We know that density = mass /volume
so
mass=volume*density
volume=mass/density
so c is wrong
Answer:
T2 =21.52°C
Explanation:
Given data:
Specific heat capacity of sample = 1.1 J/g.°C
Mass of sample = 385 g
Initial temperature = 19.5°C
Heat absorbed = 885 J
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = Final temperature - initial temperature
885J = 385 g× 1.1 J/g.°C×(T2 - 19.5°C )
885 J = 423.5 J/°C× (T2 - 19.5°C )
885 J / 423.5 J/°C = (T2 - 19.5°C )
2.02°C = (T2 - 19.5°C )
T2 = 2.02°C + 19.5°C
T2 =21.52°C
Hey there!:
Given the mass of PbCl(OH) :
0.135 Kg = 0.135 Kg*(1000g / 1Kg) = 135 g
Molecular mass of PbCl(OH) = 207+35.5+16+1 = 259.5 g / mol
Atomic mass of Pb = 207 g/mol
Hence mass of Pb in 135 g PbCl(OH) :
(207 g Pb / 259.5 g PbClOH) * 135g PbClOH =
0.79768 * 135 => 107.68 g of Pb
For Pb2Cl2CO3 :
Given the mass of Pb2Cl2CO3 :
0.135 Kg = 0.135 Kgx(1000g / 1Kg) = 135 g
Molecular mass of Pb2Cl2CO3 = 2*207+2*35.5+12+3*16 = 545 g / mol
Mass of Pb present in 1 mol (=545 g / mol) of Pb2Cl2CO3 = 2*207 = 414 g
Hence mass of Pb in 135 g Pb2Cl2CO3:
(414 g Pb / 545 g PbClOH) * 135g PbClOH =
0.75963 * 135 => 102.55 g of Pb2Cl2CO3
Hope that helps!
Hence, concentration of base is 1.17 M
Answer:
6 atm
Explanation:
Using the formula P1V1=P2V2
P1= Initial Pressure
V1= Initial Volume
P2= Final Pressure
V2= Final Volume
And knowing that at stp gas will always be at 1 atm
250L(P2) = 1500
P2= 6 atm
