Answer:
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Explanation:
Since sulfuric acid, H₂SO₄, is a diprotic acid and potassum hydroxide, KOH, contains one OH⁻ in the formula, the number of moles of potassium hydroxide must be twice the number of moles of sulfuric acid.
<u>1. Determine the number of moles of KOH in 47mL of 0.39M potassium hydroxide solution</u>
- number of moles = molarity × volume in liters
- number of moles = 0.39M × 47mL × 1liter/1,000 mL = 0.1833mol
<u>2. Determine the number of moles of sulfuric acid needed</u>
- number of moles of H₂SO₄ = number of moles of KOH/2 = 0.1833/2 = 0.009165mol
<u>3. Determine the concentration that contains 0.009165 mol in 25mL of the acid.</u>
- Molarity = number of moles / volume in liters
- M = 0.009165mol/(25mL) × (1,000mL/liter) = 0.3666M
Round to two significant figures: 0.37M
Its called the articular cartilage where <span>white tissue covers the ends of bones where they come together. It makes </span><span>our joints makes it easier to move and allows the bones to glide over each other not causing a lot of friction. </span>
The ionic compound of FePO4 is the iron III phosphate. Or you can call it ferric phosphate or ferric orthophosphate.
Just for general informations, It's normally used in steel and metal manufacturing processes, in organic farming, and many other uses.
Hope this Helps :)
The answer is atomic radii; the size or radii of an atom increases from left to right, versus the ionization energies and electronegativities of atoms which increase from right to left.
The question incomplete , the complete question is:
A student dissolves of 18.0 g urea in 200.0 mL of a solvent with a density of 0.95 g/mL . The student notices that the volume of the solvent does not change when the urea dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to significant digits.
Answer:
The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.
Explanation:
Moles of urea = 
Volume of the solution = 200.0 mL = 0.2 L (1 mL = 0.001 L)
Molarity of the urea solution ;
Mass of solvent = m
Volume of solvent = V = 200.0 mL
Density of the urea = d = 0.95 g/mL
(1 g = 0.001 kg)
Molality of the urea solution ;
The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.
