9.13
Molarity = moles of solute/Liters of solution 0.75 M KCl = moles KCl/2.25 Liters = 1.6875 moles KCl (74.55 grams/1 mole KCl) = 126 grams of KCl needed
Learn spelling first, then come back... KCl Degrees Celcius
1. 4.8 g of KCL 2. 2.1 g of KCL
8g
In a 3.4 M solution, there are 3.4 moles per liter. If you want to make 3 liters of solution, you'll need 3 liter * 3.4 moles/liter = 10.2 moles The molar mass of KCl is 39.098 g/mole K + 35.453 g/mole Cl = 74.551 g/mole KCl To get the number of grams, multiply the number of moles by the molar mass: 10.2 moles * 74.551 g/mole KCl = 760.4202 g = 0.760 kg
Molarity = moles of solute/Liters of solution 0.75 M KCl = moles KCl/2.25 Liters = 1.6875 moles KCl (74.55 grams/1 mole KCl) = 126 grams of KCl needed
12.57 grams KCl (1mol KCl/74.55g ) = 0.1686 moles
148g
The answer is 6,71 g dried KCl.
279.56
Learn spelling first, then come back... KCl Degrees Celcius
1. 4.8 g of KCL 2. 2.1 g of KCL
For this you need the atomic (molecular) mass of KCl. Take the number of grams and divide it by the atomic mass. Multiply by one mole for units to cancel. KCl= 74.6 grams50.0 grams KCl / (74.6 grams) = .670 moles KCl
can someone hurry up and answer this
2 KClO3 -> KCL + 3O2 Molar weight of O2 = 32 grams/mole (so close it doesn't matter) 30 grams/32grams/mole = 0.9375 moles Molar weight of KCL = 39+35.5 = 74.5 grams/mole (Want more accuracy? Do it yourself?) now if we have 3 moles of O2 then we have 2 moles of KCl. If we have one mole of O2 then we have 2/3 moles of KCL What ever moles we have of O2 we must multiply it by 2/3 to get the moles of KCl So we have 0.9375moles of O2 x 2/3 = 0.625 moles of KCl So 0.625 moles of KCl x 74.5 grams/mole KCl = 46.5625 grams KCl
8g
You have2KClO3 ==> 2KCl + 3O2 as the balanced equation 25 g KClO3 x 1 mole/123 g = 0.20 moles moles KCl formed = 0.20 moles KClO3 x 2 moles KCl/2 moles KClO3 = 0.20 moles KCl formed grams KCl = 0.20 moles x 74.5 g/mole = 14.9 g = 15 grams of KCl formed