Answer:
X₂O₃ , Y₂O
Explanation:
Valency of an element tells how well that element can form compounds with other atoms.
It sometimes, is also determined by the number of hydrogen atoms that it combines with.
Oxygen usually has a valency of 2, or oxidation state of -2.
That means oxygen will need 2 more electrons in it's outer shell to form a configuration of a noble gas(octet configuration).
The oxidation state of an atom in a molecule gives the number of valence electrons it has gained or lost.
Assuming both X and Y are less electronegative than oxygen, X has a valency of 3 and OS of +3.
While forming a neutral compound, we need no net charge on the compound. In X₂O₃,
total charge on X is 2 x (OS of X)= 2 x ( +3) = +6
total charge on O is 3 x (OS of O)= 3 x ( -2) = -6
Net charge is 0.
Similarly, in Y₂O
total charge on Y is 2 x (OS of Y)= 2 x ( +1) = +2
total charge on O is 1 x (OS of O)= 1 x ( -2) = -2
Net charge is 0.
Answer:
1.125 moles of hydrogen are needed.
Explanation:
Given data:
Number of moles of ammonia formed = 0.75 mol
Number of moles of hydrogen needed = ?
Solution:
Chemical equation:
N₂ + 3H₂ → 2NH₃
Now we will compare the moles of ammonia and hydrogen.
NH₃ : H₂
2 : 3
0.75 : 3/2×0.75 = 1.125 mol
1.125 moles of hydrogen are needed.
Answer:
This solution acts as an efficient buffer
Explanation:
the pH of a buffer solution can be described like this: ![pH=pKa+log\frac{[base]}{[acid]}](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D)
[acid]=[acetic acid]=
[base]=[sodium acetate]=
replacing, 
If we add an acid, pH will decrease a little bit and if we add a base, pH wil increase a little bit.
lets supose that we change the rate by increasing [base] to 0.1, then
and now lets supose that we increase [acid] to 0.1 
Big changes in concentration of base or acid doesn´t produce big changes in pH, in that way the mix of sodium acetate with acetic acid is a good buffer solution.
Answer is: <span>A. 18.02 g/mol.
At standard temperature and pressure 1 mol of gas occupied 22.4 liters:
V(H</span>₂O) = 22.4 L; volume of water.
Vm = 22.4 L/mol; molar volume at STP.
n(H₂O) = V ÷ Vm.
n(H₂O) = 22.4 L ÷ 22.4 L/mol.
n(H₂O) = 1 mol; amount of substance (water).
M(H₂O) = Ar(O) + 2Ar(H) · g/mol.
M(H₂O) = 16 + 2 ·1.01 · g/mol.
M(H₂O) = 18.02 g/mol; molar mass of water.
Answer:
1.67g H2CO3 are produced
Explanation:
Based on the reaction:
2NaHCO3 → Na2CO3 + H2CO3
<em>2 moles of NaHCO3 produce 1 mole of Na2CO3 and 1 mole of H2CO3</em>
To solve this question we need to find the moles of Na2CO3 = Moles of H2CO3. With their moles we can find the mass of H2CO3 as follows:
<em>Moles Na2CO3 -Molar mass: 105.99g/mol-</em>
2.86g Na2CO3 * (1mol/105.99g) = 0.02698 moles Na2CO3 = Moles H2CO3
<em>Mass H2CO3 -Molar mass: 62.03g/mol-</em>
0.02698 moles * (62.03g/mol) =
<h3>1.67g H2CO3 are produced</h3>
