Alkali Metals (Group 1) elements experience an increase in the vigour of their reaction in water as they go down the group (as the atomic number increase). As such the most reactive Alkali Metal would be
FRANCIUM, which is at the base of Group One.
Quite frankly, you do not want Francium to react with water- that's a huge explosion on your hand.
<h3><u>Answer</u>;</h3>
Actual yield = 46.44 g
<h3><u>Explanation;</u></h3>
1 mole of water = 18 g/mol
Therefore;
The experimental yield = 2.58 moles
equivalent to ; 2.58 × 18 = 46.44 g
The theoretical value is 47 g
Percentage yield = 46.44/47 × 100%
= 98.8%
The questions asks for actual yield = 46.44 g
Answer: 373 mL
Explanation:
Since there is no change in pressure, the formula: V / T = V / T can be used.
However, you must first convert the temperatures to Kelvin by adding 273 to them:
(19 + 273) = 292K and (90 + 273) = 363K.
Now, plug in: V / 292 = 464 / 363 → V = 373 mL :)
Stoichiomety:
1 moles of C + 1 mol of O2 = 1 mol of CO2
multiply each # of moles times the atomic molar mass of the compund to find the relation is weights
Atomic or molar weights:
C: 12 g/mol
O2: 2 * 16 g/mol = 32 g/mol
CO2 = 12 g/mol + 2* 16 g/mol = 44 g/mol
Stoichiometry:
12 g of C react with 32 g of O2 to produce 44 g of CO2
Then 18 g of C will react with: 18 * 32/ 12 g of Oxygen = 48 g of Oxygen
And the result will be 12 g of C + 48 g of O2 = 60 g of CO2.
You cannot obtain 72 g of CO2 from 18 g of C.
May be they just pretended that you use the law of consrvation of mass and say that you need 72 g - 18g = 54 g. But it violates the proportion of C and O2 in the CO2 and is not possible.
The natural environment or natural world encompasses all living and non-living things occurring naturally, meaning in this case not artificial.
