Select the correct answer.

Processes that increase the density of seawater include evaporation and _____.

Elena L [17]

Correct Answer: option C: Formation of sea ice

Reason:
 In cold regions, changes in salinity alters the water present in ocean. Further, water density also changes with temperature. In general, water density in ocean water increases with decreasing temperature. This is because, when salt is ejected into the ocean as sea ice forms, the water's salinity increases. Since, salt water is heavier, the density of the water increases.

3 0

3 years ago

Read 2 more answers

Can you help me with acrostic poem?

kkurt [141]

An acrostic poem is a type of poetry where the first, last or other letters in a line spell out a particular word or phrase.

8 0

3 years ago

At equilibrium, the concentrations of the products and reactants for the reaction, H2 (g) + I2 (g)  2 HI (g), are [H2] = 0.106

lana [24]

Answer:

The new equilibrium concentration of HI: [HI] = 3.589 M

Explanation:

Given: Initial concentrations at original equilibrium- [H₂] = 0.106 M; [I₂] = 0.022 M; [HI] = 1.29 M

Final concentrations at new equilibrium- [H₂] = 0.95 M; [I₂] = 0.019 M; [HI] = ? M

Given chemical reaction: H₂(g) + I₂(g) → 2 HI(g)

The equilibrium constant ( $K_{c}$ ) for the given chemical reaction, is given by the equation:

$K_{c} = \frac {[HI]^{2}}{[H_{2}]\: [I_{2}]}$

At the original equilibrium state:

$K_{c} = \frac {(1.29\: M)^{2}}{(0.106\: M) \times (0.022\: M)}$

$K_{c} = \frac {1.6641}{0.002332} = 713.59$

Therefore, at the new equilibrium state:

$K_{c} = \frac {[HI]^{2}}{(0.95\: M) \times (0.019\: M)}$

$\Rightarrow K_{c} = 713.59 = \frac {[HI]^{2}}{0.01805}$

$\Rightarrow [HI]^{2} = 713.59 \times 0.01805 = 12.88$

$\Rightarrow [HI] = \sqrt {12.88} = 3.589 M$

Therefore, the new equilibrium concentration of HI: [HI] = 3.589 M

6 0

3 years ago

Consider the following balanced redox reaction: 2CrO2-(aq) + 2H2O(l) + 6ClO-(aq) LaTeX: \longrightarrow⟶ 2CrO42-(aq) + 3Cl2(g) +

frutty [35]

Answer:

1. Chromium

2. Chlorine.

3. Chlorine.

4. Chromium.

5. 12 electrons.

Explanation:

Hello,

In this case, the given reaction with the appropriate oxidation states turns out:

$2(Cr^{+3}O^{-2}_2)^-(aq) + 2H_2O(l) + 6(Cl^{+1}O^{-2})^-(aq)\longrightarrow 2(Cr^{+6}O^{-2}_4)^{2-}(aq) + 3Cl^0_2(g) + 4OH^-(aq)$

In such a way, the oxidation half-reaction is written for chromium as the reducing agent so it is oxidized from +3 to +6, nonetheless, since there are two chromiums undergoing such change, 6 electrons are being transferred as shown below:

$2(Cr^{+3}O^{-2}_2)^-(aq) \longrightarrow 2(Cr^{+6}O^{-2}_4)^{2-}(aq)+6e^-$

On the other hand, chlorine's reduction half-reaction as the oxidizing agent result from the transfer of 6 electrons as well from +1 to 0, nonetheless, there are 6 chlorines undergoing such change:

$6(Cl^{+1}O^{-2})^-+6e^-\longrightarrow 3Cl^0_2(g)$

Therefore, there are 12 electrons that are being transferred, 6 for chromium and 6 for chlorine.

Best regards.

5 0

3 years ago

A 6.55 g sample of aniline (C₆H₅NH₂), molar mass = 93.13 g/mol) was combusted in a bomb calorimeter with a heat capacity of 14.2

riadik2000 [5.3K]

Answer: The heat of combustion in bomb calorimeter problems isgenerally calculated by the following formula H = -C T

Explanation:

7 0

3 years ago