Answer:
scientific approach
Explanation:
Method of investigation in which a problems is First identified and observations, experiment are done
Answer:
The answer to your question is (C). I would work out the question for you lad, but I don't know how to do it on here.
Explanation:
Answer:
0.22 grams of magnesium
This seems too low, so check the calculations/
Explanation:
How is the heat delivered? Is the magnesium hot and added to the water? If so, at what temperature.
Is the heat generated from an exothermic chemical reaction?
=================
I'll assume here that the heat comes from the reaction of magnesium with water. The balanced equation is:
Mg + 2H2O = Mg(OH)2 + H2
It has a heat of reaction of −924.7 kJ/mol.
We need enough Mg to heat 30 ml of water from 22°C to 90°C. The specific heat of water is needed. It is 4.184 J/g-K. It tells us that 4.184 Joules are needed to raise the temperature of water by 1 degree K.
30 ml of water with density 1 gram/ml means we have 30 grams of water. (We'll ignore the water that is added from the chemical reaction.).
Lets calculate the Joules required to raise 30 grams of water from 22°C to 90°C. Note that the specific heat has units of g and Kelvin. Since we need a temperature change, the number value for ΔT is the same for both °C and °K. So the temperature change is +68°K.
We can now calculate the Joules required:
(30 grams H2O)*(4.184 J/g-K)*(+68°K) = 8535.4 Joules or 8.5 kJ to 2 sig figs.
The Mg/H2O heat of reaction of −924.7 kJ/mol. will allow us to calculate the amount of Mg needed to supply 8.5 kJ. The minus sign tells us that the reaction RELEASES energy (the energy leaves the "system" of Mg and H2O).
Calculate the moles of Mg needed to release 8.5 kJ:
(924.7 kJ/mole)*(x moles) = 8.5 kJ
x moles = 0.0092 moles
Whoa. That is only (0.0092 moles)*(24.03 g/mole) = 0.22 grams of magnesium
This seems low to me, so check on the heat of reaction figure I used. And don't let the hydrogen get away.
(2) 35.0% (4) 60.0%
Answer:
The correct option is (2) 35.0%
Explanation:
If we want to calculate the percent composition by mass of nitrogen in NH4NO3 we first need to calculate the mass of nitrogen in one mole of NH4NO3.
Because of the molecular form of NH4NO3 (it has 2 ''N'') we know that one mole of NH4NO3 contains 2 moles of N.
The molar mass of N (nitrogen) is
This means that 1 mole of N contains 14.0067 grams of N.
Because we have 2 moles of N ⇒
We know that in 1 mole of NH4NO3 we have 2 moles of N(28.0134 g of N).
To calculate the percent composition by mass of nitrogen we need to divide the mass of nitrogen in one mole of NH4NO3 by the mass of one mole of NH4NO3 and then multiply it by 100.
We know that in 1 mole of NH4NO3 there are 80.0 g of NH4NO3 (Because of the molar mass data) ⇒
% ≅ 35.0%
We find that the percent composition by mass of nitrogen in NH4NO3 is 35.0%
We conclude that the correct option is (2) 35.0%
-> Hence, option B is correct.
A balanced equation is an equation for a chemical reaction in which the number of atoms for each element in the reaction and the total oxidation numbers is the same for both the reactants and the products.
-> represents a balanced chemical equation.
Learn more about the balanced equation here:
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