Answer : The half life of 28-Mg in hours is, 6.94
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 = 48.0 hr
a = initial amount of the reactant disintegrate = 53500
a - x = amount left after decay process disintegrate = 53500 - 10980 = 42520
Now put all the given values in above equation, we get


Now we have to calculate the half-life.



Therefore, the half life of 28-Mg in hours is, 6.94
Answer: The mole ratio of hydrogen to nitrogen is 3 mole: 1 mole, 3:1
Explanation:
•Mole ratios are determined using the coefficients of the substances in the balanced chemical equation. •Each coefficient represents the number of mole of each substance in the chemical reaction.
•The mole ratio can be determined by first writing out a balanced chemical equation for the reaction.
For this reaction the balanced chemical equation is
N2(g) + 3H2(g) ----> 2NH3(g)
1mol:3mol : 2mol
From the equation we can see that 1 mole of N2(g) reacts with 3 moles of H2(g) or 3 moles of H2(g) react with 1 mole of N2(g) to produce 2 moles of NH3(g).
Therefore, the mole ratio of hydrogen to nitrogen is 3 mole: 1 mole, 3:1
Ionic bonds are a metal and a non metal bond and a covalent bond is two no metals banded together.
Answer:
The mass of C2H2 in the mixture is 0.56gram using the ratio of carbon in the products contributed by the C2H2.
Explanation:
The balanced equation for the reaction is: C3H8 + 2C2H2 + 10O2 >> 7CO2 + 6H2O.
From the reaction, we know that the oxygen was in excess, this will make the Carbon sources the limiting agents in the reaction. The details of the reaction showed that the ratio of water to the carbon dioxide is 1.6:1. This also means that the expected mole of carbon dioxide will be 7/1.6, which is 3.75moles.
The individual balanced equation of reaction is:
C3H3 +5O2 >> 3CO2 + 4H2O
and 2C2H2 + 5O2 >>4CO2 + 2H2O. From this one can quickly tell that the propane is in sufficient supply as it produces 3 moles of CO2 out of the expected 3.75 moles obtained above. Leaving 0.75moles of CO2 to the ethyne.
The mass of ethyne in the mixture will therefore be: 0.75/3.75 X 2.8 = 0.56g.