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pickupchik [31]
2 years ago
4

A sample of an unknown volatile liquid was injected into a Dumas flask (mass of flask=

Chemistry
1 answer:
denis23 [38]2 years ago
8 0

Answer:

Ethyl acetate.

Explanation:

  • We can use the general law of ideal gas: <em>PV = nRT</em>.

where, P is the pressure of the gas in atm (P = 0.976 atm).

V is the volume of the gas in L (V = volume of the flask = 0.1040 L).

n is the no. of moles of the gas in mol (n = ??? mol).

R is the general gas constant (R = 0.0821 L.atm/mol.K),

T is the temperature of the gas in K (T = 18.0°C + 273 = 291 K).

∴ n = PV/RT = (0.976 atm)(0.1040 L)/(0.0821 L.atm/mol.K)(291 K) = 4.25 x 10⁻³ mol.

∵ n = mass/molar mass,

mass of the volatile liquid = (mass of flask + vapor) - (mass of flask) = 27.4593 g - 27.0928 g = 0.3665 g.

∴ Molar mass of the volatile liquid = (mass)/(n) = (0.3665 g)/(4.25 x 10⁻³ mol) = 86.235 g/mol.

  • The volatile liquid may be <em>ethyl acetate </em>which has a molar mass of (88.11 g/mol).
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An atom of beryllium (m = 8.00 u) splits into two atoms of helium (m = 4.00 u) with the release of 92.2 kev of energy. if the or
galben [10]
The kinetic energy of the products is equal to the energy liberated which is 92.2 keV. But let's convert the unit keV to Joules. keV is kiloelectro volt. The conversion that we need is: 1.602×10⁻¹⁹ <span>joule = 1 eV

Kinetic energy = 92.2 keV*(1,000 eV/1 keV)*(</span>1.602×10⁻¹⁹ joule/1 eV) = 5.76×10²³ Joules

From kinetic energy, we can calculate the velocity of each He atom:
KE = 1/2*mv²
5.76×10²³ Joules = 1/2*(4)(v²)
v = 5.367×10¹¹ m/s
4 0
2 years ago
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Rank the following acids in order of increasing acid strength. Key: Weakening of hydrogen bond and stability of resulting anion.
raketka [301]

Explanation:

The given compounds are oxyacids and in these compounds more is the electronegativity of the central atom more will be its acidic strength.

This is because more is the electronegativity of the central atom more will be the polarity of OH bond. As a result, the compound can readily lose H^{+} ion.

Also, more is the electronegativity of central atom more will be the stability of conjugate base formed.

Thus, we can conclude that given compounds are arranged in increasing acid strength as follows.

       HOI < HOBr_{2} < HOCl_{3} < HOF

8 0
2 years ago
You wish to make a buffer with pH 7.0. You combine 0.060 grams of acetic acid and 14.59 grams of sodium acetate and add water to
aleksandr82 [10.1K]

Answer:

The pH of the buffer is 7.0 and this pH is not useful to pH 7.0

Explanation:

The pH of a buffer is obtained by using H-H equation:

pH = pKa + log [A⁻] / [HA]

<em>Where pH is the pH of the buffer</em>

<em>The pKa of acetic acid is 4.74.</em>

<em>[A⁻] could be taken as moles of sodium acetate (14.59g * (1mol / 82g) = 0.1779 moles</em>

<em>[HA] are the moles of acetic acid (0.060g * (1mol / 60g) = 0.001moles</em>

<em />

Replacing:

pH = 4.74 + log [0.1779mol] / [0.001mol]

<em>pH = 6.99 ≈ 7.0</em>

<em />

The pH of the buffer is 7.0

But the buffer is not useful to pH = 7.0 because a buffer works between pKa±1 (For acetic acid: 3.74 - 5.74). As pH 7.0 is out of this interval,

this pH is not useful to pH 7.0

<em />

7 0
2 years ago
Which conjugate pair is suited best to make this buffer? Which conjugate pair is suited best to make this buffer? Phosphoric aci
Juliette [100K]

Answer:

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Explanation:

4 0
2 years ago
Consider the following reactions and their respective equilibrium constants: NO(g)+12Br2(g)⇌NOBr(g)Kp=5.3 2NO(g)⇌N2(g)+O2(g)Kp=2
maxonik [38]

Answer:

Equilibrium constant of the given reaction is 1.3\times 10^{-29}

Explanation:

NO+\frac{1}{2}Br_{2}\rightleftharpoons NOBr....(K_{p})_{1}=5.3

2NO\rightleftharpoons N_{2}+O_{2}....(K_{p})_{2}=2.1\times 10^{30}

The given reaction can be written as summation of the following reaction-

2NO+Br_{2}\rightleftharpoons 2NOBr

N_{2}+O_{2}\rightleftharpoons 2NO

......................................................................................

N_{2}+O_{2}+Br_{2}\rightleftharpoons 2NOBr

Equilibrium constant of this reaction is given as-

\frac{[NOBr]^{2}}{[N_{2}][O_{2}][Br_{2}]}

=(\frac{[NOBr]}{[NO][Br_{2}]^{\frac{1}{2}}})^{2}(\frac{[NO]^{2}}{[N_{2}][O_{2}]})

=\frac{(K_{p})_{1}^{2}}{(K_{p})_{2}}

=\frac{(5.3)^{2}}{2.1\times 10^{30}}=1.3\times 10^{-29}

7 0
2 years ago
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