Answer:
The higher the frequency, the shorter the wavelength
Explanation:
All light waves move through a vacuum at the same speed, the number of wave crests passing by a given point in one second depends on the wavelength.
Answer:
The reactants would appear at a higher energy state than the products.
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A general equation for a combustion reaction would be expressed as follows:
CxHy + (x+y/2)O2 = xCO2 + y/2H2O
Propane would obviously would only have carbon and hydrogen in its structure. Assuming a complete combustion, all of the carbon atoms would go to carbon dioxide and all of the hydrogen atoms to water. To determine the empirical, we determine the number of carbon and hydrogen atoms present.
moles C = 2.461 g CO2 ( 1 mol / 44.01 g ) ( 1 mol C / 1 mol CO2 ) = 0.06 mol C
moles H = 1.442 g H2O ( 1 mol / 18.02 g ) ( 2 mol H / 1 mol H ) = 0.16 mol H
Then, we divide the smallest amount to the each mole of the atoms. We do as follows:
C = 0.06 / 0.06 = 1
H = 0.16 / 0.06 = 2.67
Then we multiply a number in order to obtain a whole number ratio between the atoms.
1 CH2.67
2 C2H5.34
3 C3H8 <-------- empirical formula
Answer:
i) CCl₄ and Br₂ does not react
ii) CBr₄ + Cl₂ → CCl₄ + Br₂
Explanation:
i) CCl₄ + Br₂ (no reaction)
From the given activity series, we have that chlorine gas, Cl₂, is more reactive than bromine gas, Br₂, therefore, a reaction of CCl₄ + Br₂ will not have a reaction as the propensity for the chlorine to stay combined with the carbon is higher than the ability for bromine to remain combined with or attract the carbon. Therefore, for CCl₄ + Br₂ there is no reaction
ii) CBr₄ + Cl₂
From the given activity series, we have that chlorine gas, Cl₂, is more reactive than bromine gas, Br₂, therefore, a reaction of CBr₄ + Cl₂ will give products that will have the Br in the CBr₄ replaced by the Cl₂ as follows;
CBr₄ + Cl₂ → CCl₄ + Br₂
The products of the reaction of CBr₄ and Cl₂ are therefore CBr₄ and Cl₂.
The answer is D, medical diagnosis
