Answer:
0.8162 gramos de plomo por gramo de yodo
1.633 gramos de plomo por gramo de yodo
Explanation:
Asumiendo una base de 100 gramos para cada compuesto:
Primer compuesto:
Gramos plomo: 44.94g
Gramos de yodo: 100-44.94g = 55.06g
Así, la masa de plomo por gramos de yodo para el primer compuesto es:
44.94g plomo / 55.06g Yodo =
<em>0.8162 gramos de plomo por gramo de yodo</em>
<em></em>
Segundo compuesto:
Gramos plomo: 62.02g
Gramos de yodo: 100-62.02g = 37.98g
La masa de plomo por gramos de yodo para el segundo compuesto es:
62.02g plomo / 37.98g Yodo =
<em>1.633 gramos de plomo por gramo de yodo</em>
Answer:
•Li2S is at a low melting point
•NCL3 is a covalent bond
Explanation:
Just some notes to help you:))
Covalent Bonds: A covalent bond is a chemical bond that involves the sharing of electron pairs between atoms. These electron pairs are known as shared pairs or bonding pairs, and the stable balance of attractive and repulsive forces between atoms, when they share electrons
Answer:
2.067 L ≅ 2.07 L.
Explanation:
- The balanced equation for the mentioned reaction is:
<em>CS₂(g) + 3O₂(g) → CO₂(g) + 2SO₂(g),</em>
It is clear that 1.0 mole of CS₂ react with 3.0 mole of O₂ to produce 1.0 mole of CO₂ and 2.0 moles of SO₂.
- At STP, 3.6 L of H₂ reacts with (?? L) of oxygen gas:
It is known that at STP: every 1.0 mol of any gas occupies 22.4 L.
<u><em>using cross multiplication:</em></u>
1.0 mol of O₂ represents → 22.4 L.
??? mol of O₂ represents → 3.1 L.
∴ 3.1 L of O₂ represents = (1.0 mol)(3.1 L)/(22.4 L) = 0.1384 mol.
- To find the no. of moles of SO₂ produced from 3.1 liters (0.1384 mol) of hydrogen:
<u><em>Using cross multiplication:</em></u>
3.0 mol of O₂ produce → 2.0 mol of SO₂, from stichiometry.
0.1384 mol of O₂ produce → ??? mol of SO₂.
∴ The no. of moles of SO₂ = (2.0 mol)(0.1384 mol)/(3.0 mol) = 0.09227 mol.
- Again, using cross multiplication:
1.0 mol of SO₂ represents → 22.4 L, at STP.
0.09227 mol of SO₂ represents → ??? L.
∴ The no. of liters of SO₂ will be produced = (0.09227 mol)(22.4 L)/(1.0 mol) = 2.067 L ≅ 2.07 L.
initial volume of the argon sample = 5.93L according to Boyle's law
What is Boyle's law ?
Boyle's law, also known as Mariotte's law, is a relationship describing how a gas will compress and expand at a constant temperature. The pressure (p) of a given quantity of gas changes inversely with its volume (v) at constant temperature, according to this empirical connection, which was established by the physicist Robert Boyle in 1662. In equation form, this means that pv = k, a constant.
According to Boyle's law
P1/V1 = P2/V2
P1 = initial pressure
P2 = final pressure
V1 =initial volume
V2= final volume
V1 = P1*V2/P2
V1 = 2.32*18.3/7.16 = 5.93L
initial volume of the argon sample = 5.93L according to Boyle's law
To know about Boyle's law from the link
brainly.com/question/26040104
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