Counting gives an exact number and exact numbers have infinite sig figs.
Answer:
The atomic mass of the boron atom would be <em>10.135</em>
Explanation:
This is generally known as relative atomic mass.
Relative atomic mass or atomic weight is a physical quantity defined as the ratio of the average mass of atoms of a chemical element in a given sample to the atomic mass of 1/12 of the mass of a carbon-12 atom. Since both quantities in the ratio are masses, the resulting value is dimensionless; hence the value is said to be relative and does not have a unit.
<em>Note that the relative atomic mass of atoms is not always a whole number because of it being isotopic in nature.</em>
- <em>Divide each abundance by 100 then multiply by atomic mass</em>
- <em>Do that for each isotope, then add the two result. Thus</em>
Relative atomic mass of Boron = (18.5/100 x 11) + (81/100 x 10)
= 2.035 + 8.1
= 10.135
You have to use everything that is given since you have to know which is the limiting reactant. We find the limiting reactant by calculating the number of moles of each reactant and compare the number of moles. The limiting reactant would be the one that is consumed fully by the reaction.
Answer : The resulting solution will have a pH of 7.
Explanation:
Whenever acid and base reacts with each other to form water molecule is called Neutralization reaction. This water molecules is formed from the hydronium ion (
) from acid and hydroxide ion(
) from base.
When the equal volumes of acid and base of equal strength are combined, the resulting pH of the solution becomes 7.
The pH 7 value means that the solution is neutral.
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>
