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3 years ago

At a temperature of 500 kelvins, 6 liters of an ideal gas had a pressure of 300 newtons

Chemistry

2 answers:

mr Goodwill [35]3 years ago

6 0

Answer A

Explanation:

Aleks [24]3 years ago

6 0

Answer:

(A) 100 newtons/m^2

Explanation:

Hello,

In this case, the equation defining such changes is:

$P_1V_1T_2=P_2V_2T_1$

Thus, the unknown is the pressure at the second state, which is found as shown below:

$P_2=\frac{P_1V_1T_2}{V_2T_1}=\frac{300Pa*6L*250K}{9L*500K} \\P_2=100Pa=100\frac{N}{m^2}$

Therefore, the answer is (A).

Best regards.

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HC2H3O2 (aq) + H2O (l) ⇔ C2H3O2- (aq) + H3O+ (aq) Ka = 1.8 x 10-5

marin [14]

The concentration of [H3O⁺]=2.86 x 10⁻⁶ M

<h3>Further explanation</h3>

In general, the weak acid ionization reaction

HA (aq) ---> H⁺ (aq) + A⁻ (aq)

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$\large {\boxed {\bold {Ka \: = \: \frac {[H ^ +] [A ^ -]} {[HA]}}}}$

Reaction

HC₂H₃O₂ (aq) + H₂O (l) ⇔ (aq) + H₃O⁺ (aq) Ka = 1.8 x 10⁻⁵

$\tt Ka=\dfrac{[C_2H_3O^{2-}[H_3O^+]]}{[HC_2H_3O_2]}}\\\\1.8\times 10^{-5}=\dfrac{0.22\times [H_3O^+]}{0.035}$

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3 years ago

What is the bond order of li2−? express the bond order numerically?

jeka94

Atomic Number of Lithium is 3, so it has 3 electrons in its neutral state. Also, Li₂ will have 6 electrons. But the chemical formula we are given has a negative charge on it (i.e Li₂⁻) so there is an additional electron (RED) present on this compound. So, the total number of electrons are 7. The MOT diagram for this compound is shown below. According to diagram we are having 4 electrons in Bonding Molecular Orbitals (BMO) and 3 electrons in Anti-Bonding Molecular Orbitals (ABMO). Bond Order is calculated as,

Bond Order = (# of e⁻s in BMO - # of e⁻s in ABMO) ÷ 2

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3 years ago

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Yuki888 [10]

Answer:

velocity = 29925×10⁶ m/s

Explanation:

Given data:

Frequency = 35 MHz

Wavelength = 855 m

Velocity = ?

Solution:

MHz to Hz

35×10⁶ Hz

Formula:

<em>velocity = Wavelength × Frequency</em>

velocity = 855 m × 35×10⁶ Hz

velocity = 29925×10⁶ m/s

Hz = s⁻¹

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3 years ago

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Ainat [17]

Answer:

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<h3><u>Answer;</u></h3>

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