Answer is: the volume of an irregular object is 4,00 ml.
<span>Volume is the amount of space the object occupies and can be finded immersing it in water in a container with volume markings and than see how much the level of the container changes (goes up).
</span>V(irregular object) = V(final volume) - V(initial volume).
V(irregular object) = 7,50 ml - 3,50 ml.
V(irregular object) = 4,00 ml.
Noble gases are the least<span> reactive </span>elements<span> in the periodic table because they have a full valence shell. Examples are, Argon, Neon, Xenon, and Helium</span>
Answer:
36.55kJ/mol
Explanation:
The heat of solution is the change in heat when the KNO3 dissolves in water:
KNO3(aq) → K+(aq) + NO3-(aq)
As the temperature decreases, the reaction is endothermic and the molar heat of solution is positive.
To solve the molar heat we need to find the moles of KNO3 dissolved and the change in heat as follows:
<em>Moles KNO3 -Molar mass: 101.1032g/mol-</em>
10.6g * (1mol/101.1032g) = 0.1048 moles KNO3
<em>Change in heat:</em>
q = m*S*ΔT
<em>Where q is heat in J,</em>
<em>m is the mass of the solution: 10.6g + 251.0g = 261.6g</em>
S is specififc heat of solution: 4.184J/g°C -Assuming is the same than pure water-
And ΔT is change in temperature: 25°C - 21.5°C = 3.5°C
q = 261.6g*4.184J/g°C*3.5°C
q = 3830.87J
<em>Molar heat of solution:</em>
3830.87J/0.1048 moles KNO3 =
36554J/mol =
<h3>36.55kJ/mol</h3>
<em />
We will get the molality from this formula:
Molality = no.of moles of solute / Kg of solvent
So first we need the no.of moles of KNO3 = the mass of KNO3 / molar mass of KNO3
no.of moles of KNO3 = 175 / 101.01 = 1.73 mol
By substitution in the molality formula:
∴ molality = 1.73 / (750/1000) = 2.3 Molal
<u>Answer:</u>
<u>For a:</u> The equilibrium mixture contains primarily reactants.
<u>For b:</u> The equilibrium mixture contains primarily products.
<u>Explanation:</u>
There are 3 conditions:
- When
; the reaction is product favored. - When
; the reaction is reactant favored. - When
; the reaction is in equilibrium.
For the given chemical reactions:
The chemical equation follows:

The expression of
for above reaction follows:
![K_{eq}=\frac{[CN^-][H_3O^+]}{[HCN][H_2O]}=6.2\times 10^{-10}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5BCN%5E-%5D%5BH_3O%5E%2B%5D%7D%7B%5BHCN%5D%5BH_2O%5D%7D%3D6.2%5Ctimes%2010%5E%7B-10%7D)
As,
, the reaction will be favored on the reactant side.
Hence, the equilibrium mixture contains primarily reactants.
The chemical equation follows:

The expression of
for above reaction follows:
![K_{eq}=\frac{[HCl]^2}{[H_2][Cl_2]}=2.51\times 10^{4}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5BHCl%5D%5E2%7D%7B%5BH_2%5D%5BCl_2%5D%7D%3D2.51%5Ctimes%2010%5E%7B4%7D)
As,
, the reaction will be favored on the product side.
Hence, the equilibrium mixture contains primarily products.