Kia's remaining water has a mass of 50g. You can set it up as a proportion knowing that 100ml of water has a mass of 100g and thus 50ml of water would weight 50g
Answer:
3,85 g of Fe
Explanation:
1- The first thing to do is calculate the molar mass of the Fe2O3 compound. With the help of a periodic table, the weights of the atoms are searched, and the sum is made:
Molar mass of Fe2O3 = (2 x mass of Fe) + (3 x mass of O) = 2 x 55.88 g + 3 x 15.99 g = 159.65 g / mol
Then, one mole of Fe2O3 has a mass of 159.65 grams.
2- Then, the relationship between the Fe2O3 that will react and the iron to be produced. With the previous calculation, we can say that with one mole of Fe2O3, two moles of Fe can be produced. Passing this relationship to the molar masses, it would be as follows:
1 mole of Fe2O3_____ 2 moles of Fe
159.65 g of Fe2O3_____ 111.76 g of Fe
3- Finally, the calculation of the mass that can be produced of Fe is made, starting from 5.50 g of Fe2O3
159.65 g of Fe2O3 _____ 111.76 g of Fe
5.50 g of Fe2O3 ______ X = 3.85 g of Fe
<em>Calculation: 5.50 g x 111.76 g / 159.65 g = 3.85 g
</em>
The answer is that 3.85 g of Fe can be produced when 5.50 g of Fe2O3 react
Answer:
Mostly Para
Explanation:
First, let's assume that the molecule is the toluene (A benzene with a methyl group as radical).
Now the nitration reaction is a reaction in which the nitric acid in presence of sulfuric acid, react with the benzene molecule, to introduce the nitro group into the molecule. The nitro group is a relative strong deactiviting group and is metha director, so, further reactions that occur will be in the metha position.
Now, in this case, the methyl group is a weak activating group in the molecule of benzene, and is always ortho and para director for the simple fact that this molecule (The methyl group) is a donor of electrons instead of atracting group of electrons. Therefore for these two reasons, when the nitration occurs,it will go to the ortho or para position.
Now which position will prefer to go? it's true it can go either ortho or para, however, let's use the steric hindrance principle. Although the methyl group it's not a very voluminous and big molecule, it still exerts a little steric hindrance, and the nitro group would rather go to a position where no molecule is present so it can attach easily. It's like you have two doors that lead to the same place, but in one door you have a kid in the middle and the other door is free to go, you'll rather pass by the door which is free instead of the door with the kid in the middle even though you can pass for that door too. Same thing happens here. Therefore the correct option will be mostly para.
The reaction for the combustion of methane can be expressed as follows.
CH4 + 2O2 --> CO2 + 2H2O
We solve first for the amount of carbon dioxide in moles by dividing the given volume by 22.4L which is the volume of 1 mole of gas at STP.
moles of CO2 = (5.6 L) / (22.4 L/1 mole)
moles of CO2 = 0.25 moles
Then, we can see that every mole of carbon dioxide will need 1 mole of methane
moles methane = (0.25 moles CO2) x (1 moles O2/1 mole CO2)
= 0.25 moles CH4
Then, multiply this by the molar mass of methane which is 16 g/mole. Thus, the answer is 4 grams methane.
