Seawater is typically 3.5% salt and has a density of 1.03 g/mL. How many grams of salt would be needed to prepare enough seawate
1 answer:
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:
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Answer:
A
Explanation:
Let's illustrate this; see the attachment.
We see that Mrs. Jacobson is pushing to the right with a force of 100 N and there is another opposite force pushing with a force of 15 N. Since these are in opposite directions, we can say that the force opposite to Mrs. Jacobson is pushing the fridge -15 N to the right (instead of 15 N to the left).
The net force would then be:
100 N + (-15 N) = 85 N to the right
The answer is A.
Answer:
The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol
Explanation:
The ∆H (heat of reaction) of the combustion reaction is the heat that accompanies the entire reaction. For its calculation you must make the total sum of all the heats of the products and of the reagents affected by their stoichiometric coefficient (number of molecules of each compound that participates in the reaction) and finally subtract them:
Enthalpy of the reaction= ΔH = ∑Hproducts - ∑Hreactants
In this case, you have: 2 NOCl(g) → 2 NO(g) + Cl₂(g)
So, ΔH=
Knowing:
- ΔH= 75.5 kJ/mol
= 90.25 kJ/mol
= 0 (For the formation of one mole of a pure element the heat of formation is 0, in this caseyou have as a pure compound the chlorine Cl₂)
=?
Replacing:
75.5 kJ/mol=2* 90.25 kJ/mol + 0 -
Solving
-=75.5 kJ/mol - 2*90.25 kJ/mol
-=-105 kJ/mol
=105 kJ/mol
<u><em>The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol</em></u>