The molar mass of a, b and c at STP is calculated as below
At STP T is always= 273 Kelvin and ,P= 1.0 atm
by use of ideal gas equation that is PV =nRT
n(number of moles) = mass/molar mass therefore replace n in the ideal gas equation
that is Pv = (mass/molar mass)RT
multiply both side by molar mass and then divide by Pv to make molar mass the subject of the formula
that is molar mass = (mass x RT)/ PV
density is always = mass/volume
therefore by replacing mass/volume in the equation by density the equation
molar mass=( density xRT)/P where R = 0.082 L.atm/mol.K
the molar mass for a
= (1.25 g/l x0.082 L.atm/mol.k x273k)/1.0atm = 28g/mol
the molar mass of b
=(2.86g/l x0.082L.atm/mol.k x273 k) /1.0 atm = 64 g/mol
the molar mass of c
=0.714g/l x0.082 L.atm/mol.K x273 K) 1.0atm= 16 g/mol
therefore the
gas a is nitrogen N2 since 14 x2= 28 g/mol
gas b =SO2 since 32 +(16x2)= 64g/mol
gas c = methaneCH4 since 12+(1x4) = 16 g/mol
Explanation:
kclo3- 39+35.5+16*3=122.5
molar mass = 122.5
mass = 14.0g
volume= 100cm3
solubility in mol/dm3 or molarity = ?
but we know that
no of moles = solubility in mol/dm3×vol in cm3
---------------------------------------------
1000
no of moles = mass
---------
molar mass
mass\molar mass= solubility in mol/dm3 *vol in cm3/ 1000
14.0/122.5=x * 100/1000
solubility =0.875 mol/dm3
<span>0.06355391 mol
The balanced equation for the reaction is
Na2B4O7*10H2O + 2 HNO3 = 2 NaNO3 + 4 H3BO3 + 5 H2O
So for each mole of Borax to neutralize, it takes 2 moles of HNO3.
Calculate number of moles of Borax
0.2619 g / 381.372 g/mol = 0.0006867 mol
Moles of HNO3 used = 0.0006867 mol * 2 = 0.0013734 mol
Molarity is defined as moles per liter so divide the number of moles used by the volume in liters. So
0.0013734 / 0.02161 = 0.06355391 mol</span>
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For an ideal gas, we use the ideal gas equation to relate pressure, volume and temperature changes. It is expressed as:
PV=nRT
For this case, we set V, n as constant.
P/T = nR/V = constant = k
P = kT
P1/T1 = P2/T2
P1/T1 = 2P1 / T2
T2 = 2P1(T1) / P1
T2 = 2T1
Therefore, in order to double the pressure of the system, we need to double the temperature as well at constant volume and number of moles.
