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: There are 
of gas are in a container with a volume of 9.55 mL at 35 °C and a pressure of 895 mmHg
Explanation:
According to ideal gas equation:
P = pressure of gas = 895 mm Hg= 1.18 atm (760 mm Hg= 1 atm)
V = Volume of gas = 9.55 ml = 0.00955 L (1 L=1000ml)
n = number of moles = ?
R = gas constant =
T =temperature =
Thus there are of gas are in a container with a volume of 9.55 mL at 35 °C and a pressure of 895 mmHg
Answer:
Neutron does not contain any charge because the charge of the quarks that made up the neutron balances each other out.
Hope it helps.
The formula of Iron(III) oxide is Fe2O3
In order to calculate the mass of iron in a given sample of iron(III) oxide, we must first know the mass percentage of iron in iron(III) oxide. This is calculated by:
[mass of iron in one mole of iron(III) oxide/ mass of one mole of iron(III) oxide] * 100
= [(moles of iron * Mr of iron) / (moles of Iron * Mr of Iron + moles of Oxygen * Mr of Oxygen)] * 100
= [(2 * 56) / (2 * 56 + 3 * 16)] * 100
= (112 / 160) * 100
= 70%
Thus, in a 100g sample, the weight of iron will be:
100 * 70%
= 70 grams
