Answer:
Yes because the law of conservation of matter is followed. There are equal numbers of atoms of all elements in the
reactants and products.
Explanation:
A balanced equation conform to the law of conservation of matter. While unbalanced equation suggests that matter has been created or destroyed.
Na3PO4 + 3KOH —> 3NaOH + K3PO4
A careful observation of the above equation proved that the equation is balanced is balanced because there are equal numbers of atoms of all elements in the
reactants and products.
Answer:
Hi there!
Your answer is:
Ethical concerns dictate everything about scientific research. They determine what can be researched, who it can be researched on, how the data will be presented, how long the subjects can be researched, and more. To get a research study approved by the Institutional Review Board, you need to abide by strong ethical codes. In conclusion, a researchers ethical standings determines everything about their study.
Hope this helps
Answer:
Limiting reactant = B2O3
Amount of BCl3 formed = 468 g
Explanation:
The given reaction is:
In order to identify the limiting reagent calculate the moles of B2O3, C and Cl2. The reagent with the lowest moles is the limiting reactant
Since the moles of B2O3 < C < Cl2, the limiting reactant is B2O3
Based on the reaction stoichiometry:
1 mole of B2O3 produces 2 moles of BCl3
Hence, the number of moles of BCl3 produced under the experimental conditions = 2*1.997=3.994 moles
We have to first write a balanced equation.
so2 + o2 -> so3
this is not balanced though. we have 3 oxygen on right and 4 on left
2so2 + o2 -> 2so3
now it is same on both sides. we have to figure out which is limiting reagent with the given amounts of reagents. we do this by comparing the ratio between them in terms of moles. we see that so2 has a coefficient of 2 and o2 has none which implies 1 and so3 has 2. this means that for every 2 moles of so2 reacting with 1 mole of o2, we get 2 moles of so3.
lets convert the given values to moles. to do this we know that molecular weight is measured in grams per mole. we are given grams and need to cancel out the grams to get moles. so the molecular weight:
so2 =32.1 + 2 * 16 = 64.1 g/mol
o2 = 2 * 16 = 32 g/mol
so3 = 32.1 + 3 * 16 = 80.1 g/mol
now to convert 90 g of 2so2 under ideal conditions.
90g / 64.1g/mol = 1.404 moles
convert this amount of moles of so2 to moles of o2. we have 2 moles of so2 to 1 of o2
1.404moles so2 / 2 moles so2 * 1 mole o2= 0.702 moles o2
so we see under ideal conditions that 90g of so2 would react with .702g of o2. lets see how many we actually have with 100g of o2
100g / 32g/mol =3.16 mol.
so we have a lot more o2 than needed. we are looking for how much is left in grams. we have to figure out how much was used. to do this convert our ideal moles of o2 into grams.
.702 moles o2 * 32g/mol = 22.5g o2
so what we startrd with (100g) minus what we needed (22.5g) is what we have left
100 - 22.5 = 77.5g o2
Take 23mols and multiply it by the atomic mass of Li:
23mol * 6.94g/mol = 159.62g
