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Paha777 [63]
3 years ago
13

PLEASE HELP, WILL GIVE BRAINLIEST!

Chemistry
1 answer:
Over [174]3 years ago
5 0
Hi,
The purpose of the burning splint is to provide activation energy.
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An air mattress is filled with 16.5 moles of air. The air inside the mattress has a temperature of 295 K and a gauge pressure of
postnew [5]
PV=nRT
3.5×X=16.5×0.082×295
X= 114 L
The volume of the air mattress is 114 liters.
7 0
3 years ago
Read 2 more answers
Calculate the equilibrium constant for the reaction between fe2+(aq) and zn(s) under standard conditions at 25∘c.
Hatshy [7]
Following reaction occurs in the given electrochemical system:
Fe^{2+} + Zn → Fe + Zn^{2+}
Thus, under standard conditions
E(0) = E(0) Fe2+/Fe - E(0) Zn2+/Zn
where, E^{0}Fe2+/Fe = standard reduction potential of Fe2+/Fe = -0.44 v
E^{0}Zn2+/Zn = standard reduction potential of Zn2+/Zn = -0.763 v

E(0) = 0.323 v
now, we know that, ΔG(0) =-nFE(0) ............... (1)
Also, ΔG^{0} = -RTln(K) ............ (2)

On equating and rearranging equation 1 and 2, we get
K = exp( \frac{nFE(0)}{RT} )= exp (\frac{2X96500X0.323}{8.314X298}) = 8.46 x 10^{10}

7 0
2 years ago
What would be the final volume of the new solution if the 0.2 m solution on the left were diluted to 0.04 m? ml quizley?
marin [14]
Missing question: volume of <span>solution on the left is 10 mL.
V</span>₁(solution) = 10 Ml.
c₁(solution) = 0.2 M.<span>
V</span>₂(solution) = ?.<span>
c</span>₂(solution) = 0.04 M.<span>
c</span>₁ - original concentration of the solution, before it gets diluted.<span>
c</span>₂ - final concentration of the solution, after dilution.<span>
V</span>₁ - <span>volume to be diluted.
V</span>₂ - <span>final volume after dilution.
c</span>₁ · V₁ = c₂ · V₂<span>.
</span>10 mL · 0.2 M = 0.04 M · V₂.
V₂(solution) = 10 mL · 0.2 M  ÷ 0.04 M.
V₂(solution) = 50 mL.<span>

</span>
3 0
3 years ago
A gas has a volume of 3.25 liters at 54 C and 231 kPa of pressure. At what temperature will the same gas take up 4.35 liters of
Firdavs [7]

Answer: 318 K

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 231 kPa

P_2 = final pressure of gas = 168 kPa

V_1 = initial volume of gas = 3.25 L

V_2 = final volume of gas = 4.35 L

T_1 = initial temperature of gas = 54^oC=273+54=327K

T_2 = final temperature of gas = ?

Now put all the given values in the above equation, we get:

\frac{231\times 3.25}{327}=\frac{168\times 4.35}{T_2}

T_2=318K

At 318 K of temperature will the same gas take up 4.35 liters of space and have a pressure of 168 kPa

4 0
3 years ago
Assume that 12.0 g of oxygen are reacted with 20.0 g of magnesium to produce magnesium oxide.
FinnZ [79.3K]
Write out the eqn of magnesium and oxygen. this should be under “metals” chapter. do revise.

next, find the mols of both oxygen and magnesium. compare the ratios and find the LIMITING REAGENT.

use the mols of the limiting reagent to compare with the mols of the product.

take the mols of the product/mr of the product.

this will give u the mass.
8 0
3 years ago
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