1) 2700 kg/l
2) 13.6 kg/l
3) 0.1578 kg
4) 8921.5 kg/m3
5) 1.59 kg/l
6) 1.84 kg/l
7) 0.21965 kg
8) 11331.9 kg/m3
9) 7.9167 kg/l
10) 238.095 cm3
Just divide the masses by volume to find out the density, multiply the volume with density to find out the mass and divide the mass by density to find out the volume.
To turn the result into SI unit (kg/l), divide the g by 1000 and ml by 1000.
Answer:
The mole is important because it allows chemist to work with a subatomic world with macro world units and amount. Atoms molecules and formula units are very small and very difficult to work with usually. However the mole allows a chemist to work with amount large enough to use.
Answer:
The percent by mass of water in this crystal is:
Explanation:
This exercise can be easily solved using a simple rule of three where the initial weight of the hydrated crystal (6,235 g) is taken into account as 100% of the mass, and the percentage to which the mass of 4.90 g corresponds (after getting warm). First, the values and unknown variable are established:
- 6,235 g = 100%
- 4.90 g = X
And the value of the variable X is found:
- X = (4.90 g * 100%) / 6,235 g
- X = approximately 78.6%.
The calculated value is not yet the percentage of the water, since the water after heating the glass has evaporated, therefore, the remaining percentage must be taken, which can be calculated by subtraction:
- Water percentage = Total percentage - Percentage after heating.
- <u>Water percentage = 100% - 78.6% = 21.4%</u>
Answer:
a) Limiting: sulfur. Excess: aluminium.
b) 1.56g Al₂S₃.
c) 0.72g Al
Explanation:
Hello,
In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:
a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:
Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.
b) By stoichiometry, the produced grams of aluminium sulfide are:
c) The leftover is computed as follows:
NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.
Best regards.
As the question tells you, you need to use the formula
% mass= mass of solute/ mass of solution x 100
mass solute= 30.0 g
mass of solution= 30.0 + 270.0= 300.0 g
% mass= 30.0/ 300.0 x 100= 10%
answer is B
