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tankabanditka [31]
3 years ago
13

If an adult inhales 750 L/ H of O2 at stop how many moles of O2 does an adult inhale in a 24 hour day

Chemistry
2 answers:
leva [86]3 years ago
6 0

Answer:

803.6 moles O₂

Explanation:

At STP 1 mole any gas occupies 22.4 Liters.

=> moles of O₂ in 750 Liters = (750/22.4) moles O₂ = 33.5 moles O₂/hour

moles O₂/24 hrs = 33.5 moles/hr  X  24 hr = 803.6 moles O₂

drek231 [11]3 years ago
4 0
803.6 moles, is your answer , hope this helped!!!!
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Elements listed in vertical columns of the periodic table:
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Vertical columns are called groups, or families. 

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3 years ago
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A container of N, 03(9) has a pressure of 0.490 atm. When the absolute temperature of the N, O2(g) is tripled, the gas
spayn [35]

Answer: 1.59atm

Explanation:

We have that for the Question "Calculate the final pressure of the gas mixture, assuming that the container volume does not change."

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The final pressure of the gas mixture, assuming that the container volume does not change =

From the question we are told

A container of N2O3(g) has a pressure of 0.265 atm. When the absolute temperature of the N2O3(g) is tripled, the gas completely decomposes, producing NO2(g) and NO(g).

3 0
2 years ago
You have a ballon filled with hydrogen gas which keeps it at a constant pressure, regardless the volume. The initial volume of t
abruzzese [7]

Answer:

619°C

Explanation:

Given data:

Initial volume of gas = 736 mL

Initial temperature = 15.0°C

Final volume of gas = 2.28 L

Final temperature = ?

Solution:

Initial volume of gas = 736 mL (736mL× 1L/1000 mL = 0.736 L)

Initial temperature = 15.0°C (15+273 = 288 K)

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

T₂ = T₁V₂/V₁  

T₂ = 2.28 L × 288 K / 0.736 L

T₂ = 656.6 L.K / 0.736 L

T₂ = 892.2 K

K to °C:

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7 0
3 years ago
a student prepares a dilute solution of sodium hydroxidem, NaOH (aq), starting with 6 M sodium hydroxide. She then titrates a 1.
Serjik [45]

Answer:

M = 0.3077 M

Explanation:

As I said in the comments, you are missing the required volume of the base to react with the KHP. I found this on another site, and the volume it used was 21.84 mL.

Now, KHP is a compound often used to standarize NaOH or KOH solutions. This is because it contains a mole ratio of 1:1 with the base, so it's pretty easy to use and standarize any base.

Now, as we are using an acid base titration, the general expression to use when a acid base titration reach the equivalence point would be:

n₁ = n₂   (1)

This, of course, if the mole ratio is 1:1. In the case of KHP and NaOH it is.

Now, we also know that moles can be expressed like this:

n = M * V   (2)

And according to this, we are given the volume of base and the required mass of KHP. So, if we want to know the concentration of the base, we need to get the moles of the KHP, because in the equivalence point, these moles are the same moles of base.

The reported molar mass of KHP is 204.22 g/mol, so the moles are:

n = 1.372 / 204.22 = 6.72x10⁻³ moles

Now, we will use expression (2) to get the concentration of the diluted base:

n = M * V

M = n / V

M = 6.72x10⁻³ / 0.02184

M NaOH = 0.3077 M

This is the concentration of the dilute solution of NaOH

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3 years ago
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Alina [70]
It will benefit the flower because it will always get Polonated but it can be bad for the bees because if the flower dies so will the bees
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