Answer: picture shows work for # 1,2,4,5,7
Explanation:
number 3: as the pressure in the volume decreases, the volume increases causing it to expand and eventually blow.
number 6: because the temperature and the amount of gas don’t change, these terms don’t appear in the equation. What Boyle’s law means is that the volume of a mass of gas is inversely proportional to its pressure. This linear relationship between pressure and volume means doubling the volume of a given mass of gas decreases its pressure by half.
hope this helps :))
Answer:
21.5 g.
Explanation:
Hello!
In this case, since the reaction between the given compounds is:
We can see that according to the law of conservation of mass, which states that matter is neither created nor destroyed during a chemical reaction, the total mass of products equals the total mass of reactants based on the stoichiometric proportions; in such a way, we first need to compute the reacted moles of Li3P as shown below:
Now, the moles of Li3P consumed by 15 g of Al2O3:
Thus, we infer that just 0.29 moles of 0.73 react to form products; which means that the mass of formed products is:
Therefore, the total mass of products is:
Which is not the same to the reactants (53 g) because there is an excess of Li₃P.
Best Regards!
Answer:
1.327 g Ag₂CrO₄
Explanation:
The reaction that takes place is:
- 2AgNO₃(aq) + K₂CrO₄(aq) → Ag₂CrO₄(s) + 2KNO₃(aq)
First we need to <em>identify the limiting reactant</em>:
We have:
- 0.20 M * 50.0 mL = 10 mmol of AgNO₃
- 0.10 M * 40.0 mL = 4 mmol of K₂CrO₄
If 4 mmol of K₂CrO₄ were to react completely, it would require (4*2) 8 mmol of AgNO₃. There's more than 8 mmol of AgNO₃ so AgNO₃ is the excess reactant. <em><u>That makes K₂CrO₄ the limiting reactant</u></em>.
Now we <u>calculate the mass of Ag₂CrO₄ formed</u>, using the <em>limiting reactant</em>:
- 4 mmol K₂CrO₄ *
= 1326.92 mg Ag₂CrO₄
- 1326.92 mg / 1000 = 1.327 g Ag₂CrO₄
Answer:
= 9.28 g CO₂
Explanation:
First write a balanced equation:
CH₄ + 2O₂ -> 2H₂O + CO₂
Convert the information to moles
7.50g CH₄ = 0.46875 mol CH₄
13.5g O₂ = 0.421875 mol O₂
Theoretical molar ratio CH₄:O₂ -> 1:2
Actual ratio is 0.46875 : 0.421875 ≈ 1:1
If all CH₄ is used up, there would need to be more O₂
So O₂ is the limiting reactant and we use this in our equation
Use molar ratio to find moles of CO₂
0.421875 mol O₂ * 1 mol CO₂/2 mol O₂=0.2109375 mol CO₂
Then convert to grams
0.2109375 mol CO₂ = 9.28114 g CO₂
round to 3 sig figs
= 9.28 g CO₂
