The reaction H₃PO₄ + 3 NaOH → Na₃PO₄ + 3 H₂O . 6.46 grams of Na₃PO₄ can be prepared by the reaction of 3.92 grams of H₃PO₄ with an excess of NaOH.
To determine the amount of Na₃PO₄ that can be prepared, we need to consider the balanced chemical equation and the stoichiometric ratio between H₃PO₄ and Na₃PO₄.
The balanced equation is:
H₃PO₄ + 3 NaOH → Na₃PO₄ + 3 H₂O
From the equation, we can see that 1 mole of H₃PO₄ reacts to produce 1 mole of Na₃PO₄. Therefore, the stoichiometric ratio is 1:1.
First, let's calculate the number of moles of H₃PO₄ given its mass:
Mass of H₃PO₄ = 3.92 g
Molar mass of H₃PO₄ = 97.994 g/mol
Moles of H₃PO₄ = Mass / Molar mass = 3.92 g / 97.994 g/mol
Since the stoichiometric ratio is 1:1, the moles of Na₃PO₄ produced will be equal to the moles of H₃PO₄.
Moles of Na₃PO₄ = Moles of H₃PO₄ = 3.92 g / 97.994 g/mol
Now, let's calculate the mass of Na₃PO₄ using the molar mass of Na₃PO₄:
Molar mass of Na₃PO₄ = 163.94 g/mol
Mass of Na₃PO₄ = Moles of Na₃PO₄ * Molar mass of Na₃PO₄
By substituting the calculated values into the equation, we can find the mass of Na₃PO₄ that can be prepared:
Mass of Na₃PO₄ = (3.92 g / 97.994 g/mol) * 163.94 g/mol
Calculating the result:
Mass of Na₃PO₄ ≈ 6.46 g
Therefore, approximately 6.46 grams of Na₃PO₄ can be prepared by the reaction of 3.92 grams of H₃PO₄ with an excess of NaOH.
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If the uncertainty associated with the position of an electron is 3.3×10−11 m, what is the uncertainty associated with its momentum?
The uncertainty associated with the momentum of an electron is given by the Heisenberg uncertainty principle as approximately 5.5×10^(-21) kg·m/s, which is calculated by the uncertainty in position.
According to the Heisenberg uncertainty principle, the product of the uncertainty in position (Δx) and the uncertainty in momentum (Δp) of a particle is always greater than or equal to a constant value, Planck's constant (h), divided by 4π:
Δx * Δp ≥ h / (4π)
In this case, the uncertainty in position (Δx) of the electron is given as 3.3 × 10^(-11) m. To find the uncertainty in momentum (Δp), we rearrange the equation:
Δp ≥ h / (4π * Δx)
Plugging in the values, we have:
Δp ≥ (6.626 × 10^(-34) J*s) / (4π * 3.3 × 10^(-11) m)
Simplifying the expression:
Δp ≥ 5.03 × 10^(-24) kg*m/s
Therefore, the uncertainty associated with the momentum of the electron is 5.03 × 10^(-24) kg*m/s.
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The gold foil experiment performed in Rutherford's lab ________. Group of answer choices proved the law of multiple proportions
The gold foil experiment performed in Rutherford's lab did not prove the law of multiple proportions.
The gold foil experiment, also known as the Rutherford scattering experiment, was conducted by Ernest Rutherford in 1911 to investigate the structure of the atom. In this experiment, alpha particles were directed at a thin gold foil, and their scattering patterns were observed.
The main conclusion drawn from the gold foil experiment was the discovery of the atomic nucleus. Rutherford observed that most of the alpha particles passed through the gold foil with minimal deflection, indicating that atoms are mostly empty space. However, a small fraction of alpha particles were deflected at large angles, suggesting the presence of a concentrated positive charge in the center of the atom, which he called the nucleus.
The law of multiple proportions, on the other hand, is a principle in chemistry that states that when two elements combine to form multiple compounds, the ratio of masses of one element that combine with a fixed mass of the other element can be expressed in small whole numbers. This law was formulated by John Dalton and is unrelated to Rutherford's gold foil experiment.
The gold foil experiment performed in Rutherford's lab did not prove the law of multiple proportions. Its main contribution was the discovery of the atomic nucleus and the proposal of a new atomic model, known as the Rutherford model or planetary model.
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