Well, ill keep it short but that would be 6.0 Liters under same conditions! :) Hope I helped you out!
The given unbalanced equation is as follow,
Sn + H₃PO₄ → H₂ + Sn₃(PO₄)₄
In above equation first let get start with Sn, on left side its one and on right side its 3. So, multiply Sn on left side with 3 to balance Sn, Hence,
3 Sn + H₃PO₄ → H₂ + Sn₃(PO₄)₄
Now, balance Phosphorous, on left side it counts one and on right side it counts 4, so multiply H₃PO₄ on left side with 4 to balance P, Hence,
3 Sn + 4 H₃PO₄ → H₂ + Sn₃(PO₄)₄
Oxygen on both left and right are 16 in number, Hence it is balanced,
Hydrogen on left hand side are 12 in number, while that on right hand side only 2, So, multiply H₂ on right side by 6 to balance hydrogen atoms.
<u>3</u> Sn + <u>4</u> H₃PO₄ → <u>6</u><u> </u>H₂ + Sn₃(PO₄)₄
Above equation is balanced, and coefficients are highlighted bold and underlined.
T<span>he nucleus, ribosomes, endoplasmic reticulum, cell membrane and cell wall. Also included are cytoskelteon, cytoplasm, Golgi apparatus, chloroplasts, and mitochondria; and also vacuoles, vesicles and lysosomes.</span>
Answer:
Using the periodic table of the elements to find atomic weights, we find that hydrogen has an atomic weight of 1, and oxygen's is 16. In order to calculate the molecular weight of one water molecule, we add the contributions from each atom; that is, 2(1) + 1(16) = 18 grams/mole.
Explanation:
hope this helped
The IUPAC name for a binary covalent compound will lack roman numerals.
Binary covalent compound comprises of two elements which forms a
compound through the sharing of electrons.
The sharing of electrons is referred to as covalent bonding and naming such
compounds require prefixes such as mono-, di- , tri- etc.
The standard method of naming these compounds should be with the
prefixes and not with the roman numerals.
An example is carbon(IV)oxide which is wrong in IUPAC naming.
Read more on brainly.com/question/16731560