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Keith_Richards [23]
2 years ago
8

At a temperature of __________ °c, 0.444 mol of co gas occupies 11.8 l at 889 torr.

Chemistry
2 answers:
shtirl [24]2 years ago
4 0

Answer: 106^0C

Explanation: IDEAL GAS LAW

PV=nRT

where,

P = pressure of the gas  = 889 torr = 1.17 atm    (1 torr=0.0013atm)

V = volume of the gas  = 11.8 L

T = temperature of the gas  = ? K

n = number of moles of gas  = 0.444

R = Gas constant = 0.0821 Latm/moleK

1.17\times 11.8=0.444\times 0.0821\times T

T=379K

T=(379-273)^0C=106^0C

Thus temperature of gas is 106^0C

Novay_Z [31]2 years ago
3 0
<span>ideal gas law is: PV = nRT P = pressure (torr) = 889 torr V = volume (Liters) = 11.8 L n = moles of gas = 0.444 mol R = gas constant = 62.4 (L * torr / mol * k) solve for T (in kelvin) T = PV/nR T = (889*11.8)/(.444*62.4) T = 378.6 K convert to C (subtract 273) T = 105.6 deg C</span>
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To determine the heat or energy needed for the process, we use the equation,
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6 0
1 year ago
Based on the products obtained, rank the functional groups (acetamido, amino, and methoxy) in order of increasing ability to act
I am Lyosha [343]

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Amino >Methoxy > Acetamido

Explanation:

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The functional group which withdraws electron to the benzene ring through inductive effect or resonance effect deactivates the ring towards electrophilic substitution reaction.

Among given, methoxy and amino are electron donating group. Amino group are stronger electron donating group than methoxy group. Acetamido group because of presence of carbonyl group becomes electron withdrawing group.

Therefore, decreasing order will be as follows:

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2 years ago
A few drops of a mixture of sodium hydroxide(NaOH) solution and copper (II) tetraoxosulphate (VI) solution were added to a sampl
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Answer:

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5 0
1 year ago
certain alcohol contains only three elements, carbon, hydrogen, and oxygen. Combustion of a 40.00 gram sample of the alcohol pro
seraphim [82]

Answer:

The answer to your question is:    C₂H₆O₁   = C₂H₆O

Explanation:

Data

CxHyOz

mass = 40 g  produced 76.40 g of CO2

                                       46.96 g of H2O

Empirical formula = ?

Process

                         CxHyOz    + O2    ⇒    CO2  + H2O

                         44g of CO2 --------------------  12 g of Carbon

                         76.40 g of CO2 --------------- x

                          x = 20.84 g of Carbon

                         12 g of Carbon ---------------  1 mol

                         20.84 g of C    ---------------   x

                         x = (20.84 x 1) / 12

                         x = 1.74 mol of Carbon

                        18 g of H2O --------------------  2 g of H

                        46.96 g of H2O --------------   x

                        x = (46.96 x 2) / 18

                        x = 5.22 g of H

                        1 g of H ------------------------  1 mol of H

                        5.22 g of H -------------------   x

                        x = 5.22 mol of H

Mass of Oxygen = 40 - 20.84 - 5.22g

                           = 13.94 g

                         16 g of Oxygen ----------------  1 mol

                         13.94 g of O --------------------   x

                         x = 0.87 mol of O

Divide by the lowest number of moles

Carbon = 1.74 / 0.87 = 2

Hydrogen = 5.22 / 0.87 = 6

Oxygen = 0.87 / 0.87 = 1

                           

Empirical formula

                                 C₂H₆O₁   = C₂H₆O

                         

3 0
1 year ago
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