For a first order reaction, the half life is inversely proportional to the rate constant.
The formula is
half life = ln(2)/k = 0.693/k
where k is the rate constant
t = 5.50 minutes
k = ln(2)/5.50 = 0.126 min^-1
Your rate constant is 0.126 min^-1.
Answer:
3) Neither created nor destroyed. Only converted.
Explanation:
We can use the ideal gas law equation to find the volume occupied by oxygen gas
PV = nRT
where ;
P - pressure - 52.7 kPa
V - volume
n - number of oxygen moles - 12.0 g / 32 g/mol = 0.375 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature - 25 °C + 273 = 298 K
substituting the values in the equation
52 700 Pa x V = 0.375 mol x 8.314 Jmol⁻¹K⁻¹ x 298 K
V = 17.6 L
volume of the gas is 17.6 L
Read the paper more carefully if you don't understand it
Answer:
0.85 mole
Explanation:
Step 1:
The balanced equation for the reaction of CaCl2 to produce CaCO3. This is illustrated below:
When CaCl2 react with Na2CO3, CaCO3 is produced according to the balanced equation:
CaCl2 + Na2CO3 -> CaCO3 + 2NaCl
Step 2:
Conversion of 85g of CaCO3 to mole. This is illustrated below:
Molar Mass of CaCO3 = 40 + 12 + (16x3) = 40 + 12 + 48 = 100g/mol
Mass of CaCO3 = 85g
Moles of CaCO3 =?
Number of mole = Mass /Molar Mass
Mole of CaCO3 = 85/100
Mole of caco= 0.85 mole
Step 3:
Determination of the number of mole of CaCl2 needed to produce 85g (i.e 0. 85 mole) of CaCO3.
This is illustrated below :
From the balanced equation above,
1 mole of CaCl2 reacted to produced 1 mole of CaCO3.
Therefore, 0.85 mole of CaCl2 will also react to produce 0.85 mole of CaCO3.
From the calculations made above, 0.85 mole of CaCl2 is needed to produce 85g of CaCO3