Answer: 16.32 g of 
as excess reagent are left.
Explanation:
To calculate the moles :
According to stoichiometry :
2 moles of 
require = 1 mole of
Thus 0.34 moles of will require=
of
Thus is the limiting reagent as it limits the formation of product and is the excess reagent.
Moles of left = (0.68-0.17) mol = 0.51 mol
Mass of 
Thus 16.32 g of as excess reagent are left.
Answer:
A. percentage mass of iron = 5.17%
percentage mass of sand = 8.62%
percentage mass of water = 86.205%
B. (Iron + sand + water) -------> ( iron + sand) ------> sand
C. The step of separation of iron and sand
Explanation:
A. Percentage mass of the mixtures:
Total mass of mixture = (15.0 + 25.0 + 250.0) g =290.0 g
percentage mass of iron = 15/290 * 100% = 5.17%
percentage mass of sand = 25/290 * 100% = 8.62%
percentage mass of water = 250/290 * 100% = 86.205%
B. Flow chart of separation procedure
(Iron + sand + water) -------> separation by filtration using filter paper and funnel to remove water --------> ( iron + sand) -----------> separation using magnet to remove iron ------> sand
C. The step of separation of iron and sand by magnetization of iron will have the highest amount of error because during the process, some iron particles may not readily be attracted to the magnet as they may have become interlaced in-between sand grains. Also, some sand particle may also be attracted to the magnet as they are are borne on iron particles.
Answer:
0.07172 L = 7.172 mL.
Explanation:
- We can use the general law of ideal gas: <em>PV = nRT.
</em>
where, P is the pressure of the gas in atm (P = 1.0 atm, Standard P).
V is the volume of the gas in L (V = ??? L).
n is the no. of moles of the gas in mol (n = 3.2 x 10⁻³ mol).
R is the general gas constant (R = 0.0821 L.atm/mol.K),
T is the temperature of the gas in K (T = 273 K, Standard T).
<em>∴ V = nRT/P =</em> (3.2 x 10⁻³ mol)(0.0821 L.atm/mol.K)(273 K)/(1.0 atm) = <em>0.07172 L = 7.172 mL.</em>
Answer:
1st answer: A. nitrogen
2nd answer: A. Sodium carbonate
Explanation:
