Common salt
sulfur dioxide
Answer:
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
Therefore, since the masses of both of the reactants are given, one computes the available moles of sulfuric acid and those moles of it consumed by the sodium hydroxide as shown below:
In such a way, since there is more available sulfuric acid than it that is consumed, the sodium hydroxide is the limiting reagent, consequently, the maximum mass of sodium sulfate turns out:
Best regards.
Answer:
Whenever a force is applied to an object, causing the object to move, work is done by the force. ... Work can be either positive or negative: if the force has a component in the same direction as the displacement of the object, the force is doing positive work
Blank 1: nothing (to keep 2 total nitrogen)
blank 2: 3 (to make 6 total hydrogen)
blank 3: 2 (to make 2 total nitrogen and 6 total hydrogen)
hope this helps!! :)
Answer:
0.29mol/L or 0.29moldm⁻³
Explanation:
Given parameters:
Mass of MgSO₄ = 122g
Volume of solution = 3.5L
Molarity is simply the concentration of substances in a solution.
Molarity = number of moles/ Volume
>>>>To calculate the Molarity of MgSO₄ we find the number of moles using the mass of MgSO₄ given.
Number of moles = mass/ molar mass
Molar mass of MgSO₄:
Atomic masses: Mg = 24g
S = 32g
O = 16g
Molar mass of MgSO₄ = [24 + 32 + (16x4)]g/mol
= (24 + 32 + 64)g/mol
= 120g/mol
Number of moles = 122/120 = 1.02mol
>>>> From the given number of moles we can evaluate the Molarity using this equation:
Molarity = number of moles/ Volume
Molarity of MgSO₄ = 1.02mol/3.5L
= 0.29mol/L
IL = 1dm³
The Molarity of MgSO₄ = 0.29moldm⁻³
