Answer:
Empirical formula: BH3
Molecular Formula: B2H6
Explanation:
To solve the exercise, we need to know how many boron atoms and how many hydrogen atoms the compound has. We know that of the total weight of the compound, 78.14% correspond to boron and 21.86% to hydrogen. As the weight of the compound is between 27 g and 28 g, using the above percentages we can solve that the compound has between 21.1 g and 21.8 g of boron, and between 5.9 g and 6.1 g of hydrogen:
100% _____ 27 g
78.14% _____ x = 78.14% * 27g / 100% = 21.1 g boron
100% ______27 g
21.86% ______ x = 21.86% * 27g / 100% = 5.9 g hydrogen
100% _____ 28 g
78.14% _____ x = 78.14% * 28g / 100% = 21.8 g boron
100% _____ 28g
21.86% _____ x = 21.86% * 28g / 100% = 6.1 g hydrogen
So, if the atomic weight of boron is 10.8 g, there must be two boron atoms in the compound that sum 21.6 g. The weight of hydrogen is 1 g, so the compound must have six hydrogen atoms.
The molecular formula represents the real amount of atoms that form a compound. Therefore, the molecular formula of the compound is B2H6.
The empirical formula is the minimum expression that represents the proportion of atoms in a compound. For example, ethane has 2 carbon atoms and 6 hydrogen atoms, so its molecular formula is C2H6, however, its empirical formula is CH3. Therefore, the empirical formula of the boron compound is BH3.
Because that solid is frozen liquid, so when it gets heated up, it will melt. Like Ice.
Hope I Helped! :)
The answer is C due to the rain making contact with the iron it creates moisture which cause the iron to begin oxidizing which causes the iron to rust which is a chemical reaction then the blade which was once malleable because it was heat becomes harden and rust is created as I said before the physical aspect of the iron chances due to rust developing on it it's no longer has the luster it once had
repartExiplanation:cionde atomos
Answer:
643g of methane will there be in the room
Explanation:
To solve this question we must, as first, find the volume of methane after 1h = 3600s. With the volume we can find the moles of methane using PV = nRT -<em>Assuming STP-</em>. With the moles and the molar mass of methane (16g/mol) we can find the mass of methane gas after 1 hour as follows:
<em>Volume Methane:</em>
3600s * (0.25L / s) = 900L Methane
<em>Moles methane:</em>
PV = nRT; PV / RT = n
<em>Where P = 1atm at STP, V is volume = 900L; R is gas constant = 0.082atmL/molK; T is absolute temperature = 273.15K at sTP</em>
Replacing:
PV / RT = n
1atm*900L / 0.082atmL/molK*273.15 = n
n = 40.18mol methane
<em>Mass methane:</em>
40.18 moles * (16g/mol) =
<h3>643g of methane will there be in the room</h3>
