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xeze [42]
2 years ago
7

Hydrogen bromide is found in volcanic gas. What is the chemical formula for this compound?

Chemistry
2 answers:
ruslelena [56]2 years ago
3 0

Answer:

Hydrogen bromide, anhydrous appears as a colorless gas with a pungent irritating odor. Corrosive. Heavier than air. Prolonged exposure to fire or intense heat may result in the violent rupture and rocketing of the container. Long-term exposure to low concentrations or short-term exposure to high concentrations can result in adverse health effects from inhalation. Used to make other chemicals and as a catalyst in the manufacture of chemicals. Rate of onset: Immediate Persistence: Minutes to hours Odor threshold: 2 ppm Source/use/other hazard: Chemical manufacturing industry; very corrosive.

Explanation:

HBr

DIA [1.3K]2 years ago
3 0
The chemical formula is HBr
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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

The pressure is constant, so:  w=P(v_{2}-v_{1} )=0.5*100*1.5=75\frac{kJ}{kg} (There is a multiplication by 100 due to the conversion of bar to kPa)

So, the internal energy change may be calculated from the energy balance (don't forget to multiply by the mass):

U_{2}-U_{1}=500-(3*75)=275kJ

On the other hand, due to the low pressure the ideal gas law may be appropriate. The ideal gas law is written for both states:

P_{1}V_{1}=nRT_{1}

P_{2}V_{2}=nRT_{2}\\V_{2}=2.5V_{1}\\P_{2}=P_{1}\\2.5P_{1}V_{1}=nRT_{2}  

Subtracting the first from the second:

1.5P_{1}V_{1}=nR(T_{2}-T_{1})

Isolating T_{2}:

T_{2}=T_{1}+\frac{1.5P_{1}V_{1}}{nR}

Assuming that it is water steam, n=0.1666 kmol

V_{1}=\frac{nRT_{1}}{P_{1}}=\frac{8.314*0.1666*373.15}{500} =1.034m^{3}

T_{2}=100+\frac{1.5*500*1.034}{0.1666*8.314}=659.76 ºC

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