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3 years ago

How many moles are in 3.4 x 10^-7 molecules of silicon dioxide, SiO2?

Chemistry

1 answer:

avanturin [10]3 years ago

7 0

Answer:

Approximately 5.646 * 10^-17 moles

Explanation:

Avagardo's number is approximately 6.022 * 10^23 molecules. Therefore, dividing 3.4 * 10^7 by avagadro's number yields approximately 5.646 * 10^-17 moles. Hope this helps

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Suppose you had a mixture of 1 moles of methanol and 1 mole of ethanol at a particular temperature. The vapor pressure of pure m

Natasha_Volkova [10]

The total pressure is given by:
P = ∑xiPi
P = 1/2 x 81 + 1/2 x 45
P = 63 kPa

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3 years ago

I need help with 1,2,3, and 4

Schach [20]

Answer:

Problem 1: 1.85atm

Problem 2: 110mL

Problem 3: 290 mL

Problem 4: 1.14 atm

Explanation:

Problem 1

1. Data

a) P₁ = 3.25atm

b) V₁ = 755mL

c) P₂ = ?

d) V₂ = 1325 mL

r) T = 65ºC

2. Formula

Since the temeperature is constant you can use Boyle's law for idial gases:

$PV=constant\\\\P_1V_1=P_2V_2$

3. Solution

Solve, substitute and compute:

$P_1V_1=P_2V_2\\\\P_2=P_1V_1/V_2$

$P_2=3.25atm\times755mL/1325mL=1.85atm$

Problem 2

1. Data

a) V₁ = 125 mL

b) P₁ = 548mmHg

c) P₁ = 625mmHg

d) V₂ = ?

2. Formula

You assume that the temperature does not change, and then can use Boyl'es law again.

$P_1V_1=P_2V_2$

3. Solution

This time, solve for V₂:

$P_1V_1=P_2V_2\\\\V_2=P_1V_1/P_2$

Substitute and compute:

$V_2=548mmHg\times 125mL/625mmHg=109.6mL$

You must round to 3 significant figures:

$V_2=110mL$

Problem 3

1. Data

a) V₁ = 285mL

b) T₁ = 25ºC

c) V₂ = ?

d) T₂ = 35ºC

2. Formula

At constant pressure, Charle's law states that volume and temperature are inversely related:

$V/T=constant\\\\\\\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

The temperatures must be in absolute scale.

3. Solution

a) Convert the temperatures to kelvins:

T₁ = 25 + 273.15K = 298.15K

T₂ = 35 + 273.15K = 308.15K

b) Substitute in the formula, solve for V₂, and compute:

$\dfrac{V_1}{T_1`}=\dfrac{V_2}{T_2}\\\\\\\\\dfrac{285mL}{298.15K}=\dfrac{V_2}{308.15K}\\\\\\V_2=308.15K\times285mL/298.15K=294.6ml$

You must round to two significant figures: 290 ml

Problem 4

1. Data

a) P = 865mmHg

b) Convert to atm

2. Formula

You must use a conversion factor.

1 atm = 760 mmHg

Divide both sides by 760 mmHg

$\dfrac{1atm}{760mmHg}=\dfrac{760mmHg}{760mmHg}\\\\\\1=\dfrac{1atm}{760mmHg}$

3. Solution

Multiply 865 mmHg by the conversion factor:

$865mmHg\times \dfrac{1atm}{760mmHg}=1.14atm\leftarrow answer$

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3 years ago

True or false: atoms are electrically neutral; that is they do not have a charge.

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Its true, By definition, an atom is electrically neutral it has the same number of protons as it does electrons, plus some number of neutrons depending on the isotope

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3 years ago

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I think that the answers is true

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2 years ago

A pure copper penny contains approximately 2.9×1022 copper atoms. Use the following definitions to determine how many ______ of

hammer [34]

Complete question is;

A pure copper penny contains approximately 2.9 × 10^(22) copper atoms.

1 doz = 12

1 gross = 144

1 ream = 500

1 mol = 6.022 × 10^(23)

Use these definitions to determine the following:

A) How many dozens of copper atoms are in a penny.

B) How many gross of copper atoms are in a penny

C) How many reams of copper atoms are in a penny.

D) how many moles of copper atoms are in a penny?

All answers can be rounded to two significant figures

Answer:

A) 2.4 × 10^(21) dozens

B) 2.01 × 10^(20) gross

C) 5.8 × 10^(19) reams

D) 0.048 mol

Explanation:

A) A dozen contains 12.

Therefore, 2.9 × 10^(22) copper atoms will contain;

(2.9 × 10^(22))/12 dozens = 2.42 × 10^(21).

In 2 significant figures, we have;

2.4 × 10^(21) dozens

B) 1 gross = 144

Therefore, 2.9 × 10^(22) copper atoms will contain;

(2.9 × 10^(22))/144 gross ≈ 2.01 × 10^(20) gross

C) 1 ream = 500

Therefore, 2.9 × 10^(22) copper atoms will contain;

(2.9 × 10^(22))/500 reams = 5.8 × 10^(19) reams

D) 1 mol = 6.022 × 10^(23)

Therefore, 2.9 × 10^(22) copper atoms will contain;

(2.9 × 10^(22))/(6.022 × 10^(23)) = 0.048 mol

6 0

3 years ago

How many moles are in 3.4 x 10^-7 molecules of silicon dioxide, SiO2?​

How many moles are in 3.4 x 10^-7 molecules of silicon dioxide, SiO2?