A balloon is filled with 0.400 moles of gas. How many liters of gas at STP were pumped into the balloon?

A 50.0 g sample of liquid water at 25.0 degree C is mixed with 29.0 g of water at 45 degree C. The final temperature of the wate

kotegsom [21]

<u>Answer:</u> The final temperature of water is 32.3°C

<u>Explanation:</u>

When two solutions are mixed, the amount of heat released by solution 1 (liquid water) will be equal to the amount of heat absorbed by solution 2 (liquid water)

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)]$ ......(1)

where,

q = heat absorbed or released

$m_1$ = mass of solution 1 (liquid water) = 50.0 g

$m_2$ = mass of solution 2 (liquid water) = 29.0 g

$T_{final}$ = final temperature = ?

$T_1$ = initial temperature of solution 1 = 25°C = [273 + 25] = 298 K

$T_2$ = initial temperature of solution 2 = 45°C = [273 + 45] = 318 K

c = specific heat of water= 4.18 J/g.K

Putting values in equation 1, we get:

$50.0\times 4.18\times (T_{final}-298)=-[29.0\times 4.18\times (T_{final}-318)]\\\\T_{final}=305.3K$

Converting this into degree Celsius, we use the conversion factor:

$T(K)=T(^oC)+273$

$305.3=T(^oC)+273\\T(^oC)=(305.3-273)=32.3^oC$

Hence, the final temperature of water is 32.3°C

7 0

3 years ago

Write a balanced half-reaction for the oxidation of aqueous hydrazine N2H4 to gaseous nitrogen N2 in basic aqueous solution

Zigmanuir [339]

Answer:

N2H2(aq) + 2OH^-(aq) ----------> N2(g) + 2H2O(l) + 2e

Explanation:

Hydrazine is mostly used in thermal engineering as an anticorrosive agent. Hydrazine can be oxidized in aqueous solution as shown in the equation above. Oxidation has to do with loss of electrons and increase in oxidation number.

The oxidation number of nitrogen in the equation increased from -1 in hydrazine on the lefthand side of the reaction equation to zero in nitrogen on the right hand side of the reaction equation. Two electrons were lost in the process as shown.

7 0

2 years ago

The cell potential of a redox reaction occurring in an electrochemical cell under any set of temperature and concentration condi

avanturin [10]

Answer : The actual cell potential of the cell is 0.47 V

Explanation:

Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

The given redox reaction is :

$Ni^{2+}(aq)+Zn(s)\rightarrow Ni(s)+Zn^{2+}(aq)$

The balanced two-half reactions will be,

Oxidation half reaction : $Zn\rightarrow Zn^{2+}+2e^-$

Reduction half reaction : $Ni^{2+}+2e^-\rightarrow Ni$

The expression for reaction quotient will be :

$Q=\frac{[Zn^{2+}]}{[Ni^{2+}]}$

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

$Q=\frac{(0.0141)}{(0.00104)}=13.6$

The value of the reaction quotient, Q, for the cell is, 13.6

Now we have to calculate the actual cell potential of the cell.

Using Nernst equation :

$E_{cell}=E^o_{cell}-\frac{RT}{nF}\ln Q$

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature = 316 K

n = number of electrons in oxidation-reduction reaction = 2 mole

$E^o_{cell}$ = standard electrode potential of the cell = 0.51 V

$E_{cell}$ = actual cell potential of the cell = ?

Q = reaction quotient = 13.6

Now put all the given values in the above equation, we get:

$E_{cell}=0.51-\frac{(8.314)\times (316)}{2\times 96500}\ln (13.6)$

$E_{cell}=0.47V$

Therefore, the actual cell potential of the cell is 0.47 V

4 0

3 years ago

How many molecules of H2O are equivalent to 97.2g H2O

IgorLugansk [536]

como se denomina el proceso utilizado para descomponer el agua

6 0

2 years ago

Which notation designates a possible value for a principal energy level in an atom?

Nata [24]

Answer:

2

Explanation:

There are four quantum numbers:

Principal quantum number (n)

Azimuthal quantum number (l)

Magnetic quantum number (ml)

Principal quantum number (n)

It tell about the energy levels. It is designated by n.

For example,

If n =2

It means there are two energy level present.

Azimuthal quantum number (l)

The azimuthal quantum number describe the shape of orbitals. Its value for s, p, d, f... are 0, 1, 2, 3. For l=3

(n-1)

4-1 = 3

it means principle quantum is 4 and electron is present in f subshell.

Magnetic quantum number (ml)

It describe the orientation of orbitals. Its values are -l to +l. For l=3 the ml will be -3 -2 -1 0 +1 +2 +3.

Spin quantum number (ms)

The spin quantum number tells the spin of electron either its clock wise (+1/2) or anti clock wise (-1/2).

If the electron is added in full empty orbital its spin will be +1/2 because it occupy full empty. If electron is already present and another electron is added then its spin will be -1/2.

3 0

2 years ago