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a. t=0.553 s
b. vox(horizontal speed) = 3.62 m/s
<h3>Further explanation</h3>Given
h = 1.5 m
x = 2 m
Required
a. time
b. vo=initial speed
Solution
Free fall motion
a. h = 1/2 gt²(vertical motion=h=voyt+1/2gt²⇒voy = 0)
t = √2h/g
t = √2.1.5/9.8
t=0.553 s
b. x=vox.t(horizontal motion)
vox=x/t
vox=2/0.553
vox=3.62 m/s
Answer:
0.329 g
Explanation:
In the context of this problem, we have a chemical reaction between hydrochloric acid and calcium hydroxide. HCl is the acid here and calcium hydroxide is the base. Hence, we have an acid-base reaction, also known as neutralization reaction.
In a neutralization reaction, water is produced as a product, as well as a salt that we obtain after we exchange the cations: calcium bonds to chloride and hydrogen bonds to hydroxide (the latter is the formation of water). This means we also produce calcium chloride as a product. The overall reaction represents this as:
Firslt of all, we wish to find the number of moles of HCl present. Having molarity and volume, this is done by applying the molarity formula. It states that molarity is equal to the rate between moles and volume:
Rearranging for moles of HCl, n:
Based on stoichiometry of the balanced chemical equation, notice that 1 mole of calcium hydroxide reacts with 2 moles of HCl, meaning:
Now that we have the expression for moles, we may also express moles of calcium hydroxide as the ratio between its mass and molar mass:
Using the last two equations, we see that:
Substitute the given data, as well as the molar mass of calcium hydroxide:
Taking into account the reaction stoichiometry, 102 grams of Al₂O₃ are formed when 48 grams of O₂ react.
<h3>Reaction stoichiometry</h3>In first place, the balanced reaction is:
4 Al + 3 O₂ → 2 Al₂O₃
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- Al: 4 moles
- O₂: 3 moles
- Al₂O₃: 2 moles
The molar mass of the compounds is:
- Al: 27 g/mole
- O₂: 32 g/mole
- Al₂O₃: 102 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- Al: 4 moles ×27 g/mole= 108 grams
- O₂: 3 moles ×32 g/mole= 96 grams
- Al₂O₃: 2 moles ×102 g/mole= 204 grams
The following rule of three can be applied: if by reaction stoichiometry 96 grams of O₂ form 204 grams of Al₂O₃, 48 grams of O₂ form how much mass of Al₂O₃?
<u><em>mass of Al₂O₃= 102 grams</em></u>
Finally, 102 grams of Al₂O₃ are formed when 48 grams of O₂ react.
Learn more about the reaction stoichiometry:
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