(C) 0.1 mole of NaCl dissolved in 1,000. mL of water
<u>Explanation:</u>
The conductivity of 0.1 mole of NaCl dissolved in 1000 mL of water will be greatest as the number of ions in 0.1 mole of NaCl will be more than 0.001, 0.05 and 0.005 moles of NaCl. Greater the number of ions in the solution, greater will be the conductivity. Specific Conductivity decreases with a decrease in concentration. Since the number of ions per unit volume that carry current in a solution decrease on dilution. Hence, concentration and conductivity are directly proportional to each other.
9 grams of hydrogen gas (H2) will SC Johnson need to react in order to make 1 bottle of Windex.
Explanation:
Balance equation for the formation of ammonia from H2 gas.
N2 + 3H2 ⇒ 2 
Given
mass of ammonia in 1 bottle of windex = 51 gram
atomic mass of ammonia 17.01 gram/mole
number of moles = 
number of moles = 
= 3 moles of ammonia is formed.
in 1 bottle of windex there are 3 moles of ammonia 0r 51 grams of ammonia.
From the equation it can be found that:
3 moles of hydrogen reacted to form 2 moles of ammonia
so, x moles of hydrogen will react to form 3 moles of ammonia.
= 
x = 4.5 moles of hydrogen will be required.
to convert moles into gram formula used:
mass = atomic mass x number of moles (atomic mass of H2 is 2grams/mole)
= 2 x 4.5
= 9 grams of hydrogen.
Answer:
Electronegativity in group 1 decreases as we go from Lithium to Francium.
Explanation:
Electronegativity is defined as the tendency of an element to attract an electron pair towards itself.
In a group generally this tendency decreases from top to bottom as the size of the atom increases and hence the positive nucleus get far from the outer orbital.
In the same way group 1 elements i.e. from Lithium to Francium electronegativity decreases.
Answer:
The molarity of the formed CaBr2 solution is 0.48 M
Explanation:
Step 1: Data given
Number of moles CaBr2 = 0.72 moles
Volume of water = 1.50 L
Step 2: Calculate the molarity of the solution
Molarity of CaBr2 solution = moles CaBr2 / volume water
Molarity of CaBr2 solution = 0.72 moles / 1.50 L
Molarity of CaBr2 solution = 0.48 mol / = 0.48 M
The molarity of the formed CaBr2 solution is 0.48 M
