a. What is the formula weight of magnesium iodide? (in amu) b. How many magnesiu

Question

a. What is the formula weight of magnesium iodide? (in amu)
b. How many magnesium ions are present in 2.86 x 10-4 mol of magnesium iodide? Enter your answer in scientific notation.

c. How many iodide ions are present in 2.86 x 10-4 mol of magnesium iodide? Enter your answer in scientific notation

d. How many magnesium ions are present in 97 mg of magnesium iodide? Enter your answer in scientific notation.

e. How many iodide ions are present in 97 mg of magnesium iodide? Enter your answer in scientific notation

Explanation / Answer

MgI2 --> 278.1139 amu

b)

MgI2

since 1 to 1 ratio (1 MG in 1 MgI2) then

M = 2.86*10^-4 mol of I-

# atoms --> 6.022*10^23 * 2.86*10^-4 = 1.72*10^20

c)

in the same sample, there is twice I ions per MG, therefore

2X1.72*10^20 = 3.44*10^20

d)

fin amount of ions in 97 mg of MgI2

mol = mass/MW = 0.097 g /278 = 3.489*10^-4 mol of MgI2

now change to molecules of MgI2 --> 6.022*10^23 * 3.489*10^-4 = 2.10*10^20 molecules of Mg

e)

Once again , in that sample there is twice molecules of I- than Mg, therefore

2*2.10*10^20 = 4.2 *10^20 molecules of I-

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