The ___ has many functions; one is to regulate growth.

A mixture containing 0.477 mol he(g), 0.265 mol ne(g), and 0.115 mol ar(g) is confined in a 7.00-l vessel at 25 ∘c. part a calcu

enot [183]

Q1)
we can use the ideal gas law equation to find the total pressure of the system ;
PV = nRT
where P - pressure
V - volume - 7 x 10⁻³ m³
n - number of moles
total number of moles - 0.477 + 0.265 + 0.115 = 0.857 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in K - 273 + 25 °C = 298 K
substituting the values in the equation
P x 7 x 10⁻³ m³ = 0.857 mol x 8.314 Jmol⁻¹K⁻¹ x 298 K
P = 303.33 kPa
1 atm = 101.325 kPa
Therefore total pressure - 303.33 kPa / 101.325 kPa/atm = 2.99 atm

Q2)
partial pressure is the pressure exerted by the individual gases in the mixture.
partial pressure for each gas can be calculated by multiplying the total pressure by mole fraction of the individual gas.

total number of moles - 0.477 + 0.265 + 0.115 = 0.857 mol
mole fraction of He - $\frac{0.477}{0.857} = 0.557$
mole fraction of Ne - $\frac{0.265}{0.857} = 0.309$
mole fraction of Ar - $\frac{0.115}{0.857} = 0.134$
partial pressure - total pressure x mole fraction
partial pressure of He - 2.99 atm x 0.557 = 1.67 atm
partial pressure of Ne - 2.99 atm x 0.309 = 0.924 atm
partial pressure of Ar - 2.99 atm x 0.134 = 0.401 atm

6 0

3 years ago

A 1.25-g sample contains some of the very reactive compound Al(C6H5)3. On treating the compound with aqueous HCl, 0.951 g of C6H

Oksanka [162]

<span>83.9% First, determine the molar masses of Al(C6H5)3 and C6H6. Start by looking up the atomic weights of the involved elements. Atomic weight aluminum = 26.981539 Atomic weight carbon = 12.0107 Atomic weight hydrogen = 1.00794 Molar mass Al(C6H5)3 = 26.981539 + 18 * 12.0107 + 15 * 1.00794 = 258.293239 g/mol Molar mass C6H6 = 6 * 12.0107 + 6 * 1.00794 = 78.11184 g/mol Now determine how many moles of C6H6 was produced Moles C6H6 = 0.951 g / 78.11184 g/mol = 0.012174851 mol Looking at the balanced equation, it indicates that 1 mole of Al(C6H5)3 is required for every 3 moles of C6H6 produced. So given the number of moles of C6H6 you have, determine the number of moles of Al(C6H5)3 that was required. 0.012174851 mol / 3 = 0.004058284 mol Then multiply by the molar mass to get the number of grams that was originally present. 0.004058284 mol * 258.293239 g/mol = 1.048227218 g Finally, the weight percent is simply the mass of the reactant divided by the total mass of the sample. So 1.048227218 g / 1.25 g = 0.838581775 = 83.8581775% And of course, round to 3 significant digits, giving 83.9%</span>

8 0

3 years ago

A chemist designs a galvanic cell that uses these two half-reactions:

Nonamiya [84]

Answer :

(a) Reaction at anode (oxidation) : $4Fe^{2+}\rightarrow 4Fe^{3+}+4e^-$

(b) Reaction at cathode (reduction) : $O_2+4H^++4e^-\rightarrow 2H_2O$

(c) $O_2+4H^++4Fe^{2+}\rightarrow 2H_2O+4Fe^{3+}$

(d) Yes, we have have enough information to calculate the cell voltage under standard conditions.

Explanation :

The half reaction will be:

Reaction at anode (oxidation) : $Fe^{2+}\rightarrow Fe^{3+}+e^-$ $E^0_{anode}=+0.771V$

Reaction at cathode (reduction) : $O_2+4H^++4e^-\rightarrow 2H_2O$ $E^0_{cathode}=+1.23V$

To balance the electrons we are multiplying oxidation reaction by 4 and then adding both the reaction, we get:

Part (a):

Reaction at anode (oxidation) : $4Fe^{2+}\rightarrow 4Fe^{3+}+4e^-$ $E^0_{anode}=+0.771V$

Part (b):

Reaction at cathode (reduction) : $O_2+4H^++4e^-\rightarrow 2H_2O$ $E^0_{cathode}=+1.23V$

Part (c):

The balanced cell reaction will be,

$O_2+4H^++4Fe^{2+}\rightarrow 2H_2O+4Fe^{3+}$

Part (d):

Now we have to calculate the standard electrode potential of the cell.

$E^o=E^o_{cathode}-E^o_{anode}$

$E^o=(1.23V)-(0.771V)=+0.459V$

For a reaction to be spontaneous, the standard electrode potential must be positive.

So, we have have enough information to calculate the cell voltage under standard conditions.

4 0

3 years ago

Sumit and Rohan were playing in bath tub of water. While playing Sumit accidently dropped his iron key into the bath tub and Roh

Sauron [17]

Answer:

The wood

Explanation:

The block of wood shall float in the water while the iron key would sink due to the weight.

5 0

3 years ago

How to make 100 ml of 0.001 mM solution with 0.0405mM solution?

alexgriva [62]

Answer:

Measure 2.47 mL of the stock solution (i.e 0.0405 mM) and dilute it to the 100 mL mark with water

Explanation:

To make 100 mL of 0.001 mM solution from 0.0405mM solution, we need to determine the volume of 0.0405mM solution needed. This can be obtained as follow:

Molarity of stock (M₁) = 0.0405 mM

Volume of diluted (V₂) = 100 mL

Molarity of diluted solution (M₂) = 0.001 mM

Volume of stock solution needed (V₁) =?

M₁V₁ = M₂V₂

0.0405 × V₁ = 0.001 × 100

0.0405 × V₁ = 0.1

Divide both side by 0.0405

V₁ = 0.1 / 0.0405

V₁ = 2.47 mL

Therefore, to make 100 mL of 0.001 mM solution from 0.0405mM solution, measure 2.47 mL of the stock solution (i.e 0.0405 mM) and dilute it to the 100 mL mark with water.

3 0

3 years ago