Answer:
How many hydrogen MOLECULES are needed to produce 2 molecules of water (H2O)?
Explanation:
Answer:
166 g
Explanation:
Step 1: Write the reaction for the obtaining of Fe from magnetite
Fe₃O₄ ⇒ 3 Fe + 2 O₂
Step 2: Calculate the moles corresponding to 120 g of Fe
The molar mass of Fe is 55.85 g/mol.
120 g × (1 mol/55.85 g) = 2.15 mol
Step 3: Calculate the moles of Fe₃O₄ required to produce 2.15 moles of Fe
The molar ratio of Fe₃O₄ to Fe is 1:3. The moles of Fe₃O₄ required are 1/3 × 2.15 mol = 0.717 mol
Step 4: Calculate the mass corresponding to 0.717 moles of Fe₃O₄
The molar mass of Fe₃O₄ is 231.53 g/mol.
0.717 mol × 231.53 g/mol = 166 g
Answer:
See explanation
Explanation:
Salt makes ice to melt faster than it usual. Sugar is another substance that influences the melting time of ice.
If i have a sample of pure ice and another sample of ice treated with sugar. I have to maintain the both initially at 0 degrees temperature and have a stop clock to measure the melting time.
I will ultimately notice that the sample of ice treated with sugar will melt faster than the pure sample of ice due to the presence of an impurity.
а о
Explanation:
The given cation:
(Rf₂Al₂F₃)³⁺
The oxidation number gives the extent to which a specie is oxidized in a reaction.
This number is assigned based on some rules:
- Elements in combined state whose atoms combines with themselves have an oxidation number of zero.
- The charge carried on simple ions gives their oxidation number.
- Algebraic sum of all the oxidation numbers of atoms in neutral compound is zero. In an ion with more than one kind of atom, the charge on it is the oxidation number.
for the specie given;
Known:
oxidation number of Al = +3
F = -1
charge = +3
let the oxidation number of Rf = k
2k + 2(3) + 3(-1) = +3
2k + 6 - 3 = 3
2k = 0
k = 0
The oxidation state of rutherfodium is 0
learn more:
Oxidation state brainly.com/question/10017129
#learnwithBrainly
Since you didn't give the actual volume (or any of the experimental values) I can only tell you how to do it. Do the calculation using the real (determined) volume of the flask. Then, re-do the calculation with v = 125ml. Take the two values and calculate % error; m = measured vol; g = guessed vol.
<span>[mW (m) - mW (g)]/mW (m) x 100% </span>
<span>(they want % error so, if it is negative, just get rid of the sign) </span>
