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tia_tia [17]
2 years ago
7

Upon combustion, a 0.8009 g sample of a compound containing only carbon, hydrogen, and oxygen produces 1.6004 g co2 and 0.6551 g

h2o. determine the empirical formula of the compound.
Chemistry
1 answer:
dlinn [17]2 years ago
3 0
The compound contains Carbon, Hydroxide and Oxide
1 mole of carbon iv oxide contains 44 g, out of which 12 g are carbon.
Therefore, 1.6004 g of CO2 will contain;
    1.6004 ×12/44 = 0.4365 g of carbon
1 mole of water contains 18 g of which 2 g is hydrogen,
Therefore, 0.6551 g of H2O will hace ;
         0.6551 × 2/18 = 0.0728 g of hydrogen.
The total mass of the compound is 0.8009 g,
Thus the mass of oxygen = 0.8009 -(0.4365 +0.0728)
                                         = 0.2916 g
To get the empirical formula we first get the number of moles of each element;\
 Carbon = 0.4365/12= 0.036375 moles
Hydrogen = 0.0728/1 = 0.0728 moles
Oxygen = 0.2916/16 = 0.018225 moles
Then, to get the smallest ratio we divide each with the smallest value;
Carbon :                        Hydrogen :                    Oxygen
= (0.036375/0.018225) : (0.0728/0.018225) : ( 0.018225/0.018225)
= 1.996                         :   3.995                    : 1
≈ 2 : 4 : 1
Therefore, the empirical formula is C2H4O


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An equimolar mixture of N2(g) and Ar(g) is kept inside a rigid container at a constant temperature of 300 K. The initial partial
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Answer:

The final pressure of the gas mixture after the addition of the Ar gas is P₂= 2.25 atm

Explanation:

Using the ideal gas law

PV=nRT

if the Volume V = constant (rigid container) and assuming that the Ar added is at the same temperature as the gases that were in the container before the addition, the only way to increase P is by the number of moles n . Therefore

Inicial state ) P₁V=n₁RT

Final state )  P₂V=n₂RT

dividing both equations

P₂/P₁ = n₂/n₁ → P₂= P₁ * n₂/n₁

now we have to determine P₁ and n₂ /n₁.

For P₁ , we use the Dalton`s law , where p ar1 is the partial pressure of the argon initially and x ar1 is the initial molar fraction of argon (=0.5 since is equimolar mixture of 2 components)

p ar₁ = P₁ * x ar₁  →  P₁ = p ar₁ / x ar₁ = 0.75 atm / 0.5 = 1.5 atm

n₁ = n ar₁ + n N₁ =  n ar₁ + n ar₁ = 2 n ar₁

n₂ = n ar₂ + n N₂ = 2 n ar₁ + n ar₁ = 3 n ar₁

n₂ /n₁ = 3/2

therefore

P₂= P₁ * n₂/n₁ = 1.5 atm * 3/2  = 2.25 atm

P₂= 2.25 atm

8 0
1 year ago
Small beads of iridium-192 are sealed in a plastic tube and inserted through a needle into breast tumors. If an Ir-192 sample ha
Stells [14]

Answer:

296.1 day.

Explanation:

  • The decay of radioactive elements obeys first-order kinetics.
  • For a first-order reaction: k = ln2/(t1/2) = 0.693/(t1/2).

Where, k is the rate constant of the reaction.

t1/2 is the half-life time of the reaction (t1/2 = 1620 years).

∴ k = ln2/(t1/2) = 0.693/(74.0 days) = 9.365 x 10⁻³ day⁻¹.

  • For first-order reaction: <em>kt = lna/(a-x).</em>

where, k is the rate constant of the reaction (k = 9.365 x 10⁻³ day⁻¹).

t is the time of the reaction (t = ??? day).

a is the initial concentration of Ir-192 (a = 560.0 dpm).

(a-x) is the remaining concentration of Ir-192 (a -x = 35.0 dpm).

<em>∴ kt = lna/(a-x)</em>

(9.365 x 10⁻³ day⁻¹)(t) = ln(560.0 dpm)/(35.0 dpm).

