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3 years ago

NEED HELP ASAP!!! really confused. can someone explain how to do it? (Stoichiometry)

Chemistry

1 answer:

Katena32 [7]3 years ago

8 0

Answer:

The answer to your question is below

Explanation:

Data

mass of Gold = 35.12 g

mass number of gold = 197 g (from the periodic table)

1.- To calculate the number of moles use proportions

197 g of gold ----------- 1 mol

35.12 g of gold -------- x

x = (35.12 x 1)/197

<u>x = 0.18 moles of gold</u>

2.- To calculate the number of atoms also use proportions and the Avogadro's number.

1 mol of gold ------------ 6.023 x 10 ²³ atoms

0.18 moles of gold ----- x

x = (0.18 x 6.023 x 10²³) / 1

<u> x = 1.084 x 10²³ atoms</u>

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A substance has a volume of 20mL and a density of 2.5 g/mL. what is it’s mass?

DochEvi [55]

2.5X20=50g
50g should be the right answer
Mass=volumeXdenisty.

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3 years ago

H₃C-Ç=CCH₃<br>сна сHз (what is the name of this <br> chemical formula)

algol [13]

Answer:2,3 -dimethyl-2-butenr

Explanation:

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3 years ago

Choose the aqueous solution that has the highest boiling point. These are all solutions of nonvolatile solutes and you should as

UNO [17]

Answer: 0.100 m $K_2SO_4$

Explanation:

Elevation in boiling point is given by:

$\Delta T_b=i\times K_b\times m$

$\Delta T_b=T_b-T_b^0$ = Elevation in boiling point

i= vant hoff factor

$K_b$ = boiling point constant

m= molality

1. For 0.100 m $K_2SO_{4}$

$K_2SO_4\rightarrow 2K^{+}+SO_4^{2-}$

, i= 3 as it is a electrolyte and dissociate to give 3 ions. and concentration of ions will be $3\times 0.100=0.300$

2. For 0.100 m $LiNO_3$

$LiNO_3\rightarrow Li^{+}+NO_3^{-}$

, i= 2 as it is a electrolyte and dissociate to give 2 ions, concentration of ions will be $2\times 0.100=0.200$

3. For 0.200 m $C_2H_8O_3$

, i= 1 as it is a non electrolyte and does not dissociate, concentration of ions will be $1\times 0.200=0.200$

4. For 0.060 m $Na_3PO_4$

$Na_3PO_4\rightarrow 3Na^{+}+PO_4^{3-}$

, i= 4 as it is a electrolyte and dissociate to give 4 ions. and concentration of ions will be $4\times 0.060=0.24m$

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2 years ago

Determine the mass (in grams) of 8.02 x 1024 atoms of mercury (Hg).

Virty [35]

<h3>Answer:</h3>

2670 g Hg

<h3>General Formulas and Concepts: </h3>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation: </h3>

<u>Step 1: Define</u>

8.02 × 10²⁴ atoms Hg

<u>Step 2: Identify Conversions</u>

Avogadro's Number

Molar Mass of Hg - 200.59 g/mol

<u>Step 3: Convert</u>

Set up: $\displaystyle 8.02 \cdot 10^{24} \ atoms \ Hg(\frac{1 \ mol \ Hg}{6.022 \cdot 10^{23} \ atoms \ Hg})(\frac{200.59 \ g \ Hg}{1 \ mol \ Hg})$
Divide/Multiply: $\displaystyle 2671.42 \ g \ Hg$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

2671.42 g Hg ≈ 2670 g Hg

3 0

3 years ago

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Leni [432]

Answer : The molar mass of solute is, 89.9 g/mol

Explanation : Given,

Mass of solute = 5.8 g

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Formula used :

$\Delta T_f=K_f\times m\\\\T_f^o-T_f=T_f\times\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of water}}$

where,

$\Delta T_f$ = change in freezing point

$T_f^o$ = temperature of pure solvent (water) = $0^oC$

$T_f$ = temperature of solution = $1.20^oC$

$K_f$ = freezing point constant of water = $1.86^oC/m$

m = molality

Now put all the given values in this formula, we get

$(0^oC)-(1.20^oC)=1.86^oC/m\times \frac{5.8g\times 1000}{\text{Molar mass of solute}\times 100g}$

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3 years ago