The statement that correctly describes the reaction is: Tungsten (W) changes oxidation numbers from +6 to zero, so it undergoes reduction.
<h3>WHAT IS
OXIDATION AND REDUCTION:</h3>
- Oxidation is the process whereby electrons are lost in a reaction while reduction involves the gain of electrons.
- According to this question, the following reaction is given: WO3(s) + 3H2(g) → W(s) + 3H2O(g)
- Tungsten has an oxidation number of +6 in WO3 and it is changed to 0 in W, hence, there is a gain of electrons by tungsten (W).
Therefore, the statement that correctly describes the reaction is: Tungsten (W) changes oxidation numbers from +6 to zero, so it undergoes reduction.
Learn more about oxidation-reduction reaction at: brainly.com/question/3867774
Enzymes Transcription: Transcription factors (eukaryotes), RNA polymerase
<span>Enzymes Translation: initiation factors, elongation factors, termination factors </span>
<span>Molecules transcription: mRNA , DNA </span>
<span>Molecules translation: ribosomes, tRNA, mRNA, GTP (energy source)</span>
A) Limiting reactant
You need the molar ratios (from the balanced chemical equation) and the molar masses of each compound (from the atomic masses)
a) Molar ratios:
6 mol HF : 1 mol SiO2 : 1 mol H2SiF6
2) Molar masses:
Atomic masses:
H: 1 g/mol
F: 19 g/mol
Si: 28 g/mol
O: 16g/mol
=>
HF:1g/mol + 19 g/mol = 20 g/mol
SiO2: 28g/mol + 2*16g/mol = 60 g/mol
H2SiF6: 2*1g/mol + 28g/mol + 6*19g/mol = 144g/mol
3) convert data in grams to moles
21.0 g SiO2 / 60 g/mol = 0.35 mol SiO2
70.5 g HF / 20 g/mol = 3.525 mol HF
4) Use the theorical ratios to deduce which is in excess and which is the limiting reactant.
6 mol HF / 1mol SiO2 < 3.525 mol HF / 0.35 mol SiO2 ≈ 10
=> There is more HF than the needed to react with 0.35mol of SiO2 =>
SiO2 is the limiting reactant (HF is in excess)
b) Mass of excess reactant.
1) Calculate how many grams reacted, which requires to calculate first the number of moles that reacted
0.35 mol SiO2 * 6 mol HF / 1 mol SiO2 = 2.1 mol of HF
2.1 mol HF * 20 g/mol = 42 gram of HF
2) Subtract the quantity that reacted from the original quantity:
70.5 g - 42 g = 28.5 g of HF in excess
c) Theoretical yield of H2SiF6
1 mol of SiO2 ; 1 mol of H2SiF6 => 0.35 mol SiO2 : 0.35 mol H2SiF6
Convert those moles to grams: 0.35 mol * 144 g/mol = 50.4 grams
d) % yield
% yield = actual yield / theoretical yield * 100 = 45.8 / 50.4 * 100 = 90.87%
+2, because Beryllium is in the Group II of the periodic table.
Hope this helps!
Answer:Soluble
Explanation:Soluble:(of a substance) able to be dissolved, especially in water.
