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Vlada [557]
3 years ago
5

Find the rms speed of the molecules of a sample of n2 (diatomic nitrogen) gas at a temperature of 32.5°c.

Chemistry
1 answer:
Korvikt [17]3 years ago
5 0
The rms speed of a gas can be calculated using the following rule:
Vrms = sqrt[(3RT) / M] where
R is the gas constant = 8.314 <span>J K^−1 mol^−1
T is the temperature = 32.5 + 273 = 305.5 degree kelvin
M is the molar mass = 2*14 = 28 grams = 0.028 kg

Substitute with the givens in the equation to get the rms speed as follows:
Vrms = sqrt [(3*8.314*305.5) / 0.028]
Vrms = 521.665 m/sec</span>
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A(g) + 2B(g) → C(g) + D(g)
WITCHER [35]

Answer:

0.169

Explanation:

Let's consider the following reaction.

A(g) + 2B(g) ⇄ C(g) + D(g)

We can find the pressures at equilibrium using an ICE chart.

       A(g) + 2 B(g) ⇄ C(g) + D(g)

I       1.00     1.00        0        0

C       -x        -2x        +x       +x

E    1.00-x  1.00-2x     x         x

The pressure at equilibrium of C is 0.211 atm, so x = 0.211.

The pressures at equilibrium are:

pA = 1.00-x = 1.00-0.211 = 0.789 atm

pB = 1.00-2x = 1.00-2(0.211) = 0.578 atm

pC = x = 0.211 atm

pD = x = 0.211 atm

The pressure equilibrium constant (Kp) is:

Kp = pC × pD / pA × pB²

Kp = 0.211 × 0.211 / 0.789 × 0.578²

Kp = 0.169

6 0
3 years ago
What is the molar mass of magnesium chlorite (Mg(CIO2)2)?
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3 years ago
Is Lilium lancifolium Heterotroph, Autotroph, or Both
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4 0
3 years ago
The Ka for formic acid (HCO2H) is 1.8 × 10-4. What is the pH of a 0.35 M aqueous solution of sodium formate (NaHCO2)?
anzhelika [568]

Answer:

9.36

Explanation:

Sodium formate is the conjugate base of formic acid.

Also,

K_a\times K_b=K_w

K_b for sodium formate is K_b=\frac {K_w}{K_a}

Given that:

K_a of formic acid = 1.8\times 10^{-4}

And, K_w=10^{-14}

So,

K_b=\frac {10^{-14}}{1.8\times 10^{-4}}

K_b=5.5556\times 10^{-11}

Concentration = 0.35 M

HCOONa    ⇒     Na⁺ +    HCOO⁻

Consider the ICE take for the formate  ion as:

                                   HCOO⁻ + H₂O   ⇄   HCOOH + OH⁻

At t=0                              0.35                            -              -

At t =equilibrium           (0.35-x)                          x           x            

The expression for dissociation constant of sodium formate is:

K_{b}=\frac {[OH^-][HCOOH]}{[HCOO^-]}

5.5556\times 10^{-11}=\frac {x^2}{0.35-x}

Solving for x, we get:

x = 0.44×10⁻⁵  M

pOH = -log[OH⁻] = -log(0.44×10⁻⁵) = 4.64

pH + pOH = 14

So,

<u>pH = 14 - 4.64 = 9.36</u>

5 0
3 years ago
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