Chemistry, 08.02.2021 04:50 Kykebailey2356

How many moles is 287.7 grams of Iridium (Ir)?

Answers: 3

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Answers

Answer from: ALEXMASTERS64

1.497 mol Ir

General Formulas and Concepts:

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to Right

Chemistry

Atomic Structure

Reading a Periodic Table

Stoichiometry

Using Dimensional AnalysisExplanation:

Step 1: Define

287.7 g Ir

Step 2: Identify Conversions

[PT] Molar Mass of Ir - 192.22 g/mol

Step 3: Convert

Set up: $\displaystyle 287.7 \ g \ Ir(\frac{1 \ mol \ Ir}{192.22 \ g \ Ir})$ Multiply/Divide: $\displaystyle 1.49672 \ mol \ Ir$

Step 4: Check

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

1.49672 mol Ir ≈ 1.497 mol Ir

Answer from: erikloza12pdidtx

$\boxed {\boxed {\sf About \ 1.497 \ mol \ Ir }}$

Explanation:

To convert from grams to moles, we must the molar mass of iridium, which can be found on the Periodic Table.

Ir: 192.22 g/mol

Use the molar mass as a ratio.

$\frac{ 192.22 \ g \ Ir}{ 1 \ mol \ Ir}$

Multiply by the given number of grams: 287.7

$287.7 \ g Ir *\frac{ 192.22 \ g \ Ir}{ 1 \ mol \ Ir}$

Flip the fraction so the grams of Iridium will cancel.

$287.7 \ g Ir *\frac{1 \ mol \ Ir }{ 192.22 \ g \ Ir}$

$287.7 *\frac{1 \ mol \ Ir }{ 192.22 }$

$\frac {287.7 \ mol \ Ir }{192.22} \\$

Divide.

$1.49672251 \ mol \ Ir$

The original measurement has 4 significant figures, so we must round our answer to 4 sig figs as well. For the answer we calculated, that is the thousandth place.

The 7 in the ten thousandth place tells us to round the 6 up to a 7.

$1.497 \ mol \ Ir$

There are about 1.497 moles of iridium in 287.7 grams.

Answer from: Quest

Yes it would be a physical change.

Answer from: Quest

(newtons first law)

conservation of momentum law states :

momentum before collision = momentum after collision

momentum p (kg-m/s)= mass * velocity

say moving car=1 000 kg, velocity = 10 m/s then p = 1 000*10= 10 000 kg-m/s

say still car = 2 000 kg, velocity = 0 m/s then p = 2 000* 0 = 0 kg-m/s

total momentum prior to collision = 10 000 + 0 = 10 000 kg-m/s

momentum after collision = 10 000 kg-m/s

mass now = 1 000 kg+2 000 kg = 3 000 kg

10 000 = 3 000 kg* velocity m/s

10 000/3 000 = velocity

3.33 m/s = velocity after collision

an impulse is a force applied over time , momentum is exclusive of external forces and a perfect collision is implied

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