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AlexFokin [52]
2 years ago
13

How many moles are in 5.25 L of oxygen gas as stp

Chemistry
1 answer:
Masteriza [31]2 years ago
8 0

Answer:

The correct answer is 0, 235 mol

Explanation:

We use the formula PV =nRT. The normal conditions of temperature and pressure are 273K and 1 atm, we use the gas constant = 0, 082 l atm / K mol:

1 atm x 5, 25l = n  x 0, 082 l atm / K mol x 273 K

n= 1 atm x 5, 25l /0, 082 l atm / K mol x 273 K

n= 0, 235 mol

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Sladkaya [172]

Answer:

2 Mg + O2 -> 2 MgO

is a synthesis reaction.

6 0
3 years ago
An aqueous basic solution has a concentration of 0. 050 m and kb is 4. 4 × 10^-4. What is the concentration of hydronium ion in
Alex_Xolod [135]

An aqueous basic solution has a concentration of 0. 050 m and kb is 4. 4 × 10^-4, the concentration of hydronium ion in this solution (m) is 2.234 × 10⁻¹² M.

Methylamine is an amine which is an organic weak base. Its chemical formula is CH₃NH₂. When it undergoes hydrolysis wherein water is acting as an acid, the reaction would be

CH₃NH₂ + H₂O ⇆  CH₃NH₃ + OH⁻

Then, we use the ICE analysis which stands for Initial-Change-Equilibrium.

CH₃NH₂ + H₂O ⇆  CH₃NH₃ + OH⁻

Initial                    0.05          -                 0         0

Change                 -x                              +x       +x

----------------------------------------------------------------------------

Equilibrium         0.05-x                           x          x

Then, we use the equation for the equilibrium constant of basicity.

Kb = [CH₃NH₃][OH⁻]/[CH₃NH₂] = 4.4×10⁻⁴

4.4×10⁻⁴ = [x][x]/[0.05-x]

[x] = 4.4756×10⁻³

Here, variable x denotes the number of moles of the substance which is involved in the reaction. Since the equilibrium amount of OH⁻ is equal to x, then the concentration of OH⁻ is also 4.4756×10⁻³. Thus,

pOH = -log[OH⁻]

pOH = -log[4.4756×10⁻³] = 2.35

The relationship between pOH and pH is that pH + pOH = 14. Thus,

pH = 14 - 2.35 = 11.65

pH = -log[H⁺]

11.65 = -log[H⁺]

[H⁺] = 2.234 × 10⁻¹² M

Thus, we find the concentration of solution is 2.234 × 10⁻¹² M.

Learn more about aqueous solution: brainly.com/question/11097800

#SPJ4

5 0
1 year ago
A sample of xenon gas occupies a volume of 6.80 L at 52.0°C and 1.05 atm. If it is desired to increase the volume of the gas sam
Pavlova-9 [17]

Answer:

207.03°C

Explanation:

The following data were obtained from the question:

V1 (initial volume) = 6.80 L

T1 (initial temperature) = 52.0°C = 52 + 273 = 325K

P1 (initial pressure) = 1.05 atm

V2 (final volume) = 7.87 L

P2 (final pressure) = 1.34 atm

T2(final temperature) =?

Using the general gas equation P1V1/T1 = P2V2/T2, the final temperature of the gas sample can be obtained as follow:

P1V1/T1 = P2V2/T2

1.05 x 6.8/325 = 1.34 x 7.87/T2

Cross multiply to express in linear form as shown below:

1.05 x 6.8 x T2 = 325 x 1.34 x 7.87

Divide both side by 1.05 x 6.8

T2 = (325 x 1.34 x 7.87) /(1.05 x 6.8)

T2 = 480.03K

Now, let us convert 480.03K to a number in celsius scale. This is illustrated below:

°C = K - 273

°C = 480.03 - 273

°C = 207.03°C

Therefore, the final temperature of the gas will be 207.03°C

5 0
3 years ago
Consider the following balanced equation:
algol [13]

Moles of PF₃ : 4

<h3>Further explanation</h3>

A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products.

Reaction

\tt P_4(s)+6F_2(g)\rightarrow 4PF_3(g)

1.25 moles of P₄(s) is reacted with 6 moles of F₂(g)

Limiting reactant : the smallest ratio (mol divide by coefficient)

P₄ : F₂ =

\tt \dfrac{1.25}{1}\div \dfrac{6}{6}=1.25\div 1\rightarrow F_2~limiting~reactant(smallest~ratio)

mol PF₃ based on mol of limiting reactant(F₂), so mol PF₃ :

\tt \dfrac{4}{6}\times 6~moles=4~moles

8 0
3 years ago
What Type of reaction is this equation <br> C2H5S + O2 ---&gt; CO2 + H2O + SO2
sweet-ann [11.9K]
I think it might be a decomposition.
4 0
3 years ago
Read 2 more answers
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