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otez555 [7]
3 years ago
8

How many grams of oxygen will be required for sulfur to be burned to give 100.0 g of so2

Chemistry
1 answer:
klemol [59]3 years ago
4 0

Answer:

50.05 g.

Explanation

you welcome ;)

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Pls help me will mark Brianliest!!! I know the answer choice is a bit cut off. Topic is chem equilibrium
Svetlanka [38]

Answer:

See notes on LeChatlier's Principle I gave you yesterday.

Explanation:

Remember chemical see-saw => Removing Fe⁺³ makes the reactant side of the see-saw lighter causing the balance to tilt right then shift left to establish a new equilibrium with new concentration values. Such would result in a decrease in FeSCN⁺² concentration and increases in Fe⁺³ and SCN⁻ concentrations to replace the original amount of ppt'd Fe⁺³. => Answer Choice 'B' ... Also, see attached => Concentration effects on stability of chemical equilibrium .

6 0
3 years ago
An observation balloon was filled with 0.50 atmospheres of Helium, 54.0 mm Hg of Nitrogen, and 0.400 atmospheres of Hydrogen. Wh
noname [10]

Answer:

the pressure would be 0.9 atmospheres

Explanation:

you just gotta add the presures for each of the gases that are added

7 0
3 years ago
Hydrogen peroxide decomposes spontaneously to yield water and oxygen gas according to the reaction equation 2 H 2 O 2 ( aq ) ⟶ 2
Ede4ka [16]

Answer:

The value of temperature for unanalysed reaction is 456.1 K

Explanation:

Step 1: Data given

The activation energy of uncatalyzed reaction is 75.0 kJ/mol.

The activation energy of catalyzed reaction is 49.0 kJ/mol.

The temperature for catalyzed reaction is  25 °C

<u>Step 2:</u>  The balanced equation

2 H2O2(aq) ⟶ 2H2O(l) + O2(g)

<u>Step 3:</u> According to Arrhenius equation:

k = Ae ^ (-Ea/RT)

⇒ with k = the rate constant

⇒ with Ea = the activation energy = 49 kJ/mol

⇒ with R = the universal gas constant = 8.314 J/mol*K

⇒ with T = the temperature  = 298 K

For uncatalyzed reaction, the rate constant is calculated as:

k(uncatalyzed) = Ae ^ (-75000/8.314 J/mol*K * T)

k(uncatalyzed) = Ae ^ (-9020.93/T)

For catalyzed reaction, the rate constant is calculated as:

k(catalyzed) = Ae ^ (-Ea(catalyzed)/RT))

 ⇒ with Ea(catalyzed) = 49000 J/mol

 ⇒ with T = 298 K

k(catalyzed) = Ae ^ (-49000J/(8.314J/mol *K * 298K))

k(catalyzed) = Ae ^(-19.78)

k(catalyzed) = k(uncatalyzed)

Ae^(-19.78) = = Ae ^ (-9020.93/T)

-19.78 = -9020.93/T

T = -9020.93/-19.78

T = 456.1 K

The value of temperature for unanalysed reaction is 456.1 K

8 0
3 years ago
Read 2 more answers
A scientist has two solutions, which she has labeled solution a and solution
klio [65]
Let's make ;
A = ounces solution A
B = ounces solution B


total number of ounces: 
A + B = 120

concentration:
0.65A + 0.80B = 0.7(120)

 Substitute A for 120 - B in the second equation to combine them.

0.65(120 - B) + 0.80B = 84
78 - 0.65B + 0.80B = 84
0.15B = 84 - 78
0.15B = 6
B = 6/0.15
 
B = 40 ounces
Then... A = 120 - 40
A = 60 ounces

8 0
3 years ago
Calculation: If you have a pH of 5.5 for a weak acid with a pKa of 4.76, then is there more A- or more HA in the solution? Expla
devlian [24]

Answer:

lets set the ratio -A/HA as R:

pH = pKa + log(R,10) => pKa + log10(R)

pH = 5.5

pKa = 4.76

R => 10^(pH - 4.76)

10^(pH - 4.76) => 5.4954

Given R (-A/HA) a number bigger than 1, then the concentration of  -A is bigger than HA

Explanation:

6 0
3 years ago
Read 2 more answers
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