Answer:
- <u><em>Ratio of the mass carbon that combines with 1.00 g of oxygen in compound 2 to the mass of carbon that combines with 1.00 g of oxygen in compound 1 = 2</em></u>
Explanation:
First, detemine the mass of oxygen in the two samples by difference:
- mass of oxygen = mass of sample - mass of carbon
Item Compound 1 Compound 2
Sample 80.0 g 80.0 g
Carbon 21.8 g 34.3 g
Oxygen: 80.0 g - 21.8g = 58.2 g 80.0 g - 34.3 g = 45.7 g
Second, determine the ratios of the masses of carbon that combine with 1.00 g of oxygen:
- For each sample, divide the mass of carbon by the mass of oxygen determined above:
Sample Mass of carbon that combines with 1.00 g of oxygen
Compound 1 21.8 g / 58.2 g = 0.375
Compound 2 34.3 g / 45.7 g = 0.751
Third, determine the ratio of the masses of carbon between the two compounds.
- Divide the greater number by the smaller number:
- Ratio = 0.751 / 0.375 = 2.00 which in whole numbers is 2
Explanation:
Vapor pressure is defined as the pressure exerted by vapors or gas on the surface of a liquid.
It is known that at standard condition, vapor pressure is 760 mm Hg.
And, it is given that methanol vapor pressure in air is 88.5 mm Hg.
Hence, calculate the volume percentage as follows.
Volume percentage = 
= 
= 11.65%
Thus, we can conclude that the maximum volume percent of Methanol vapor that can exist at standard conditions is 11.65%.
<span>The problem has to do with oxidation states of the matter. The oxidation state of oxygen will always be -2 with the exception of peroxides which will have a state of -1. The overall balanced state of chemical compounds will be 0, so the oxidation state of Mn in MnO2 will be +4. The oxidation state of MnO4- will then be +7 to balance out to the negative one charge. The state change from +4 to +7 is 3, thus three electrons have to be lost in order for this to happen; a loss of a charge of -3 results in an increase of charge of 3. Oxidation is always the process of 'losing' electrons.
</span><span>E] MnO2(s) MnO4-(aq</span>
Answer:
2.03 moles of Gold
Explanation:
Gold is one of the most precious metal metal used in many applications and mainly as a jewellery. In terms of purity it is categorized in Karats. 24 Karat is considered the purest Gold (i.e. 100 % Gold) while other Karats (14, 18, 22 e.t.c) are alloys with other metals and gyms.
Data Given:
Mass of Gold = 400 g
A.Mass of Gold = 196.97 g.mol⁻¹
Calculate Moles of Gold as,
Moles = Mass ÷ M.Mass
Putting values,
Moles = 400 g ÷ 196.97 g.mol⁻¹
Moles = 2.03 moles of Gold
Answer:-
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Explanation:
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