(9.365 x 10⁻³ day⁻¹)(t) = 2.773.

<em>∴ t </em>= (2.773)/(9.365 x 10⁻³ day⁻¹) =<em> 296.1 day.</em>

5 0
2 years ago
A flask contains 0.180 mol of liquid bromine, br2. determine the number of bromine molecules present in the flask.
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<span>Avogadro's number represents the number of units in one mole of any substance. This has the value of 6.022 x 10^23 units / mole. This number can be used to convert the number of atoms or molecules into number of moles. We calculate as follows:

0.180 mol Br2 ( </span>6.022 x 10^23 molecules / mole ) = 1.084x10^23 molecules Br2
5 0
2 years ago
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The answer to this is A i think.
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1 year ago
Consider the standard galvanic cell based on the following half-reactions The electrodes in this cell are and . Does the cell po
Free_Kalibri [48]

Question: The question is incomplete and can't be comprehended. See the complete question below and the answer.

Consider a galvanic cell based upon the following half reactions: Ag+ + e- → Ag 0.80 V Cu2+ + 2 e- → Cu 0.34 V

How would the following changes alter the potential of the cell?

a) Adding Cu2+ ions to the copper half reaction (assuming no volume change).

b) Adding equal amounts of water to both half reactions.

c) Removing Cu2+ ions from solution by precipitating them out of the copper half reaction (assume no volume change).

d) Adding Ag+ ions to the silver half reaction (assume no volume change)

Explanation:

Nernst equation relates the reduction potential of an electrochemical reaction (half-cell or full cell reaction) to the standard electrode potential, temperature, and activities (often approximated by concentrations) of the chemical species undergoing reduction and oxidation.

Reaction under consideration:

Ag+ + e- → Ag 0.80 V

Cu+2 + 2 e- → Cu 0.34 V

Clearly, Ag reduction potential is high and this indicates that it gets reduced readily which leaves Cu to oxidize. Cu+2 ions are products of reaction and Ag+ ions are reactant ions.

Nernst equation : Ecell = E°cell­ – (2.303 RT / n F) log Q    

where                            

             Ecell = actual cell potential

             E°cell­ ­​ = standard cell potential

             R = the universal gas constant = 8.314472(15) J K−1 mol−1

             T = the temperature in kelvins

              n = the number of moles of electrons transferred                                    

 F = the Faraday constant, the number of coulombs per mole of electrons:

  (F = 9.64853399(24)×104 C mol−1)

 Q = [product ion]y / [reactant ion] x

Accordingly when applied to above reaction one will get the following

= E°cell­ – (2.303 x RT / 6 F) log [Cu+2] / [Ag+]

Now the given variables can be studied according to Le Chatelier's principle which states when any system at equilibrium is subjected to change in its concentration, temperature, volume, or pressure, then the system readjusts itself to (partially) counteract the effect of the applied change and a new equilibrium is established.

a)        Adding Cu2+ ions to the copper half reaction (assuming no volume change).

Addition of Cu+2 ions increases its concentration and consequently increases the Q value which results in reduction of Ecell. In other words the addition of Cu+2 ions favors the backward reaction to maintain the equilibrium of reaction and hence the forward reaction rate decreases.

b)       Adding equal amounts of water to both half reactions.

Addition of water increases the dilution of the electrochemical cell. For weak electrolytes such as Ag+/ Cu+2 with increase in dilution, the degree of dissociation increases and as a result molar conductance increases.

c)        Removing Cu2+ ions from solution by precipitating them out of the copper half reaction (assume no volume change).

Based on Le Chatelier's principle when Cu+2 ions amount is decreased by its continuous removal from  the system the forward reaction is favored. As the Cu+2 ions is removed the system attempts to generate more Cu+2 ions to counter the affect of its removal.

d)       Adding Ag+ ions to the silver half reaction (assume no volume change)

Addition of reactant ions, i.e. Ag+ ions, will favour the forward reaction, which results in more product formation.

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