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2 years ago

For the reaction

Chemistry

1 answer:

iVinArrow [24]2 years ago

5 0

Answer:

16 mol NaCl.

Explanation:

Do the train track method to cancel out all the units except moles of NaCl on top. Remember one mole of any gas occupies 22.4 L at STP.

179.2 L CO2 x 1 mol CO2/22.4 L CO2 x 2 mol NaCl/1 mol CO2

= 16 mol NaCl

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The molecule ch3-ch2-ch2-ch2-ch2-ch2-ch3 is best described as a(n) ________.

Gwar [14]

Given molecule:

CH3-CH2-CH2-CH2-CH2-CH2-CH3

- The molecule contain only C and H atoms, hence it is a hydrocarbon

- The C atoms in the linear chain are linked together by carbon-carbon single bonds, hence it is a saturated compound

- It has 7 C atoms and 16 H atoms, hence the molecular formula is C7H16

-The molecular formula can be generally represented as CnH2n+2 where n = 7. This corresponds to the general molecular formula for alkanes

The given molecule is best described as an alkane

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3 years ago

What volume of nitrogen dioxide is formed at 735 torr and 28.2 °C by reacting 3.56 cm3 of copper (d = 8.95 g/cm3) with 200 mL of

weqwewe [10]

Answer:

25.76 L

Explanation:

Given, Volume of Copper = 3.56 cm³

Density = 8.95 g/cm³

Considering the expression for density as:

$Density=\frac {Mass}{Volume}$

So,

So, Mass= Density * Volume = 8.95 g/cm³ * 3.56 cm³ = 31.862 g

Mass of copper = 31.862 g

Molar mass of copper = 63.546 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{31.862\ g}{63.546\ g/mol}$

Moles of copper = 0.5014 moles

Given, Volume of nitric acid solution = 200 mL = 200 cm³

Density = 1.42 g/cm³

Considering the expression for density as:

$Density=\frac {Mass}{Volume}$

So,

So, Mass= Density * Volume = 1.42 g/cm³ * 200 cm³ = 284 g

Also, Nitric acid is 68.0 % by mass. So,

Mass of nitric acid = $\frac {68}{100}\times 284\ g$ = 193.12 g

Molar mass of nitric acid = 63.01 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{193.12\ g}{63.01\ g/mol}$

Moles of nitric acid = 3.0649 moles

According to the reaction,

$Cu_{(s)}+4HNO_3_{(aq)}\rightarrow Cu(NO_3)_2_{(aq)} + 2NO_2_{(g)} + 2H_2O_{(l)}$

1 mole of copper react with 4 moles of nitric acid

Thus,

0.5014 moles of copper react with 4*0.5014 moles of nitric acid

Moles of nitric acid required = 2.0056 moles

Available moles of nitric acid = 3.0649 moles

Limiting reagent is the one which is present in small amount. Thus, nitric acid is present in large amount, copper is the limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of copper on reaction forms 2 moles of nitrogen dioxide

So,

0.5014 mole of copper on reaction forms 2*0.5014 moles of nitrogen dioxide

Moles of nitrogen dioxide = 1.0028 moles

Given:

Pressure = 735 torr

The conversion of P(torr) to P(atm) is shown below:

$P(torr)=\frac {1}{760}\times P(atm)$

So,

Pressure = 735 / 760 atm = 0.9632 atm

Temperature = 28.2 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (28.2 + 273.15) K = 301.35 K

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

0.9632 atm × V = 1.0028 mol × 0.0821 L.atm/K.mol × 301.35 K

⇒V = 25.76 L

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3 years ago

Using the correct answer from Part B, calculate the volume of a rectangular prism with a length of 5.6 cm, a width of 2.1 cm, an

son4ous [18]

So the question is asking to compute the volume of a rectangular prism with a length of 5.6 cm, width of 2.1 cm and a height of 6.6 cm, and base on the said given and using the formula in getting the volume of a rectangular prism, the volume 77.62 cm^3. I hope this would help.

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3 years ago

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How much heat is required, in calories, to raise the temperature of 57.8 g of silver from

Stella [2.4K]

Answer:

87.3 calories of heat is required.

Explanation:

Heat = mcΔT

m= mass, c = specific heat of silver, T = temperature

H= 57.8 g * 0.057 cal/g°C * ( 43.5 - 17 °C)

H = 57.8 * 0.057 * 26.5

H = 87.3069 cal.

The heat required to raise the temperature of 57.8 g of silver from 17 °C to 43.5 °C is 87.3 calories.

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3 years ago

The prophet Isaac was chosen to lead the Israelites after the death of Abraham.

Softa [21]

False. . . . . . . .

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3 years ago

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