The combustion of ethane yields carbon dioxide, and with 5.90 moles of ethane being reacted, it results in the production of 11.8 moles of CO2.
The question pertains to the concept of stoichiometry in chemistry, and the chemical reaction in question is a combustion reaction involving ethane (C2H6). From the balanced reaction, it is evident that 2 moles of ethane (C2H6) produce 4 moles of carbon dioxide (CO2). Therefore, if we have 5.90 moles of ethane reacting, it's a straightforward calculation to determine that this would yield twice that many moles of CO2. We simply multiply the moles of ethane by the stoichiometric ratio (4/2) to get the moles of CO2.
Example Calculation: 5.90 moles of ethane x (4 moles CO2 / 2 moles C2H6) = 11.8 moles CO2
So, when 5.90 moles of ethane are burned in an excess of oxygen, 11.8 moles of CO2 are produced.
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In the combustion of ethane, for every mole of ethane burned, two moles of carbon dioxide are produced. Hence, when 5.90 moles of ethane are burned, 11.8 moles of carbon dioxide are produced.
The chemical reaction given, 2C2H6(g) + 7O2(g) ⟶ 4CO2(g) + 6H2O(g), states that 2 moles of ethane (C2H6) produce 4 moles of carbon dioxide (CO2). Thus, the mole-to-mole ratio of ethane to carbon dioxide is 2:4, or simplified, 1:2. So, for every mole of ethane burned, two moles of carbon dioxide are produced.
Given that 5.90 moles of ethane are burned, we can calculate the quantity of carbon dioxide produced by multiplying 5.90 moles by 2. Hence, when 5.90 moles of ethane are burned in an excess of oxygen, 11.8 moles of carbon dioxide are produced.
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Answer:
Explanation:
H₂O(ℓ) ⟶ H₂O(g)
Data:
T = 25 °C
ΔG° = 8.6 kJ·mol⁻¹
Calculations:
T = (25 + 273.15) K = 298.15 K
Standard pressure is 1 bar.
c) potassium, K(s)
b) manganese, Mn(s)
d) boron, B(s)
Answer:
The molar mass of:
Helium = 4.00 g/mol
Potassium = 39.0983 g/mol
Manganese = 54.94 g/mol.
Boron = 10.81 g / mol
Explanation:
Helium = 4.00 g/mol
Potassium = 39.0983 g/mol
Manganese = 54.94 g/mol.
Boron = 10.81 g / mol
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Answer:
Endangered. The meaning of endangered is when a species declines in number- this might be because they are being eaten by another animal which has a higher popupayion, or perhaps because they are being hunted by humans for their meat or fur. Neither reason might be that their food or habitat has been destroyed b humans to create land for farming or housing. one example of name endangered species is Polar Bears: they like in the Antarctic where there is no land, simply ice. Due to humans causing climate change, the world is warming up and the ice is melting. This means polar bears are dying out because it would be too tiring to swim to find ice. For the same reason, other animals are dying in the Antarctic meaning that Poplar Bears have less to eat (being at the top of the food chain -no animal eats polar bears).
Answer:
Explanation:
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In this case, when ribose (C₅H₁₀O₅) yields carbon dioxide (CO₂) we write:
Which needs to be balanced by adding water and hydrogen ions:
You can also see that there are 20 transferred electrons, since the carbon atoms in the ribose have 0 as their oxidation state and the carbon atoms in the carbon dioxide have +4 as the oxidation state, thus, each carbon transfers 4 electrons, a five carbon atoms transfer 20 electrons overall.
In such a way, since the carbon is increasing its oxidation state, such half reaction is an oxidation half reaction.
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The mass percentage of the water in hydrated magnesium sulfate (MnSO4 . H2O) is 10.6%.
The percentage mass is the ratio of the mass of the element or molecule in the given compound.
The percentage can be given as:
The mass of the water is 18.02 g/mol and the molar mass of hydrated magnesium sulfate (MnSO4 . H2O) is 169.03 g/mol.
Thus,
Therefore, the mass percentage of the water in hydrated magnesium sulfate (MnSO4 . H2O) is 10.6%.
Learn more about percentage mass:
Answer:
10.6%
Explanation:
The determined percent mass of water can be calculated from the formula of the hydrate by
dividing the mass of water in one mole of the hydrate by the molar mass of the hydrate and
multiplying this fraction by 100.
Manganese(ii) sulphate monohydrate is MnSO4 . H2O
1. Calculate the formula mass. When determining the formula mass for a hydrate, the waters of
hydration must be included.
1 Manganes 52.94 g = 63.55 g
1 Sulphur 32.07 g =
32.07 g 2 Hydrogen is = 2.02 g
4 Oygen =
64.00 g 1 Oxygen 16.00 = 16.00 g
151.01 g/mol 18.02 g/mol
Formula Mass = 151.01 + (18.02) = 169.03 g/mol
2. Divide the mass of water in one mole of the hydrate by the molar mass of the hydrate and
multiply this fraction by 100.
Percent hydration = (18.02 g /169.03 g) x (100) = 10.6%
The final result is 10.6% after the two steps calculations