Answer:
because it melting point is low
Answer:
Amount of salt needed is around 2.3*10³ g
Explanation:
The salt content in sea water = 3.5 %
This implies that there is 3.5 g salt in 100 g sea water
Density of seawater = 1.03 g/ml
Volume of seawater = volume of tank = 62.5 L = 62500 ml
Therefore, the amount of seawater required is:

The amount of salt needed for the calculated amount of seawater is:

Answer: The most likely partial pressures are 98.7MPa for NO₂ and 101.3MPa for N₂O₄
Explanation: To determine the partial pressures of each gas after the increase of pressure, it can be used the equilibrium constant Kp.
For the reaction 2NO₂ ⇄ N₂O₄, the equilibrium constant is:
Kp = 
where:
P(N₂O₄) and P(NO₂) are the partial pressure of each gas.
Calculating constant:
Kp = 
Kp = 0.0104
After the weights, the total pressure increase to 200 MPa. However, at equilibrium, the constant is the same.
P(N₂O₄) + P(NO₂) = 200
P(N₂O₄) = 200 - P(NO₂)
Kp = 
0.0104 = ![\frac{200 - P(NO_{2}) }{[P(NO_{2} )]^{2}}](https://tex.z-dn.net/?f=%5Cfrac%7B200%20-%20P%28NO_%7B2%7D%29%20%20%7D%7B%5BP%28NO_%7B2%7D%20%29%5D%5E%7B2%7D%7D)
0.0104
+
- 200 = 0
Resolving the second degree equation:
=
= 98.7
Find partial pressure of N₂O₄:
P(N₂O₄) = 200 - P(NO₂)
P(N₂O₄) = 200 - 98.7
P(N₂O₄) = 101.3
The partial pressures are
= 98.7 MPa and P(N₂O₄) = 101.3 MPa
Answer:
NaOH₂2CO3
Explanation:yes i know it is not a Balinese equation
Answer:
=> 1366.120 g/mL.
Explanation:
To determine the formula to use in solving such a problem, you have to consider what you have been given.
We have;
mass (m) = 25 Kg
Volume (v) = 18.3 mL.
From our question, we are to determine the density (rho) of the rock.
The formula:

First let's convert 25 Kg to g;
1 Kg = 1000 g
25 Kg = ?

= 25000 g
Substitute the values into the formula:

= 1366.120 g/mL.
Therefore, the density (rho) of the rock is 1366.120 g/mL.