Answer:
c =0.2 J/g.°C
Explanation:
Given data:
Specific heat of material = ?
Mass of sample = 12 g
Heat absorbed = 48 J
Initial temperature = 20°C
Final temperature = 40°C
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
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 = 40°C -20°C
ΔT = 20°C
48 J = 12 g×c×20°C
48 J =240 g.°C×c
c = 48 J/240 g.°C
c =0.2 J/g.°C
Answer: has properties similar to other elements in group 18, does not react readily with other elements, is part of the noble gas group
Explanation: I’ve done on edg before
Answer:
Number of moles = 0.042 mol
Explanation:
Given data:
Number of moles = ?
Mass of calcium carbonate = ?
Solution:
Formula:
Number of moles = mass/ molar mass
now we will calculate the molar mass of calcium carbonate.
atomic mass of Ca = 40 amu
atomic mass of C = 12 amu
atomic mass of O = 16 amu
CaCO₃ = 40 + 12+ 3×16
CaCO₃ = 40 + 12+48
CaCO₃ = 100 g/mol
Now we will calculate the number of moles.
Number of moles = 4.15 g / 100 g/mol
Number of moles = 0.042 mol
Answer:
no it will have no charge...it would be electrically neutral! because the number of protons and electrons are equal!
Answer:
1) 1.15 mol
2) M=0.45
3) 22.5 mL
4) 6.25 mL
Explanation:
1)
550 mL= 0.55 L
M= mol solute/ L solution
mol solute= M * L solution
mol solute= (2.1 M * 0.55 L ) M=1.15 mol solute
2)
155 mL = 0.155 L
80 g -> 1 mol NH4NO3
5.61 g -> x
x= (5.61 g * 1 mol NH4NO3)/80 g x= 0.07 mol NH4NO3
M=(0.07 mol NH4NO3)/0.155 L M=0.45
3) M1V1=M2V2
V1= M2V2/M1
V1= (0.500 M * 0.225 L)/5.00 M V1=0.0225 L =22.5 mL
4) M1V1=M2V2
V1= M2V2/M1
V1= (0.25 M * 0.45 L)/ 18.0 M
V1=6.25 x 10^-3 L = 6.25 mL
