Chemistry, 05.01.2021 05:40, axonify

How many moles are in 1.806 x 1024 molecules of bromine?

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answered: ineedhelpwhomework

2.999 mol Br

General Formulas and Concepts:

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to Right

Chemistry

Atomic Structure

Using Dimensional AnalysisAvogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.Explanation:

Step 1: Define

1.806 × 10²⁴ molecules Br

Step 2: Identify Conversions

Avogadro's Number

Step 3: Convert

$\displaystyle 1.806 \cdot 10^{24} \ molecules \ Br(\frac{1 \ mol \ Br}{6.022 \cdot 10^{23} \ molecules \ Br} )$ = 2.999 mol Br

Step 4: Check

We are given 4 sig figs. Follow sig fig rules and round.

Our final answer is already in 4 sig figs, so there is no need to round.

answered: chrisraptorofficial

$\boxed {\boxed {\sf About \ 2.999 \ moles \ of \ bromine}}$

Explanation:

To convert from moles to molecules, we must use Avogadro's Number.

$6.022*10^{23}$

This tells us the number of particles (atoms, molecules, ions, etc) in 1 mole. In this problem, the particles are molecules of bromine in 1 mole of bromine.

$6.022*10^{23} \ molecules \ Br / 1 \ mol \ Br$

1. Convert from moles to molecules

Use Avogadro's number as a fraction or ratio.

$\frac{6.022 *10^{23} \ molecules \ Br}{1 \ mol \ Br}$

Multiply this fraction by the given number of bromine molecules.

$1.806 *10^{24} molecules \ Br *\frac{6.022 *10^{23} \ molecules \ Br}{1 \ mol \ Br}$

Flip the fraction so the molecules of bromine can cancel out.

$1.806 *10^{24} \ molecules \ Br* \frac{1 \ mol \ Br}{6.022 *10^{23} \ molecules \ Br}$

$1.806 *10^{24} * \frac{1 \ mol \ Br}{6.022 *10^{23}}$

Multiply and condense the expression into 1 fraction.

$\frac{1.806 *10^{24} \ mol \ Br}{6.022 *10^{23} }$

Divide.

$2.999003653 \ mol \ Br$

2. Round

The original measurement, 1.806*10^24 has 4 significant figures (1, 8, 0, and 6). We must round our answer to 4 sig figs. For the answer we found, that is the thousandth place.

$2.999003653 \ mol \ Br$

The 0 in the ten-thousandth place tels us to leave the 9 in the thousandth place.

$\approx 2.999 \ mol \ Br$

There are about 2.999 moles of bromine in 2.806 *10^23 molecules.

answered: Guest

The correct answer is b

answered: Guest

The activity series is shown in the attached picture. if the element is below the other, then you can displace the other. following the activity series as a guide, the reactions are as follows: nickel and aluminum chloride3 ni + 2 alcl₃ → 3 nicl₂ + 2 albarium and copper bromideba + cubr → no reaction because cu is less oxidizable than bapotassium nitrate and aluminum3 kno₃ + al → al(no₃)₃ + 3 ksodium phosphate and zincnaso₄ + zn → znso₄ + na

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