What is the theoretical volume of 10mL of an ideal gas at absolute zero?

Which is the correctly balanced equation for the reaction of rust, Fe2O3, and hydrochloric acid, HCl?

Mariana [72]

Answer: Option (c) is correct answer.

Explanation:

The given reaction equation is as follows.

$Fe_{2}O_{3} + HCl \rightarrow FeCl_{3} + H_{2}O$

The number of atoms on reactant side are as follows.

Fe = 2
O = 3
H = 1
Cl = 1

The number of atoms on product side are as follows.

Fe = 1
O = 1
H = 2
Cl = 3

Therefore, in order to balance the equation we will balance the number of atoms in both reactant and product side.

Since number of Fe atoms in reactant side is 2, therefore, multiply $FeCl_{3}$ by 2 in the product side.

Number of O atoms in reactant side is 3, therefore, multiply $H_{2}O$ by 3 in the product side.

Now, the number of H atoms on product side is 6, therefore, multiply HCl by 6 in the reactant side.

The number of chlorine atoms on both reactant and product side equals 6.

Thus, the equation will become as follows.

$Fe_{2}O_{3} + 6HCl \rightarrow 2FeCl_{3} + 3H_{2}O$

Hence, the equation is balanced.

5 0

3 years ago

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A car speeds upu as it rolls down a hill. Which is this an example of?

anygoal [31]

Positive acceleration

4 0

3 years ago

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My name is scooter hey fat boy you waa fight

zavuch27 [327]

Answer:

HEHE

YES

Explanation:

6 0

3 years ago

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A thick steel sheet of area 100 in.2 is exposed to air near the ocean. After a one-year period it was found to experience a weig

vladimir1956 [14]

Answer:

Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.

Explanation:

The corrosion rate is the rate of material remove.The formula for calculating CPR or corrosion penetration rate is

$CPR=\frac{KW}{DAT}$

K= constant depends on the system of units used.

W= weight =485 g

D= density =7.9 g/cm³

A = exposed specimen area =100 in² =6.452 cm²

K=534 to give CPR in mpy

K=87.6 to give CPR in mm/yr

mpy

$CPR=\frac{KW}{DAT}$

$=\frac{534\times( 485g)\times( 10^3mg/g)}{(7.9g/cm^3) \times (100in^2)\times (24h/day)\times (365day/yr)\times 1yr}$

=37.4mpy

mm/yr

$CPR=\frac{KW}{DAT}$

$=\frac{87.6\times (485g)\times (10^3 mg/g)}{(7.9g/cm^3)\times (100in^2)\times(2.54cm/in)^2\times (24h/day)\times (365day/yr)\times 1yr}$

=0.952 mm/yr

Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.

3 0

3 years ago

Consider a well-insulated rigid container with two chambers separated by a membrane. The total volume is 5.0 cubic meters. The f

mamaluj [8]

Answer:

The Entropy generated by the steam = 2.821 kJ/K

Explanation:

Total volume of container = 5m³

Heat transfer does not exist between system and surrounding, dQ = 0

At the first chamber, temperature of water at saturated liquid is 300°C

From the steam table:

Specific enthalpy of saturated liquid at 300°C , $h_{f} = 1344.8 kJ/kg$

Specific internal energy of saturated liquid at 300°C, $U_{f1} =$ 1332.7 kJ/kg

For closed system, the first law of thermodynamics state that:

dQ = dw + dU..................(1)

work done for free expansion, dw =0

0 = 0 + dU

dU = 0 , i.e. U₁ = U₂

At the second chamber,

The final pressure, P₂ = 50 kPa

From the steam table, at P₂ = 50 kPa, $U_{f2}$ = 340.49 kJ/kg

$(U_{fg} )_{2} = 2142.7 kJ/kg$

Let the dryness fraction at the second chamber = x

$U_{2} = U_{f2} + U_{fg2}$

$U_{2} = 340.49 + x2140.7$ Since U₁ = U₂

1332.7 = 340.49 + x2140.7

Dryness fraction, x = 0.463

From steam table, the specific volume is, $u_{f2} = 0.00103 m^{3} /kg\\$

$u_{2} = u_{f2} + xu_{fg2}$

$u_{2} = 0.00103 + 0.463(3.2393)\\u_{2} = 1.5 m^{3} /kg\\$

$u_{2} = \frac{v_{2} }{m_{2} }$

V₂ = 5 m³

1.5 = 5/m₂

m₂ = 3.33 kg

At 300°C $S_{1} = S_{f} = 3.2548 kJ/kg-k\\$

$S_{2} = S_{f2} + xS_{fg2}$

From the steam table,

$S_{f2} = 1.0912 kJ/kg-k\\S_{fg2} = 6.5019 kJ/kg-k\\S_{2} = 1.0912 + 0.463(6.5019)\\S_{2} = 4.102 kJ/kg-k$

Therefore the entropy generated will be :

Entropy = mass* (S₂ - S₁)

Entropy = 3.33* (4.102 - 3.2548)

Entropy = 2.821 kJ/K

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3 years ago

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