So in order for us to know the percentage of sugar present in a 12.00 g of milk chocolate, what we are going to do is that, we just have to divide 8 by 12 and multiply in by 100 and we get 66.67. Therefore, the percentage of sugar present in 12.00 g of milk chocolate bar is 66.67%. Hope this answers your question. Have a great day!
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:
Final volume 30.513 L.
Explanation:
According to general gas equation:
P₁V₁/T₁ = P₂V₂/T₂
Given data:
Initial volume = 17 L
Initial pressure = 2.3 atm
Initial temperature = 299 K
Final temperature = 350 K
Final volume = ?
Final pressure = 1.5 atm
Solution:
P₁V₁/T₁ = P₂V₂/T₂
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 2.3 atm × 17 L × 350 K / 299 K × 1.5 atm
V₂ = 13685 atm .L. K / 448.5 K . atm
V₂ = 30.513 L
A student determines that 23.1 J of heat are required to raise the temperature of 6.67 g of an
Answer:
The resulting solution is basic.
Explanation:
The reaction that takes place is:
First we <u>calculate the added moles of HNO₃ and KOH</u>:
- HNO₃ ⇒ 12.5 mL * 0.280 M = 3.5 mmol HNO₃
- KOH ⇒ 5.0 mL * 0.920 M = 4.6 mmol KOH
As <em>there are more KOH moles than HNO₃,</em> the resulting solution is basic.
