To answer the problem above first we need to find the difference of molar mass of NI3 from I2, 394.71 g/mol - 253.80 g/mol = 140.91 g/mol. Knowing the molar mass of the difference of NI3 from I2, in equation mass (g) / moles (mol) = molar mass, then we substitute. 3.58g / moles = 140.91 g/mol.
moles = 3.58 / 140.91 = 0.025 moles.
Ionic Bond, which is the transfer of electrons of a metal to a non-metal.
HNO3+NaOH ----> H2O
H⁺ +NO3⁻+Na⁺+OH⁻ ---> Na⁺ +NO3⁻ +H2O
H⁺ (aq)+OH⁻(aq)----> H2O(l)
Answer:
6.9 ml of concentrate
Explanation:
100 ml of .1 M will require .01 moles
from a 1.45 M solution, .01 mole would be
.01 mole / ( 1.45 mole / liter) = 6.9 ml of the concentrate then dilute to 100 ml
