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miv72 [106K]
3 years ago
7

In the industrial synthesis of hydrogen, mixtures of CO and H2 are enriched in H2 by allowing the CO to react with steam. The ch

emical equation for this so-called water-gas shift reaction is
Chemistry
1 answer:
saw5 [17]3 years ago
8 0

Answer:

CO(g) + H2(g) + H2O(g) ==> CO2(g) + 2H2(g)

Explanation:

In the industry, hydrogen is prepared from water and hydrocarbons. Water gas being the major method of preparation of hydrogen industrially.

The water-gas reaction is an industrial process in which steam is passed over red-hot coke giving a gaseous mixture of carbon monoxide and hydrogen:

C + H2O(g) → CO + H2.

The mixture of CO and H2 is Futher passed through steam according to the equation:

CO(g) + H2(g) + H2O(g) ==> CO2(g) + 2H2(g) to give hydrogen and carbon dioxide.

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lukranit [14]

Answer:

A I think

Explanation:

4 0
3 years ago
Read 2 more answers
What is the mass percent of oxygen in sodium bicarbonate (NaHCO3)?
Alenkasestr [34]
The answer is 57.14%.

First we need to calculate molar mass of <span>NaHCO3. Molar mass is mass of 1 mole of a substance. It is the sum of relative atomic masses, which are masses of atoms of the elements.

Relative atomic mass of Na is 22.99 g
</span><span>Relative atomic mass of H is 1 g
</span><span>Relative atomic mass of C is 12.01 g
</span><span>Relative atomic mass of O is 16 g.
</span>
Molar mass of <span>NaHCO3 is:
22.99 g + 1 g + 12.01 g + 3 </span>· <span>16 g = 84 g

Now, mass of oxygen in </span><span>NaHCO3 is:
3 </span>· 16 g = 48 g

mass percent of oxygen in <span>NaHCO3:
48 g </span>÷ 84 g · 100% = 57.14%

Therefore, <span>the mass percent of oxygen in sodium bicarbonate is 57.14%.</span>
8 0
3 years ago
An energy of 6.8 x 10^-19 J/atom is required to cause an aluminum atom on a metal surface to lose an electron.
Nana76 [90]

Wavelength of the light is 2.9 × 10⁻⁷ m.

<u>Explanation:</u>

Planck - Einstein equation shows the relationship between the energy of a photon and its frequency, and they are directly proportional to each other and  it is given by the equation as E = hν,

where E is the energy of the photon

h is the Planck's constant = 6.626 × 10⁻³⁴ J s

ν is the frequency

From the above equation, we can find the frequency by rearranging the equation as,

ν = $ \frac{E}{h} = $ \frac{6.8 \times 10^{-19}}{6.626\times10^{-34}} = 1.03\times10^{15} s^{-1}

Now the frequency and the wavelength are in inverse relationship with each other.

ν × λ = c

It can be rearranged to get λ as,

λ = c / ν

 = $\frac{3\times 10^{8} ms^{-1}}{1.03\times10^{15}s^{-1}} = 2.9\times 10^{-7} m

So wavelength is 2.9 × 10⁻⁷ m.

6 0
4 years ago
Which physical method can be used for obtaining a sample of salt from a small beaker of salt water?
kipiarov [429]
The answer is evaporation<span>.Evaporation is the vaporization of the liquid from its surface into gaseous phase, without boiling the liquid. When all the liquid has passed to gaseous phase the salt dissolved in the salt water will remain as solid crystals.</span><span />
5 0
3 years ago
The normal boiling point of bromine is 58.8°C, and its enthalpy of vaporization is 30.91 kJ/mol. What is the approximate vapor p
saul85 [17]

Answer : The vapor pressure of bromine at 10.0^oC is 0.1448 atm.

Explanation :

The Clausius- Clapeyron equation is :

\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1 = vapor pressure of bromine at 10.0^oC = ?

P_2 = vapor pressure of propane at normal boiling point = 1 atm

T_1 = temperature of propane = 10.0^oC=273+10.0=283.0K

T_2 = normal boiling point of bromine = 58.8^oC=273+58.8=331.8K

\Delta H_{vap} = heat of vaporization = 30.91 kJ/mole = 30910 J/mole

R = universal constant = 8.314 J/K.mole

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

\ln (\frac{1atm}{P_1})=\frac{30910J/mole}{8.314J/K.mole}\times (\frac{1}{283.0K}-\frac{1}{331.8K})

P_1=0.1448atm

Hence, the vapor pressure of bromine at 10.0^oC is 0.1448 atm.

4 0
3 years ago
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