The empirical formula of the compound obtained from the question given is NaBrO₃
<h3>Data obtained from the question </h3>
- Sodium (Na) = 15.24%
- Bromine (Br) = 52.95%
- Oxygen (O) = 31.81%
<h3>How to determine the empirical formula </h3>
The empirical formula of the compound can be obtained as illustrated below:
Divide by their molar mass
Na = 15.24 / 22.99 = 0.663
Br = 52.95 / 79.90 = 0.663
O = 31.81 / 16 = 1.988
Divide by the smallest
Na = 0.663 / 0.663 = 1
Br = 0.663 / 0.663 = 1
O = 1.988 / 0.663 = 3
Thus, the empirical formula of the compound is NaBrO₃
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In a particular experiment, the per cent yield is 79.0%. This means that in this experiment, a 7.90-g sample of fluorine yields is 7g of SF6.
<h3>How is Sulphur hexafluoride formed?</h3>
Sulfur Hexafluoride is a disparity agent formed of an inorganic fluorinated inert gas comprised of six fluoride atoms bound to one sulfur atom, with possible diagnostic activity upon imaging.
Thus, a sample of fluorine yields 7g of SF6.
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Answer:
14 mol e⁻
Explanation:
Step 1: Write the balanced half-reaction for the reduction of permanganate to manganese
8 H⁺(aq) + 7 e⁻ + MnO₄⁻(aq) ⇒ Mn(s) + 4 H₂O(l)
Step 2: Calculate the moles corresponding to 110 g of manganese
The molar mass of Mn is 55 g/mol.
110 g × 1 mol/55 g = 2 mol
Step 3: Calculate the number of moles of electrons needed to produce 2 moles of Mn
According to the half-reaction, 7 moles of electrons are required to produce 1 mole of Mn.
2 mol Mn × 7 mol e⁻/1 mol Mn = 14 mol e⁻
The N2H4 bond angle will be probably 109 degrees. Since, well,<span> it has a bent </span>trigonal pyramidal<span> geometry.</span>