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A filter? Lead (II) chloride isn't very soluble (and lead (IV) chloride isn't very stable, tending to decompose into lead (II) chloride and chlorine gas), and you could use HCl to raise the chloride concentration (and therefore lower the lead concentration) even further.
1750 C is the higher temperature. To convert C to Kelvin add 273.16 so 1750 0C= 1750 + 273.16 = 2023.16 K which is higher than 1860 K. However, the question is wrong as the melting point for lead is 327.5 0C .
Cells can be placed in solutions with higher, lower, or equal concentration to the cell... 1. ISOTONIC: - a solution with equal concentration to the cell. - 0.9% NaCl solutions is isotonic to RBC (red blood cells). - isotonic solutions cause no net gain or loss of water to a cell. 2. HYPOTONIC: - solute concentration is greater on the inside of the cell (or: the outer solution has less concentration than inside). - >0.9% NaCl solutions is hypotonic to RBC (red blood cells). - causes swelling, could burst (lyse) - net gain of water 3. HYPERTONIC: - <0.9% NaCl solutions is hypertonic to RBC (red blood cells). - net loss of water from the cell. - solute concentration is greater on the outside of the cell (or: the outer solution is greater concentration than the inside). - causes the cell shrink (crenation in RBC)
sulfuric acid decreases
PbS is lead (II) sulfide, which contains lead and sulfur.
Lead will form Lead(II) Chloride in a HCl solution, but if exists a higher chloride ion concentration, it would form a soluble complex and a colourless solution.
Solution that has less concentration than its surroundings and which would lead the solution from the surrounding to enter the body to compensate for the low concentration
This only depends on how much water is used to make this solution:0.4 mol Pb2+ ions in 1.0 Litre has a (molar) concentration value of 0.4 mol/L or 0.4 M Pb2+
The formula of lead (II) nitrate is Pb(NO3)2. This shows that in any solution of lead nitrate only, the molar concentration of nitrate ions will be twice as much as the molar concentration of lead (II) nitrate. Molar concentration is defined as number of moles per liter of solution, and 800 mL is the same as 0.800 liters. Therefore the molar concentration of nitrate ions in the specified solution will be 2(0.027823/0.800) or 6.96 X 10-2 . Only three significant digits are justified because that is the number of digits in 800.
A filter? Lead (II) chloride isn't very soluble (and lead (IV) chloride isn't very stable, tending to decompose into lead (II) chloride and chlorine gas), and you could use HCl to raise the chloride concentration (and therefore lower the lead concentration) even further.
A car battery is a rechargeable, or secondary, wet cell battery that contains lead, lead oxide, plates and an electrolyte solution that contains a mixture of water and acid.
The concentration of a solution can be expressed in many ways. One of them is as the molarity of the solution. A solution with molarity equal to one has one mole of the solute dissolved in every liter of the solutions
Hilter did not dicide what to do to the Jews so he took them to GETTO where then they where send to the Final place concentration camps
The ionic equation would usually be written as: Pb+2 (aq) + SO4-2 (aq) -> PbSO4 (s)
1750 C is the higher temperature. To convert C to Kelvin add 273.16 so 1750 0C= 1750 + 273.16 = 2023.16 K which is higher than 1860 K. However, the question is wrong as the melting point for lead is 327.5 0C .
(This question is answered below by assuming that the lower-case "m" in the first line of the question was intended to be a capital "M" as in the second line. If the "m" was intended to mean "molality" rather than molarity, the question can not be answered without additional information, specifically the density of the Pb(NO3) solution at the temperature at which the volume specified was measured.) The formula for a "mole" [more strictly, "formula unit"] of the lead nitrate solution given in the question shows that each unit contains one lead cation. The KI solution is stated to be in excess, so that every lead cation supplied to the reaction will be precipitated. A 2.0 molar solution by definition contains 2.0 moles of lead cations in each liter of solution. The solution is homogeneous. therefore, 500ml of it will contain half as many lead cations, in this instance 1.0. Therefore, 1.0 moles of lead (II) iodide will be precipitated.
Lead sulphide (lead sulfide) contains only two elements, lead and sulfur. Lead sulfate contains lead, sulfur and oxygen.