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Crazy boy [7]
3 years ago
11

What values of c and d would make the following expression represent a real number? i(2 3i)(c di) c = 2, d = 3 c = –2, d = 3 c =

3, d = –2 c = –3, d = –2
SAT
1 answer:
Olegator [25]3 years ago
8 0

Complete question:

<em>What values of c and d would make the following expression represent a real number? i(2 + 3i)(c + di) </em>

<em>choices: A)c = 2, d = 3 B)c = –2, d = 3 C)c = 3, d = –2 D)c = –3, d = –2</em>

<em />

<em>Option D</em><em> is correct. </em> The value of c and d for the expression to be real are -3 and -2 respectively

Given the expression

i(2 + 3i)(c + di)

Expand the expression

=i(2 + 3i)(c + di)\\=(2i+3i^2)(c+di)\\=(2i+3(-1))(c+di)\\=(2i-3)(c+di)\\=2ic+2di^2-3c-3di\\=2ic-2d-3c-3di

Collect the like terms

=2ic-3di-2d-3c

For the resulting function to be real, then the imaginary part must be zero i.e.

2ic-3di=0\\2ci = 3di\\2c = 3d\\c=\frac{3d}{2}\\

If d = -2, then;

c=\frac{3(-2)}{2}\\c=\frac{-6}{2}\\c= -3

Hence the value of c and d for the expression to be real are -3 and -2 respectively.

Learn more here: brainly.com/question/4244460

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How many grams of lead product would theoretically be produced between 17. 0 g potassium iodide, ki, and 25. 0 g of lead (ii) ni
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The number of grams of lead product that would theoretically be produced from the given reaction is; <u><em>Mass of lead product = 23.6 g</em></u>

Potassium iodide reacting with Lead(II) Nitrate would yield the balanced equation;

2KI + Pb(NO₃)₂ = 2KNO₃ + PbI₂

2moles : 1 mole = 2 moles : 1 mole

From online tables;

Molar mass of KI = 166 g/mol

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We are given;

Mass of KI = 17 g

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Thus;

Number of moles of PbI₂ yielded in order to find the limiting reactant will be;

no of moles of PbI₂ from KI = (17g/166 g/mol) × (1 mole of PbI₂/2 mol of KI)

no of moles of PbI₂ from KI = 0.0512 mols

Similarly;

no of moles of PbI₂ from Pb(NO₃)₂ = (25g/331.2 g/mol) × (1 mole of PbI₂/1 mol of Pb(NO₃)₂)

no of moles of PbI₂ from Pb(NO₃)₂ = 0.0755 moles

We can see that KI produced the least amount of moles. Thus;

Mass = 0.0512 * molar mass of  PbI₂

From tables, molar mass of  PbI₂ = 461 g/mol

Thus;

Mass of lead product = 461 * 0.0512

Mass of lead product = 23.6 g

Read more about theoretical yield at; brainly.com/question/21091465

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