Answer:
distance and time are needed to find velocity
In general, we have this rate law express.:
![\mathrm{Rate} = k \cdot [A]^x [B]^y](https://tex.z-dn.net/?f=%5Cmathrm%7BRate%7D%20%3D%20k%20%5Ccdot%20%5BA%5D%5Ex%20%5BB%5D%5Ey)
we need to find x and y
ignore the given overall chemical reaction equation as we only preduct rate law from mechanism (not given to us).
then we go to compare two experiments in which only one concentration is changed
compare experiments 1 and 4 to find the effect of changing [B]
divide the larger [B] (experiment 4) by the smaller [B] (experiment 1) and call it Δ[B]
Δ[B]= 0.3 / 0.1 = 3
now divide experiment 4 by experient 1 for the given reaction rates, calling it ΔRate:
ΔRate = 1.7 × 10⁻⁵ / 5.5 × 10⁻⁶ = 34/11 = 3.090909...
solve for y in the equation
![\Delta \mathrm{Rate} = \Delta [B]^y](https://tex.z-dn.net/?f=%5CDelta%20%5Cmathrm%7BRate%7D%20%3D%20%5CDelta%20%5BB%5D%5Ey)

To this point,
![\mathrm{Rate} = k \cdot [A]^x [B]^1](https://tex.z-dn.net/?f=%5Cmathrm%7BRate%7D%20%3D%20k%20%5Ccdot%20%5BA%5D%5Ex%20%5BB%5D%5E1%20)
do the same to find x.
choose two experiments in which only the concentration of B is unchanged:
Dividing experiment 3 by experiment 2:
Δ[A] = 0.4 / 0.2 = 2
ΔRate = 8.8 × 10⁻⁵ / 2.2 × 10⁻⁵ = 4
solve for x for
![\Delta \mathrm{Rate} = \Delta [A]^x](https://tex.z-dn.net/?f=%5CDelta%20%5Cmathrm%7BRate%7D%20%3D%20%5CDelta%20%5BA%5D%5Ex)

the rate law is
Rate = k·[A]²[B]
Answer
Hi, the correct answer is,
B. It is an acid-base reaction because the sodium hydroxide base reacts with the hydrochloric acid to form a salt and water
Explanation
When an acid reacts with a base, the products are salt and water.In this case, sodium hydroxide is a base which contains hydroxide ions, OH-, whereas hydrochloric acid is an acid that has Hydrogen ions, H+.When they react, they form sodium chloride, NaCl in aqueous solution as the salt , and additional water molecules. The equation is;
NaOH (aq) + HCl (aq) → NaCl (aq) + H2O (l)
Answer:
7.568g of CO₂ will be produced
Explanation:
Based on the reaction:
C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
<em>Where 1 mole of C₃H₈ react with 5 moles of O₂ to produce 3 moles of CO₂ and 4 moles of H₂O</em>
Moles of C₃H₈ and O₂ are:
C₃H₈: 39.0g ₓ (1mol / 44.1g) = 0.884 moles
O₂: 11.0g ₓ (1mol / 32g) = 0.344 moles
For a complete reaction of 0.884 moles of C₃H₈ are necessaries:
0.884 mol C₃H₈ ₓ (6 mol O₂ / 1 mol C₃H₈) = 5.304 moles of O₂
As you have just 0.344 moles of O₂, it is the limiting reactant.
And moles of CO₂ produced are:
0.344 mol O₂ ₓ (3 mol CO₂ / 6 mol O₂) = <em>0.172 moles of CO₂</em>
Thus, mass in grams of CO₂ (44g / mol) is:
0.172 moles of CO₂ ₓ (44g / mol) = <em>7.568g of CO₂ will be produced</em>