At equivalence there is no more HA and no more NaOH, for this particular reaction. So that means we have a beaker of NaA and H2O. The H2O contributes 1 x 10-7 M hydrogen ion and hydroxide ion. But NaA is completely soluble because group 1 ion compounds are always soluble. So NaA breaks apart in water and it just so happens to be in water. So now NaA is broken up. The Na+ doesn't change the pH but the A- does change the pH. Remember that the A anion is from a weak acid. That means it will easily attract a hydrogen ion if one is available. What do you know? The A anion is in a beaker of H+ ions! So the A- will attract H+ and become HA. When this happens, it leaves OH-, creating a basic solution, as shown below.
Answer:
The percent composition of a component in a compound is the percent of the total mass of the compound that is due to that component. To calculate the percent composition of a component in a compound: Find the molar mass of the compound by adding up the masses of each atom.
Answer:
I think so henterogeneous
Answer:
Number of moles = 2.8 mol
Explanation:
Given data:
Number of moles of water = ?
Volume of water = 50 mL
Density of water = 1.00 g/cm³
Solution:
1 cm³ = 1 mL
Density = mass/ volume
1.00 g/mL = mass/ 50 mL
Mass = 1.00 g/mL× 50 mL
Mass = 50 g
Number of moles of water:
Number of moles = mass/molar mass
Number of moles = 50 g / 18 g/mol
Number of moles = 2.8 mol
Answer:
Elements that fall between those on the left and right sides of the periodic table
Explanation:
Transition metals:
These are present at the center of periodic table.
These are d-block elements.
They include the elements of group 3 to 12 in periodic table.
They have large charge to radius ratio.
They mostly form paramagnetic compounds.
They shoes more than one oxidation state.
They form colored compounds.
They all have high melting and boiling point.
They have high densities.
They form stable complexes.
The elements of f-block are also transition but they are called inner transition.These are consist of two series lanthanide and actinides.