<h3>
Answer:</h3>
298.15 K
<h3>
Explanation:</h3>
W e are supposed to calculate the Value of K at 25°C
Assuming the value of K represent K, the question wants us to convert degree Celsius to Kelvin.
- To convert degrees Celsius to kelvin scale, we use the relationship;
- Kelvin (K) = Degrees Celsius + 273.15 ; 273.5 is a constant
- That is, to convert temperature from °C to Kelvin we add a constant of 273.15 to the °C given.
In this case;
Temperature is 273.15 °c
Thus, to Kelvin scale temperature will be;
= 25°C + 273.15
= 298.15 K
Therefore, the value of K, at 25°C is 298.15 K
Answer:
1kg/L
Explanation:
1.) convert grams to kilograms
1000g÷1000=1kg
2.)use formula to find density
= 1kg/1.0L
=1kg/L
Equation for Half life :
A = a(0.5)^(t/h)
A is current amount, "a" is initial amount, h is halflife, t is time
5 = 40(0.5)^(t/1.3x10^9)
5/40 = (0.5)^(t/1.3x10^9)
take the log of both sides , power rule
Log(5/40) = (t/1.3x10^9) * Log(0.5)
(1.3x10^9) * Log(5/40) / Log(0.5) = t
3.9x10^9 years = t
And if you think about what a half life is, the time it take for the amount to reduce to half.
40/2 = 20
20/2 = 10
10/2 = 5
It went through 3 half-lifes
3 * 1.3x10^9 = 3.9x10^9 years
Answer:
% purity of limestone = 96.53%
Explanation:
Question (4).
Weight of impure CaCO₃ = 25.9 g
Molecular weight of CaCO₃ = 40 + 12 + 3(16)
= 100 g per mole
We know at S.T.P. number of moles of CO₂ = 1 and volume = 22.4 liters
From the given reaction, 1 mole of CaCO₃ reacts with 1 mole or 22.4 liters of
CO₂.
∵ 22.4 liters of CO₂ was produced from CaCO3 = 100 g
∴ 1 liter of CO₂ will be produced by CaCO₃ = 
∴ 5.6 liters of CO₂ will be produced by CaCO₃ = 
= 25 g
Therefore, % purity of CaCO₃ = 
= 
= 96.53 %
