Bronsted lowry bases
NO2- amd OH-
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:
d
Explanation:
from you know the atomic number it should be easy to find
Answer:
See explanation
Explanation:
Let us recall that the basic rule in writing balanced chemical reaction equations is that the number of atoms of each element on the right hand side of the reaction equation is the same of the number of atoms of the same element on the left hand side of the reaction equation.
The reaction of red hot iron and steam is written as follows;
3Fe + 4H2O → Fe3O4 + 4H2.
The decomposition reaction of ammonium dichromate is written as;
(NH4)2Cr2O7 → N2 + Cr2O3 + 4H2O
Reaction of aluminium, sodium hydroxide and water is as follows,
2Al + 2NaOH + 2H2O ----> 2NaAlO2 + 3H2
Reaction of potassium bicarbonate with sulphuric acid;
2KHCO3 + H2SO4 -------> K2SO4 + 2H2O + 2CO2
Reaction of zinc and sodium hydroxide is as follows;
Zn + 2NaOH→Na2ZnO2 + H2
Answer:
33.95 grams of NaN3
Explanation:
Number of moles of NaN3 = mass (m)/MW = m/65 mole
I mole of NaN3 requires 22.4L air bag
m/65 mole of NaN3 required 11.7L
22.4m/65 = 11.7
22.4m = 65×11.7
22.4m = 760.5
m = 760.5/22.4 = 33.95grams of NaN3
