To determine the molar mass of the unknown gas, we use Graham's Law of Effusion where it relates the effusion rates of two gases with their molar masses. It is expressed as r1/r2 = √M2/M1. We calculate as follows:
Let 1 = argon gas 2 = unknown gas
r2 = 0.91r1r1/r2 = 1/0.91
1/0.91 = √M2/M1 = √M2/40M2 = 48.30 g/mol
C2H4(g) + 2O2(g) → 2CO(g) + 2H2O(g) ∆H1 = −758 kJ mol−1 ....1)
2C(s) + 2H2(g) → C2H4(g) ∆H2 = +52 kJ mol−1 ....2)
H2(g) + O2(g) → H2O(g) ∆H3 = −242 kJ mol−1 ....3)
Now, enthalpy of formation of carbon monoxide is given by :
∆H = ∆H1 + ∆H2 - ∆H3
∆H = ( -758 + 52 - ( -242 ) ) kJ mol−1
∆H = −464 kJ mol−1
Therefore, the enthalpy of formation of carbon monoxide is -464 kJ mol−1.
Hence, this is the required solution.
c. physical change
Explanation:
The separation of ink into its different pigments during chromatography is a physical change.
Ink is a mixture of many different pigments.
During chromatography, a mixture is separated based on the relative migration of constituents of mixture over an adsorbent.
- Separation techniques are used to celebrate mixtures based on their physical properties.
- During a physical change only the physical properties of matter like their form and state is changed.
- The process is easily reversible and no new kinds of matter are formed.
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We are given with the following pairs:
<span>carbon and oxygen
hydrogen and helium
gold and silver
and we are asked if there is a pair that will produce the same spectrum. The answer is
</span>No two elements produce the same spectrum.This is because a light spectrum is unique to each element.
Answer:
we would need to see questions 35-37 in order to answer the question