Answer:
Explanation:
Firstly, we have to determine the mass of metal X. We can do that by interpreting the first and second statement mathematically.
Metal X can form 2 oxides (A and B).
A + B = 3g
The mass of oxygen in A is 0.72g and the mass of oxygen in B is 1.16g.
The mass of metal X in the two oxides will be the same because it's the same metal.
Thus, we represent the mass of the metal in the two oxides as 2X.
2X + 0.72 + 1.16 = 3
2X + 1.88 = 3
2X = 3 - 1.88
2X = 1.12
X = 0.56
<u>Thus, 0.56 g of the metal combines with 0.72g of oxygen in A and 1.16 g of oxygen in B.</u>
Thus, mass of metal (X) in 1g of oxygen in A is
0.56g ⇒ 0.72g
X ⇒ 1
X = 1 × 0.56/0.72
X = 0.78 g
Hence, 0.78g of the metal will combine with 1g of oxygen for A
Also, mass of metal (X) in 1g of oxygen in B is
0.56g ⇒ 1.16g
X ⇒ 1g
X = 1×0.56/1.16
X = 0.48 g
Thus, 0.48g of the metal will combine with 1g of oxygen for B
Somewhere in the Orange to red range
it only happensonce every 7 years
Answer:
41.45 mL
Explanation:
Applying the general gas equation,
PV/T = P'V'/T'............... Equation 1
Where P = Initial pressure of hydrogen, V = Initial volume of hydrogen, T= Initial Temperature of hydrogen, P' = Final pressure of hydrogen, V' = Final Volume of Hydrogen, T' = Final Temperature.
make V' the subject of the equation
V' = PVT'/TP'................ Equation 2
Given: P = 718 torr = (718×133.322) N/m² = 95725.196 N/m², V = 47.9 mL = 0.0479 dm³, T = 26 °C = (26+273) = 299 K, T' = 273 K, P' = 101000 N/m²
Substitute these values into equation 2
V' = ( 95725.196×0.0479×273)/(299×101000)
V' = 0.04145 dm³
V' = 41.45 mL
Answer:
Molecular formula naphthalene → C₁₀H₈
Empirical formula naphthalene → C₅H₄
Explanation:
Centesimal composition means that in 100 g of compound we have x g of the element. Therefore in 100 g of naphthalene we have:
93.7 g of C
6.3 g of H
Let's make a rule of three:
In 100 g of naphthalene we have 93.7 g of C and 6.3 g of H
In 128 g of naphthalene we would have:
128 . 93.7 / 100 = 120 g of C
128. 6.3 / 100 = 8 g of H
We convert the mass to moles, by molar mass:
120 g . 1mol / 12 g = 10 moles C
8 g . 1mol/ 1g = 8 moles H
Molecular formula naphthalene → C₁₀H₈
Empirical formula naphthalene → C₅H₄
(The sub-index of each element is divided by the largest possible number)
