Answer:
The answer is SiO2
Explanation:
Silocon dioxide is written without a 1 after the silocon and with a 2 after the oxygen.
That is a chemical change.. Hope I helped!
Answer:
- The limiting reagent is N2O4
- 14,09g
Explanation:
- First, we adjust the reaction.
+ 
⇄
- Second, we assume that the participating moles are equal to the stoichiometric ratios because we do not know the amounts of the reagents.
We can determinate what is the limiting reagent comparing of product amounts which can be formed from each reactant.
Using 
to form 
Using to form 
The limiting reagent is N2O4, because can produce only 0, 783 mol of H2O.
This is the minimum measure can be formed of each product.
∴ 
Complete Question:
A chemist adds 55.0 mL of a 1.1M barium acetate (Ba(C2H3O2)2) solution to a reaction flask. Calculate the mass in grams of barium acetate the chemist has added to the flask. Round your answer to 2 significant digits.
Answer:
15 g
Explanation:
The concentration of the barium acetate is given in mol/L (M), thus, the number of moles (n) of it is the concentrantion multiplied by the volume (55.0 mL = 0.055 L):
n = 1.1 * 0.055
n = 0.0605 mol
The molar mass of the substance can be calculated by the sum of the molar mass of each element, which can be found at the periodic table. Thus:
Ba = 137.33 g/mol
C = 12.00 g/mol
H = 1.00 g/mol
O = 16.00 g/mol
Ba(C2H3O2)2 = 137.33 + 4*12 + 6*1 + 4*16 = 255.33 g/mol
The molar mass is the mass divided by the number of moles, thus the mass (m) is the molar mass multiplied by the number of moles.
m = 255.33 * 0.0605
m = 15.45 g
Rounded by 2 significant digits, m = 15 g.
