It decreases i guess. just make sure,though i'm mostly correct.
M=f/a m=129000N/84m/s/s=1535.714286kg/m/s/s
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
