Answer:
It has two isomers; n-butane and isobutane. Here n-butane is a straight-chain compound with four carbon atoms bonded with single covalent bonds. Explanation: Butane is an alkane with four carbon atoms so molecular formula is C4H10.
Explanation:
Its hybridization would be sp because Be only has 2 covalent bonds with Cl
Answer:
what can you speak English
The full question asks to decide whether the gas was a specific gas. That part is missing in your question. You need to decide whether the gas in the flask is pure helium.
To decide it you can find the molar mass of the gas in the flask, using the ideal gas equation pV = nRT, and then compare with the molar mass of the He.
From pV = nRT you can find n, after that using the mass of gass in the flask you use MM = mass/moles.
1) From pV = nRT, n = pV / RT
Data:
V = 118 ml = 0.118 liter
R = 0.082 atm*liter/mol*K
p = 768 torr * 1 atm / 760 torr = 1.0105 atm
T = 35 + 273.15 = 308.15 K
n = 1.015 atm * 0.118 liter / [ 0.082 atm*liter/K*mol * 308.15K] =0.00472 mol
mass of gas = mass of the fask with the gas - mass of the flasl evacuated = 97.171 g - 97.129 g = 0.042
=> MM = mass/n = 0.042 / 0.00472 = 8.90 g/mol
Now from a periodic table or a table you get that the molar mass of He is 4g/mol
So the numbers say that this gas is not pure helium , because its molar mass is more than double of the molar mass of helium gas.
2 C₃H₇OH (l) + 9 O₂ (g) → 6 CO₂ (g) + 8 H₂O (g)
Explanation:
To balance the chemical equation the number of atoms of each element entering the reaction have to be equal to the number of atoms of each element leaving the reaction, in order to conserve the mass.
Bellow we have the balanced chemical equation of the complete combustion of C₃H₇OH:
C₃H₇OH (l) + (9/2) O₂ (g) → 3 CO₂ (g) + 4 H₂O (g)
to have integer coefficients we multiply the reaction with 2:
2 C₃H₇OH (l) + 9 O₂ (g) → 6 CO₂ (g) + 8 H₂O (g)
where:
(l) - liquid
(g) - gaseous
Learn more about:
combustion reaction
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