Because sodium ions have only one positive electric charge units, but sulfur ions have two negative electric charges each.
An ionic bond is formed between sodium and sulfur. Sodium has one electron to lose, while sulfur has six electrons to gain, resulting in the transfer of one electron from sodium to sulfur to form sodium ions and sulfur ions that are attracted to each other.
You would need two potassium ions to balance the charge of one sulfide ion. Potassium has a charge of +1, while sulfide has a charge of -2, so two potassium ions with a total charge of +2 would balance the charge of one sulfide ion with a charge of -2.
To find the moles of sulfate ions in the solution, you first need to determine the moles of aluminum sulfide present. Since you are not provided with the amount of aluminum sulfide, you cannot calculate the moles of sulfate ions. Additionally, oxygen is not relevant to determining the moles of sulfate ions.
There are at least three kinds of sodium sulfide, but assuming that the question refers to the most common one with the formula Na2S, its gram formula mass, the mass corresponding to molar mass for covalently bonded compounds, is 78.04. Therefore 125.00 constitutes 125.00/78.04 or 1.6017 "moles".
The equation for the reaction specified is 2 NaOH + H2S -> Na2S + H2O. Therefore, if the yield were 100 %, two formula masses of sodium hydroxide are required to produce one formula mass of sodium sulfide. The gram formula mass of NaOH is 40.00 and that of sodium sulfide is 78.04. The specified number of grams of sodium hydroxide corresponds to 2.53/40.00 or 0.06325 formula masses and therefore would provide half this many formula masses of sodium sulfide, for a mass of (0.06325)(78.04)/2.000 or 2.568 grams of sodium sulfide. Since the yield is specified as 91.0 %, the actual amount of sodium sulfide produced is 2.25 grams, to the justified number of significant digits.
An ionic bond is formed between sodium and sulfur. Sodium has one electron to lose, while sulfur has six electrons to gain, resulting in the transfer of one electron from sodium to sulfur to form sodium ions and sulfur ions that are attracted to each other.
NaS2 is an unbalanced equation. It would need to be Na2S to be a balanced equation (two sodium, one sulfide). Na2S is Sodium Sulfide.
To determine the number of moles in 73.5 g of sodium sulfide (Na2S), you first need to calculate the molar mass of Na2S, which is 78.04 g/mol. Then divide the given mass by the molar mass to find the number of moles. Therefore, 73.5 g / 78.04 g/mol = 0.942 moles of sodium sulfide.
You would need two potassium ions to balance the charge of one sulfide ion. Potassium has a charge of +1, while sulfide has a charge of -2, so two potassium ions with a total charge of +2 would balance the charge of one sulfide ion with a charge of -2.
Sodium sulfate's formula is Na2SO4 because it contains two sodium ions (Na+) for every one sulfate ion (SO4 2-). This combination of ions results in a neutral compound where the total positive charge from the sodium ions balances the total negative charge from the sulfate ion.
I think you're looking for three ... over the long run. But the trick is that K+ doesn't need to be pumped in. Membrane proteins act as variable sized pore in the membrane (channels) and the potassium flows in under electrostatic forces ... all the work is done pumping the Na+ out.
To find the moles of sulfate ions in the solution, you first need to determine the moles of aluminum sulfide present. Since you are not provided with the amount of aluminum sulfide, you cannot calculate the moles of sulfate ions. Additionally, oxygen is not relevant to determining the moles of sulfate ions.
The balanced reaction should look like this: 2Ar(NO3)3 + 3Na2S Ar2S3 + 6Na(NO3) The molar weight of arsenic sulfide is 176.091 g/mol. 1 gram of arsenic sulfide therefore constitutes 0.006 mol. The stoichiometry of the above chemical reaction indicates that 2 moles of arsenic nitrate and 3 moles of sodium sulfide are required to make 1 mole of arsenic sulfide and 6 moles of sodium nitrate. These numbers are the ones that appear in front of the chemicals and are called the stoichiometric coefficients. You therefore need 0.012 mol of arsenic nitrate and 0.018 mol of sodium sulfide. The molar weights of these compounds are 132.959 g/mol and 78.043 g/mol respectively. Multiplication then gives you the answer which is 1.596 g of arsenic nitrate, and 1.405 g of sodium sulfide.
There are at least three kinds of sodium sulfide, but assuming that the question refers to the most common one with the formula Na2S, its gram formula mass, the mass corresponding to molar mass for covalently bonded compounds, is 78.04. Therefore 125.00 constitutes 125.00/78.04 or 1.6017 "moles".
The equation for the reaction specified is 2 NaOH + H2S -> Na2S + H2O. Therefore, if the yield were 100 %, two formula masses of sodium hydroxide are required to produce one formula mass of sodium sulfide. The gram formula mass of NaOH is 40.00 and that of sodium sulfide is 78.04. The specified number of grams of sodium hydroxide corresponds to 2.53/40.00 or 0.06325 formula masses and therefore would provide half this many formula masses of sodium sulfide, for a mass of (0.06325)(78.04)/2.000 or 2.568 grams of sodium sulfide. Since the yield is specified as 91.0 %, the actual amount of sodium sulfide produced is 2.25 grams, to the justified number of significant digits.
Sodium (Na) plus sulfur (S) creates sodium sulfide (Na2S). The reason is that Na gives up it's valence electron and S takes it. It takes 2 sodiums because S need TWO electrons to complete the octet. This forms the ionic compound sodium sulfide.
To determine the ratio of ions in a compound, you first need to identify the chemical formula of the compound. The subscripts in the chemical formula indicate the ratio of ions in the compound. For example, in NaCl (sodium chloride), the ratio of sodium ions to chloride ions is 1:1.