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3 years ago

How many formulas units are in 3.11 mole of Ca(NO3)2 can you show your work by step by step correct answer

Chemistry

2 answers:

Lyrx [107]3 years ago

4 0

Answer:

18.73× 10²³ formula units

Explanation:

Given data:

Number of moles of Ca(NO₃)₂ = 3.11 mol

Number of formula units = ?

Solution:

Avogadro number:

"It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance"

For example,

18 g of water = 1 mole = 6.022 × 10²³ molecules of water

1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen

Number of formula units of Ca(NO₃)₂:

1 mole contain 6.022 × 10²³ formula units

3.11 mol × 6.022 × 10²³ formula units / 1 mol

18.73× 10²³ formula units

Fittoniya [83]3 years ago

4 0

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2 years ago

You have a 1.153 g sample of an unknown solid acid, HA, dissolved in enough water to make 20.00 mL of solution. HA reacts with K

seropon [69]

Answer:

HA + KOH → KA + H₂O

Explanation:

The unknown solid acid in water can release its proton as this:

HA + H₂O → H₃O⁺ + A⁻

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2 years ago

The freezing point of benzene C6H6 is 5.50°C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in benzene is DDT . H

Vlad1618 [11]

Answer: 0.028 grams

Explanation:

Depression in freezing point :

Formula used for lowering in freezing point is,

$\Delta T_f=k_f\times m$

or,

$\Delta T_f=k_f\times \frac{\text{ Mass of solute in g}\times 1000}{\text {Molar mass of solute}\times \text{ Mass of solvent in g}}$

where,

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$k_f$ = freezing point constant (for benzene} = $5.12^0Ckg/mol$

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Putting in the values we get:

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3 years ago

6. Cuando se oxidan en el aire 12,120 g de vapor de Zinc se obtienen 15,084 g del óxido. ¿Cuál es la fórmula empírica del óxido?

Eddi Din [679]

Answer:

12120 g + O2 = 15084 g

m Zn = 12.120 Kg

m óxido = 15.084 Kg

1. calcular la masa de cinc en gramos

g = 12,120 Kg x 1000 = 12120 g de cinc

g = 15.084 Kg x 100 = 15084 g de oxígeno

2. calcular gramos de Oxigeno

g O = 15084 g - 12120 g = 2964 g O2

3. calcular % de Zn y O

%m/m ( m soluto / m solc.) x 100

%m/m (Zn) = ( 1210 g / 15084 g ) x 100

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4. Calcular moles de cada elemento

Zn: 80.35 g / 65.38 g/mol = 1.228 mol

O: 19.65 g / 16 g/mol = 1.228 mol

5. dividir entre el menor de los elementos

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6. Fórmula empírica: ZnO

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3 years ago

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jekas [21]

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3 years ago