0.125 Molar solution!
Molarity = moles of solute/Liters of solution
Algebraically manipulated,
Moles of copper sulfate = 2.50 Liters * 0.125 M
= 0.313 moles copper sulfate needed
===========================
Mortar and pestle
885x140=123900grams of copper sulphate per hour. If you are talking about using 885 gm of copper sulphate per ton of ore in the solution then the density of the copper sulphate(penta hydrate now because it's in water) is 2.284 gm per cm3 so that's 2.284x5=11.42gm per liter of solution, so 123900/11.42=10849.387 liters of copper sulphate(pentahydrate) per hour, NOTE:this is only how much copper sulphate is being used total in the solution which is 30% of the total liters used of solution because 25% of the water is inside the copper sulphate, the other 70% is just water. If you want the liters per hour of solution total, it is 34964.62 litres per hour of your 5% solution. I hope this was what you were looking for, I saw noone had answered and decided to try and get you what you needed.
HOW LARGE IS THE CONCRETE POND?
5.50
22
Depending on the desired concentration of the solution !
If you needed to use powered copper 2 sulfate in an experiment and were only supplied with clumps of copper 2 sulfate which tool would you need to convert the copper sulfate to usable form?
i donβt know the answer
mortar and pestle
mortar and pestle
Mortar and pestle
The process of plating copper is relatively simple with the right materials and know-how. This tutorial will cover a basic copper plating procedure which is used on a large industrial scale to achieve high-purity copper.There are several materials needed for this process. The first and most obvious is the impure copper to be refined. Also needed will be copper II sulfate, water, a DC power source, and some lengths of wire.The first step is to make a saturated solution of copper II sulfate, which has a maximum solubility of 316g per liter of water. The final solution should be dark blue.Two copper electrodes are then introduced into the copper sulfate bath. One of these electrodes consists of the impure copper sample and the other is simply a copper wire.The DC power source is applied to the two electrodes. The positive side (anode) of the supply is connected to the impure copper block while the negative side (cathode) is attached to the copper wire.Copper has only positive oxidation states and is therefore attracted to the negative electrode. Pure copper is pulled from the solution of copper sulfate and attaches itself to the wire electrode, breaking the sulfate anion off and creating a weak solution of sulfuric acid. The acid then eats at the impure copper, turning it into more copper sulfate. Effectively, the copper from the impure anode is transferred to the cathode, leaving all its impurities behind. The copper plated onto the cathode will be very pure.Cell current should be monitored carefully as high current causes hydrogen bubbles to form on the cathode and will inhibit the growth of the pure copper. If the power supply does not have a current limiting adjustment, the spacing between the anode and cathode can be varied to regulate the current. The closer the electrodes, the higher the current flow and vice versa.Never dispose of the used copper sulfate down the drain as it is an environmental hazard. Simply evaporate the solution to recover the copper sulfate for later use.
885x140=123900grams of copper sulphate per hour. If you are talking about using 885 gm of copper sulphate per ton of ore in the solution then the density of the copper sulphate(penta hydrate now because it's in water) is 2.284 gm per cm3 so that's 2.284x5=11.42gm per liter of solution, so 123900/11.42=10849.387 liters of copper sulphate(pentahydrate) per hour, NOTE:this is only how much copper sulphate is being used total in the solution which is 30% of the total liters used of solution because 25% of the water is inside the copper sulphate, the other 70% is just water. If you want the liters per hour of solution total, it is 34964.62 litres per hour of your 5% solution. I hope this was what you were looking for, I saw noone had answered and decided to try and get you what you needed.
4.84
Molarity = moles of solute/liters of solution or, for our purposes moles of solute = liters of solution * Molarity moles of AgNO3 = 0,50 liters * 4.0 M = 2.0 moles of AgNO3 needed --------------------------------------
HOW LARGE IS THE CONCRETE POND?
5.50