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3 years ago

El superóxido de potasio, KO2, se emplea en máscaras de respiración para generar

Chemistry

1 answer:

Orlov [11]3 years ago

4 0

Answer:

Reactivo límite: Superóxido de potasio.

Moles de oxígeno producidas: $n_{O_2}=0.11molO_2$

Explanation:

Hola,

En este caso, considerando la reacción química llevada a cabo:

$4KO_2(s) + 2H_2O(l) \rightarrow 4KOH(s) + 3O_2(g)$

Es posible identificar el reactivo límite calculando las moles de superóxido de potasio que serían consumidas por 0.10 mol de agua por medio de la relación molar 4 a 2 que hay entre ellos:

$n_{KO_2}^{consumido\ por\ agua}=0.10molH_2O*\frac{4molKO_2}{2molH_2O} =0.2molKO_2$

Así, dado que solo hay 0.15 mol the superóxido de potasio, podemos decir que este es el reactivo límite. Luego, calculamos las moles de oxígeno producidas, considerando la relación molar 4 a 3 que hay entre el superóxido y el oxígeno:

$n_{O_2}=0.15molKO_2*\frac{3molO_2}{4molKO_2} \\\\n_{O_2}=0.11molO_2$

Best regards.

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When the ph of an aqueous solution is changed from 1 to 2, the concentration of hydronium ions in the solution is.

kiruha [24]

Answer:

decreased by a factor of 10

Explanation:

pH is defined in such a way that;

pH= −log10(H)

Where H represents the concentration of Hydronium or Hydrogen ions

Given that pH is changed from 1 to 2,

By rearranging the above formula , we get 10−pH = H

if pH=1,H=10−1=0.1M
if pH=2,H=10−2=0.01M

Therefore, 0.1/0.01 = 10 and 0.1 > 0.01

Hence, the concentration of hydronium ions in the solution is decreased by a factor of 10

7 0

2 years ago

4. When 1.00 L of 1.00 M Ba(NO3)2 solution at 25.0˚C is mixed with 1.00 L of 1.00 M Na2SO4 solution at 25.0˚C in a calorimeter,

myrzilka [38]

Answer:

The final temperature of the mixture is 28.11 °C

Explanation:

Step 1: Data given

Volume of 1.00 M Ba(NO3)2 = 1.00 L

Temperature = 25.0 °C

Volume of 1.00 M Na2SO4 = 1.00 L

enthalpy change is – 26 kJ per mol BaSO4

The specific heat of water is 4.18 J/g ·˚C

the density of water is 1.00 g/mL

Step 2: The balanced equation

Ba(NO3)2(aq) + Na2SO4(aq) → 2NaNO3(aq) + BaSO4(s)

Step 3: Calculate the total volume

Total volume = 1.00 L + 1.00 L = 2.00 L = 2000 mL

Step 4: Calculate mass

Mass = volume * density

Mass = 2000 mL * 1g/mL

Mass = 2000 grams

Step 5: Calculate moles BaSO4 formed

For 1 mol Ba(NO3)2 we need 1 mol Na2SO4 to produce 1 mol BaSO4

There is no limiting reactant, both Ba(NO3)2 and Na2SO4 will be completely be consumed (1 mol). We'll have 1.0 mol of BaSO4 produced.

Step 6: Calculate Q

Q = - ΔH

ΔH is negative so the reaction is exothermic, what means the temperature increases

Q is always positive, so Q = 26kJ = 26000 J

Step 6: Calculate the heat transfer

Q= m*c*ΔT

⇒with Q = the heat transfer = TO BE DETERMINED

⇒with m =the mass of the solution = 2000 grams

⇒with c= the specific heat of the solution = 4.18 J/g°C

⇒with ΔT = the change of temperature = T2 - T1 = T2 - 25.0

26000 = 2000 * 4.18 * (T2 - 25.0 °C)

3.11 = T2 - 25.0 °C

T2 = 25.0 + 3.11 °C

T2 = 28.11 °C

The final temperature of the mixture is 28.11 °C

7 0

3 years ago

PLEASE HELP

Marysya12 [62]

Answer:

Military

Explanation:

I may be wrong but military seems most likely

6 0

2 years ago

Read 2 more answers

How does one calculate the pH of a solution? O A. The pH is the product of [H+] and [OH-]. O B. The pH is the negative log of [O

blagie [28]

Answer:

I believe it is A

Explanation:

pH = -log[H+] and pOH = -log[OH−].

8 0

2 years ago

In an experiment, 4.14 g of phosphorus combined with chlorine to produce 27.8 g of a white solid compound. what is the empirical

Umnica [9.8K]

Grams of Phosphorus = 4.14 grams
Grams of white compound = 27.8 grams
Grams of Chlorine would be = 27.8 - 4.14 = 23.66 grams
Calculating moles which would be grams / molar mass
Molar mass of P = 30.97 grams / moles; Molar mass of Cl = 35.45 grams / moles
Moles of Phosphorus = 4.14 grams / 30.97 grams / moles = 0.1337 moles
Moles of Chlorine = 23.66 grams / 35.45 grams / moles = 0.6674 moles
Calculating the ratios by dividing with the small entity
P = 0.1337 moles / 0.1337 moles = 1
Cl = 0.6674 moles / 0.1337 moles = 5
So the empirical formula would be PCl5

3 0

3 years ago