Malleable, shiny and good conductors
A B E
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>
Covalent since they are both non metals, a metal and a non metal make an ionic bond
The question is incomplete. Complete question is attached below:
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Answer:
Given: conc. of HBr = 1.4 M
Volume of HBr = 15.4 mL
Volume of KOH = 22.10 mL
We know that, M1V1 = M2V2
(HBr) (KOH)
Therefore, M2 = M1V1/V2
= 1.4 X 15.4/22.10
= 0.9756 M
Concentration of KOH is 0.9756 M.