Answer:
7200 kPa
Explanation:
Applying,
PV/T = P'V'/T'................ Equation 1
Where P = Initial pressure of neon gas, V = Initial volume of neon gas, T = Initial temperature of neon gas, P' = Final pressure of neon gas, V' = Final volume of neon gas, T' = Final Temperature of neon gas
Make P' the subject of the equation
P' = PVT'/V'T.............. Equation 2
Given: P = 900 kPa, V = 8.0 L, T = 300 K, V' = 2.0 L, T' = 600 K
Substitute these values into equation 2
P' = (900×8×600)/(2×300)
P' = 7200 kPa
Mass of Sulphur dioxide : 256 g
<h3>Further explanation</h3>
Given
Reaction
S + O2 --> SO2 *
Required
Mass of Sulphur dioxide
Solution
mol of Sulphur (Ar=32 g/mol) :
mol = mass : Ar
mol = 128 : 32
mol = 4
From the equation, mol ratio S : SO2 = 1 : 1, so mol SO2 = 4
Mass of SO2 :
mass = mol x MW SO2
mass = 4 x 64
mass = 256 g
Hey there!
<span>Use the equation of Clapeyron:
</span>
T in kelvin :
26 + 273.15 => 299.15 K
R = 0.082
V = 10.2 L
P = 0.98 atm
number of moles :
P *V = n * R * T
0.98 * 10.2 = n * 0.082 * 299.15
9.996 = n * 24.5303
n = 9.996 / 24.5303
n = 0.4074 moles
Therefore:
Molar mass H2O = 18.01 g/mol
1 mole H2O ------------- 18.01 g
0.4074 moles ----------- m
m = 0.4074 * 18.01 / 1
m = 7.339 g of H2O
Answer:-
27.38%
Explanation:-
Molecular weight of sodium hydrogen carbonate NaHCO3
= 23 x 1 + 1 x 1 + 12 x 1 + 16 x 3
= 84 gram
Total sodium Na present in NaHCO3 = 23 x 1 = 23 gram
Percentage of sodium present=
Total sodium x 100/ (Total molecular weight)
= 23 x 100 / 84
= 27.38 %
yes, because they are combined in different ratios.
