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klio [65]
2 years ago
9

When gases are treated as real, via use of the van der Waals equation, the actual volume occupied by gas molecules ________ the

pressure exerted and the attractive forces between gas molecules ________ the pressure exerted, as compared to an ideal gas.
decreases, increases
does not affect, decreases
increases, decreases
does not affect, increases
increases, increases
Chemistry
1 answer:
Tresset [83]2 years ago
6 0

Answer:

Increases, decreases

Explanation:

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Classify each organic compound based on the functional group it contains.
bixtya [17]

Answer:

H2C2O1

Explanation:

8 0
2 years ago
II need this asap !!!!!!!
Xelga [282]

Answer:

1. B.10 mol/l

2. D.

3.A.

Explanation:

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7 0
2 years ago
What is the Ka of a weak acid (HA) if the initial concentration of weak acid is 4.5 x 10-4 M and the pH is 6.87? (pick one)
rewona [7]

Answer:

Ka = 4.04 \times 10^{-11}

Explanation:

Initial concentration of weak acid = 4.5 \times 10^{-4}\ M

pH = 6.87

pH = -log[H^+]

[H^+]=10^{-pH}

[H^+]=10^{-6.87}=1.35 \times 10^{-7}\ M

HA dissociated as:

HA \leftrightharpoons H^+ + A^{-}

(0.00045 - x)    x     x

[HA] at equilibrium = (0.00045 - x) M

x = 1.35 \times 10^{-7}\ M

Ka = \frac{[H^+][A^{-}]}{[HA]}

Ka = \frac{(1.35 \times 10^{-7})^2}{0.00045 - 0.000000135}

0.000000135 <<< 0.00045

Therefore, Ka = \frac{(1.35 \times 10^{-7})^2}{0.00045 } = 4.04 \times 10^{-11}

5 0
3 years ago
What is the empirical formula of a compound with a percent composition of 22.5% Phosphorous and 77.5% Chlorine?
sashaice [31]

Answer:

\boxed {\boxed {\sf PCl_3}}

Explanation:

We are given the percent composition: 22.5% phosphorus and 77.5% chlorine.

We can assume there are 100 grams of this compound. We choose 100 because we can simply use the percentages as the masses.

  • 22.5 g P
  • 77.5 g Cl

Next, convert these masses to moles, using the molar masses found on the Periodic Table.

  • P: 30.974 g/mol
  • Cl: 35.45 g/mol

Use the molar masses as ratios and multiply by the number of grams. 22.5 \ g \ P  * \frac {1 \ mol \ P }{30.974 \ g \ P}= \frac {22.5 \ mol \ P }{ 30.974} = 0.7264157035 \ mol \ P

77.5 \ g \ Cl  * \frac {1 \ mol \ Cl }{35.45 \ g \ Cl}= \frac {77.5 \ mol \ Cl }{ 35.45} \ =2.186177715 \ mol \ Cl

Divide both of the moles by the smallest number of moles to find the mole ratio.

\frac {0.7264157035} {0.7264157035} = 1

\frac {2.186177715}{0.7264157035}=3.009540824 \approx 3

The mole ratio is about 1 P: 3 Cl, so the empirical formula is written as:<u> PCl₃</u>

4 0
2 years ago
Assume reaction N (g )space plus O (g )rightwards arrow N O (g )occurs in high layers of the atmosphere. What is the H subscript
8090 [49]

Answer:

Kindly check the explanation section.

Explanation:

From the description given in the question above, that is '' H subscript f to the power of degree of the reaction" we have that the description matches what is known as the heat of formation of the reaction, ∆fH° where the 'f' is a subscript.

In order to determine the heat of formation of any of the species in the reaction, the heat of formation of the other species must be known and the value for the heat of reaction, ∆H(rxn) must also be known. Thus, heat of formation can be calculated by using the formula below;

∆H(rxn) = ∆fH°( products) - ∆fH°(reactants).

That is the heat of formation of products minus the heat of formation of the reaction g specie(s).

Say heat of formation for the species is known as N(g) = 472.435kj/mol, O(g) = 0kj/mol and NO = unknown, ∆H°(rxn) = −382.185 kj/mol.

−382.185 = x - 472.435kj/mol = 90.25 kJ/mol

5 0
3 years ago
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