Balance Fe3O4 + H2 = Fe + H2O

Can you guys help me with this one too

posledela

The tractor would have the most kinetic energy :D

3 0

2 years ago

Enter the appropriate symbol for an isotope of potassium-39 corresponding to the isotope notation A/Z x

Taya2010 [7]

Appropriate symbol for an isotope of potassium - 39 corresponding to the isotope notion r A ZX

7 0

3 years ago

Need help !!!!! ASAP

Ksivusya [100]

<h2>Hello!</h2>

The answer is:

We have that there were produced 0.120 moles of $CO_{2}$

$n=0.120mol$

We are asked to calculate the number of moles of the given gas, also, we are given the volume, the temperature and the pressure of the gas, we can calculate the approximate volume using The Ideal Gas Law.

The Ideal Gas Law is based on Boyle's Law, Gay-Lussac's Law, Charles's Law, and Avogadro's Law, and it's described by the following equation:

$PV=nRT$

Where,

P is the pressure of the gas.

V is the volume of the gas.

n is the number of moles of the gas.

T is the absolute temperature of the gas (Kelvin).

R is the ideal gas constant (to work with pressure in mmHg), which is equal to:

$R=62.363\frac{mmHg.L}{mol.K}$

We must remember that the The Ideal Gas Law equation works with absolute temperatures (K), so, if we are given relative temperatures such as Celsius degrees or Fahrenheit degrees, we need to convert it to Kelvin before we proceed to work with the equation.

We can convert from Celsius degrees to Kelvin using the following formula:

$Temperature(K)=Temperature(C\°) + 273K$

So, we are given the following information:

$Pressure=760mmHg\\Volume=2.965L\\Temperature=25.5C\°=25.5+273K=298.5K$

Now, isolating the number of moles, and substituting the given information, we have:

$PV=nRT$

$n=\frac{PV}{RT}$

$n=\frac{760mmHg*2.965L}{62.363\frac{mmHg.L}{mol.K}*298.5K}$

$n=\frac{760mmHg*2.965L}{62.363\frac{mmHg.L}{mol.K}*298.5K}\\\\n=\frac{2242mmHg.L}{18615.355\frac{mmHg.L}{mol.}}\\\\n=0.120mole$

Hence, we have that there were produced 0.120 moles of $CO_{2}$

$n=0.120mol$

Have a nice day!

7 0

3 years ago

a beaker of benzene is at room temperature. what is its vapor pressure. Is this vapor pressure large enough to cause the benzene

Serhud [2]

Answer:

(1). The vapor pressure is 91 mmHg at 20°C.

(2). No, benzene will not boil at sea level.

Explanation:

Benzene, C6H6 is an aromatic, liquid compound with with molar mass of 78.11 g/mol and Melting point of 5.5 °C. One of the importance or the uses of benzene is in the making of fibres and plastics.

The vapour pressure of benzene can be gotten from the table showing the vapor pressure of different liquids.

Boiling point can simply be defined as the point or the temperature in which the vapor pressure is the same with the atmospheric pressure.

The atmospheric pressure is 760mmHg, while the vapor pressure at sea level is at the temperature of 15°C which is equal to 71 mmHg( from the table showing the vapor pressure of different liquids).

71 mmHg is not equal to 760 mmHg, thus, at sea level Benzene will not boil.

8 0

2 years ago

Three kilograms of steam is contained in a horizontal, frictionless piston and the cylinder is heated at a constant pressure of

lakkis [162]

Answer:

Final temperature: 659.8ºC

Expansion work: 3*75=225 kJ

Internal energy change: 275 kJ

Explanation:

First, considering both initial and final states, write the energy balance:

$U_{2}-U_{1}=Q-W$

Q is the only variable known. To determine the work, it is possible to consider the reversible process; the work done on a expansion reversible process may be calculated as:

$dw=Pdv$