation we cans see that oxygen is present in limiting amount which will limit the reaction.Hence, the limiting reactant of this reaction is oxygen.
-273°C
-160°C
50°C
Answer: The temperature at the point of intercept is -273°C.
Explanation: Intercept is a point on a graph which crosses the x and y axes.
Here, The X-axis represents Temperature in °C and Y-axis represents the Volume in mL.
On X-axis, the major division is 100 and minor division is 10. The slope intercepts the X-axis at third minor division from the origin. Hence, the temperature of intercept corresponds to -273°C.
Copper have in its nucleus?
Answer:
Melting point of aqueous solution = -10.32 °C
Explanation:
Where,
ΔT_f = Depression in freezing point
k_f = molal depression constant
m = molality
Formula for the calculation of molality is as follows:
density of water = 1 g/mL
density = mass/volume
Therefore,
mass = density × volume
volume = 3 L = 3000 mL
Mass of water = 1 g/mL × 3000 mL
= 3000 g
van't Hoff factor (i) for MgCl2 = 3
Substitute the values in the equation (1) to calculate depression in freezing point as follows:
Melting point of aqueous solution = 0 °C - 10.32 °C
= -10.32 °C
Answer:
The melting point of the solution is - 1.953 °C
Explanation:
In an ideal solution, the freezing point depression is computed as follows:
where:
is the freezing-point depression
is the cryoscopic constant, in this case is equal to 1.86
b is the molality of the solution
i is the van't Hoff factor, number of ion particles per individual molecule of solute, in this case is equal to 3
Molality is defined as follows:
b = moles of solute/kg of solvent
Moles of solute is calculated as follows:
moles of solute = mass of solute/molecular weight of solute
In this case there are 100 g of solute and its molecular weight is 35.5*2 + 24 = 95 g/mole. So, the moles are:
moles of solute = 100 g/(95 g/mol) = 1.05 moles
The mass of solvent is computed as follows:
mass of solvent = density of solvent * Volume of solvent
Replacing with the data of the problem we get:
mass of solvent = 1 kg/L*3 L = 3 kg
Finally, the molality of the solution is:
b = 1.05/3 = 0.35 mol/kg
Then, the freezing-point depression is:
The freezing-point depression is the difference between the melting point of the pure solvent (here water) and the melting point of the solution. We know that the the melting point of water is 0 °C, then:
melting point of water - melting point of the solution = 1.953 °C
melting point of the solution = 0 °C - 1.953 °C = - 1.953 °C