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juin [17]
3 years ago
10

How many moles are in 18.3 g of CO2?

Chemistry
1 answer:
kaheart [24]3 years ago
7 0

Answer:

0.416moles

Explanation:

Using the formula;

number of moles = mass ÷ molar mass

Molar mass of CO2 = 12 + 16(2)

= 12 + 32

= 44g/mol

According to the information in this question, there are 18.3g of CO2

number of moles (n) = 18.3/44

= 0.4159

= 0.416moles of CO2

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Write the overall, balanced molecular equation and indicate which element is oxidized and which is reduced for the following rea
galina1969 [7]

Answer:

Cd(s) + AgNO₃(aq)  → Cd(NO₃)₂ (aq) + Ag(s)

Oxidized: Cd

Reduced: Ag

Explanation:

Cd(s) + AgNO₃(aq)  → Cd(NO₃)₂ (aq) + Ag(s)

Cd → Cd²⁺  +  2e⁻      Half reaction oxidation

1e⁻ + Ag⁺ → Ag           Half reaction reduction

Ag changed oxidation number from +1 to 0

Cd changed oxidation number from 0 to +2

Let's ballance the electrons

( Cd → Cd²⁺  +  2e⁻ ) .1

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Cd(s) + 2AgNO₃(aq)  → Cd(NO₃)₂ (aq) + 2Ag(s)

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3 years ago
If you place 1.0 L of ethanol (C2H5OH) in a small laboratory that is 3.0 m long, 2.0 m wide, and 2.0 m high, will all the alcoho
ankoles [38]

If you place 1.0 L of ethanol (C2H5OH) in a small laboratory that is 3.0 m long, 2.0 m wide, and 2.0 m high, will all the alcohol evaporate? If some liquid remains, how much will there be? The vapor pressure of ethyl alcohol at 25 °C is 59 mm Hg, and the density of the liquid at this temperature is 0.785g/cm^3 .

will all the alcohol evaporate? or none at all?

Answer:

Yes, all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore  be zero.

Explanation:

Given that:

The volume of alcohol which is placed in a small laboratory = 1.0 L

Vapor pressure of ethyl alcohol  at 25 ° C = 59 mmHg

Converting 59 mmHg to atm ; since 1 atm = 760 mmHg;

Then, we have:

= \frac{59}{760}atm

= 0.078 atm

Temperature = 25 ° C

= ( 25 + 273 K)

= 298 K.

Density of the ethanol = 0.785 g/cm³

The volume of laboratory = l × b × h

= 3.0 m × 2.0 m × 2.5 m

= 15 m³

Converting the volume of laboratory to liter;

since 1 m³ = 100 L; Then, we  have:

15 × 1000 = 15,000 L

Using ideal gas equation to determine the moles of ethanol in vapor phase; we have:

PV = nRT

Making n the subject of the formula; we have:

n = \frac{PV}{RT}

n = \frac{0.078 * 15000}{0.082*290}

n = 47. 88 mol of ethanol

Moles of ethanol in 1.0 L bottle can be calculated as follows:

Since  numbers of moles = \frac{mass}{molar mass}

and mass = density × vollume

Then; we can say ;

number of moles = \frac{density*volume }{molar mass of ethanol}

number of moles =\frac{0.785g/cm^3*1000cm^3}{46.07g/mol}

number of moles = \frac{&85}{46.07}

number of moles = 17.039 mol

Thus , all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore be zero.

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9.0 × 10⁻⁴ m × 8.1 × 10⁴ m = 7.3 × 10¹ m²

The significant figures rule for multiplication is that we must us the amount of significant figures in the number with the least significant figures, in this question this is 2 significant digits from both 9.0 × 10⁻⁴ m and 8.1 × 10⁴ m.

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