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2 years ago

At saturation, the further addition of water vapor or a decrease in temperature results in __________

1 answer:

7 0

The answer in the space provided is active condensation. Active condensation will likely occur if in saturation, further addition of water vapor has occurred or there is a presence of decrease in temperature. Condensation is a process that occurs when physical state is being changes such example of this when gas is formed into a liquid.

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BH+ClO4- is a salt formed from the base B (Kb = 1.00e-4) and perchloric acid. It dissociates into BH+, a weak acid, and ClO4-, w

Len [333]

Answer:

The pH of 0.1 M BH⁺ClO₄⁻ solution is 5.44

Explanation:

Given: The base dissociation constant: $K_{b}$ = 1 × 10⁻⁴, Concentration of salt: BH⁺ClO₄⁻ = 0.1 M

Also, water dissociation constant: $K_{w}$ = 1 × 10⁻¹⁴

The acid dissociation constant ( $K_{a}$ ) for the weak acid (BH⁺) can be calculated by the equation:

$K_{a}. K_{b} = K_{w}$

$\Rightarrow K_{a} = \frac{K_{w}}{K_{b}}$

$\Rightarrow K_{a} = \frac{1\times 10^{-14}}{1\times 10^{-4}} = 1\times 10^{-10}$

Now, the acid dissociation reaction for the weak acid (BH⁺) and the initial concentration and concentration at equilibrium is given as:

Reaction involved: BH⁺ + H₂O ⇌ B + H₃O+

Initial: 0.1 M x x

Change: -x +x +x

Equilibrium: 0.1 - x x x

The acid dissociation constant: $K_{a} = \frac{\left [B \right ] \left [H_{3}O^{+}\right ]}{\left [BH^{+} \right ]} = \frac{(x)(x)}{(0.1 - x)} = \frac{x^{2}}{0.1 - x}$

$\Rightarrow K_{a} = \frac{x^{2}}{0.1 - x}$

$\Rightarrow 1\times 10^{-10} = \frac{x^{2}}{0.1 - x}$

$As, x$

$\Rightarrow 0.1 - x = 0.1$

$\therefore 1\times 10^{-10} = \frac{x^{2}}{0.1 }$

$\Rightarrow x^{2} = (1\times 10^{-10})\times 0.1 = 1\times 10^{-11}$

$\Rightarrow x = \sqrt{1\times 10^{-11}} = 3.16 \times 10^{-6}$

Therefore, the concentration of hydrogen ion: x = 3.6 × 10⁻⁶ M

Now, pH = - ㏒ [H⁺] = - ㏒ (3.6 × 10⁻⁶ M) = 5.44

Therefore, the pH of 0.1 M BH⁺ClO₄⁻ solution is 5.44

5 0

2 years ago

You have 49.8 g of O2 gas in a container with twice the volume as one with CO2 gas. The pressure and temperature of both contain

IrinaVladis [17]

Answer:

34.2 g is the mass of carbon dioxide gas one have in the container.

Explanation:

Moles of $O_2$ :-

Mass = 49.8 g

Molar mass of oxygen gas = 32 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{49.8\ g}{32\ g/mol}$

$Moles_{O_2}= 1.55625\ mol$

Since pressure and volume are constant, we can use the Avogadro's law as:-

$\frac {V_1}{n_1}=\frac {V_2}{n_2}$

Given ,

V₂ is twice the volume of V₁

V₂ = 2V₁

n₁ = ?

n₂ = 1.55625 mol

Using above equation as:

$\frac {V_1}{n_1}=\frac {V_2}{n_2}$

$\frac {V_1}{n_1}=\frac {2\times V_1}{1.55625}$

n₁ = 0.778125 moles

Moles of carbon dioxide = 0.778125 moles

Molar mass of $CO_2$ = 44.0 g/mol

Mass of $CO_2$ = Moles × Molar mass = 0.778125 × 44.0 g = 34.2 g

34.2 g is the mass of carbon dioxide gas one have in the container.

5 0

2 years ago

Based on the values you obtained for δh∘rxn, which of the reactions would you expect to be thermodynamically favorable and which

QveST [7]

Following reactions are involved in present reaction
1) Ag+(aq) + Li(s) → Ag(s) + Li+(aq) − 384.4kJ
2) 2Fe(s) + 2Na+(aq) → Fe2+(aq) + 2Na(s) + 392.3kJ
3) 2K(s) + 2H2O(l) → 2KOH(aq) +H2(g) −393.1kJ

In above reaction, reaction 1 and 3 has negative value of δh∘f, while reaction 2 has posiyive value of δh∘f. As per the sign convention positive sign indicates that heat is given out during the reaction, while negative sign indicates heat is to be supplied for reaction to occur. In alternative words, product formed in reaction 2 is stable as compared to reactant. Hence, it is thermodynamically favorable.

7 0

2 years ago

Compound A is an organic compound which contains Carbon, Hydrogen and Oxygen. When 0.240g of the vapour of A is slowly passed ov

Keith_Richards [23]

Answer:

1) 0.009 61 g C; 2) 0.008 00 mol C

Step-by-step explanation:

You know that you will need a balanced equation with masses, moles, and molar masses, so gather all the information in one place.

M_r: 12.01 44.01

C + ½O₂ ⟶ CO₂

m/g: 0.352

1) Mass of C

Convert grams of CO₂ to grams of C

44.01 g CO₂ = 12.01 g C

Mass of C = 0.352 g CO₂ × 12.01 g C/44.01 g CO₂

Mass of C = 0.009 61 g C

2) Moles of C

Convert mass of C to moles of C.

1 mol C = 12.01 g C

Moles of C = 0.00961 g C × (1 mol C/12.01 g C)

Moles of C = 0.008 00 mol C

All the carbon comes from Compound A, so there are 0.008 00 mol C in Compound A.

7 0

2 years ago

In the redox conversion of Ni2+ to NiO4−, the oxidation number of Ni goes from (−2, 0, +2) to (−1, +1, +7, +9). Recall that the

inessss [21]

Oxidation number of an atom is the charge that atom would have if the compound is composed of ions. In neutral substances that contains atoms of one element the oxidation number of an atom is zero. Thus atoms in O2, Ni2, and aluminium all have oxidation number of zero.
In this case, Ni2, the oxidation number of Ni atom is zero,
for NiO4-, assuming oxidation number of Ni is x
(x ×1) + (-2 × 4) = -1
x = + 7
Therefore, the oxidation number goes from 0 to +7

7 0

2 years ago