Answer : The half-life of the compound is, 145 years.
Explanation :
First we have to calculate the rate constant.
Expression for rate law for first order kinetics is given by:
where,
k = rate constant = ?
t = time passed by the sample = 60.0 min
a = let initial amount of the reactant = 100 g
a - x = amount left after decay process = 100 - 25 = 75 g
Now put all the given values in above equation, we get
Now we have to calculate the half-life of the compound.
Therefore, the half-life of the compound is, 145 years.
Answer:
C₆H₈O₆
Explanation:
The molecular formula of given compound is C₆H₈O₆. Because there are six carbon, six oxygen and eight hydrogen atom present in this compound.
The molecular weight of following compound is,
C₆H₈O₆
C = 12 amu
H = 1 amu
O = 16 amu
C₆H₈O₆ = 12×6 + 1×8 + 16×6
C₆H₈O₆ = 72 +8 + 96
C₆H₈O₆ = 176 g/mol
Answer : The amount of energy needed to raise one gram of a substance one degree Celsius is a characteristic property known as, Specific heat capacity.
Explanation:
Heat capacity : It is defined as the heat required to raise the temperature by one degree.
Specific heat capacity : It is defined as the amount of heat required by one gram of a substance to raise its temperature by one degree Celsius.
The heat required to change the temperature of a substance is related to heat capacity of the substance by the expression as :
where,
C = specific heat capacity
m = mass of a substance
q = heat required
= change in temperature of substance
Hence, the amount of energy needed to raise one gram of a substance one degree Celsius is a characteristic property known as, Specific heat capacity.
Answer:
Option B. 4.25×10¯¹⁹ J
Explanation:
From the question given above, the following data were obtained:
Frequency (f) = 6.42×10¹⁴ Hz
Energy (E) =?
Energy and frequency are related by the following equation:
Energy (E) = Planck's constant (h) × frequency (f)
E = hf
With the above formula, we can obtain the energy of the photon as follow:
Frequency (f) = 6.42×10¹⁴ Hz
Planck's constant (h) = 6.63×10¯³⁴ Js
Energy (E) =?
E = hf
E = 6.63×10¯³⁴ × 6.42×10¹⁴
E = 4.25×10¯¹⁹ J
Thus, the energy of the photon is 4.25×10¯¹⁹ J
