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Anika [276]
2 years ago
7

Thermohaline circulation _______. a. supplies heat to polar regions b. causes salt lake formation c. causes oxbow lake formation

d. is unaffected by heat fluctuations Please select the best answer from the choices provided A B C
Chemistry
1 answer:
DerKrebs [107]2 years ago
7 0

Thermohaline circulation that occur in the ocean part that flows across sea surface help to supply heat to polar regions.

<h3>What is Thermohaline circulation?</h3>

The thermohaline circulation refer to a section and region of the ocean circulation which is majorly driven by heat gradient and freshwater that flows across the sea surface and also from the mixture of heat and salt that occur at the inside part of the ocean.

Therefore, Thermohaline circulation help to supply heat to polar regions.

Learn more about thermohaline circulation here.

brainly.com/question/1176119

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Is this right????????????
zzz [600]

Answer: yea its right

Explanation:

5 0
3 years ago
Read 2 more answers
When silver nitrate reacts with copper, copper(II) nitrate and silver are produced. The balanced equation for this reaction is:
posledela

The question is incomplete, here is the complete question:

When silver nitrate reacts with copper, copper(II) nitrate and silver are produced. The balanced equation for this reaction is:

2AgNO_3+Cu\rightarrow Cu(NO_3)_2+2Ag

Suppose 6 moles of silver nitrate react. The reaction consumes___ moles of copper. The reaction produces __ moles of copper(II) nitrate and __ moles of silver.

<u>Answer:</u> The amount of copper metal reacted is 3 moles, amount of copper (II) nitrate produced is 3 moles and amount of silver metal produced is 6 moles.

<u>Explanation:</u>

We are given:

Moles of silver nitrate = 6 moles

For the given chemical reaction:

2AgNO_3+Cu\rightarrow Cu(NO_3)_2+2Ag

  • <u>For copper metal:</u>

By Stoichiometry of the reaction:

2 moles of silver nitrate reacts with 1 mole of copper metal

So, 6 moles of silver nitrate will react with = \frac{1}{2}\times 6=3mol of copper metal

Moles of copper reacted = 3 moles

  • <u>For copper(II) nitrate:</u>

By Stoichiometry of the reaction:

2 moles of silver nitrate produces 1 mole of copper(II) nitrate

So, 6 moles of silver nitrate will produce = \frac{1}{2}\times 6=3mol of copper(II) nitrate

Moles of copper(II) nitrate produced = 3 moles

  • <u>For silver metal:</u>

By Stoichiometry of the reaction:

2 moles of silver nitrate produces 2 moles of silver metal

So, 6 moles of silver nitrate will produce = \frac{2}{2}\times 6=6mol of silver metal

Moles of silver metal produced = 3 moles

Hence, the amount of copper metal reacted is 3 moles, amount of copper (II) nitrate produced is 3 moles and amount of silver metal produced is 6 moles.

6 0
3 years ago
Given K = 3.61 at 45°C for the reaction A(g) + B(g) equilibrium reaction arrow C(g) and K = 7.19 at 45°C for the reaction 2 A(g)
Firlakuza [10]

Answer:

K = 0.55

Kp = 0.55

mol fraction B = 0.27

Explanation:

We need to calculate the equilibrium constant for the reaction:

C(g) + D(g) ⇄ 2B(g)              K₁= ?                       (1)

and we are given the following equilibria with their respective Ks

A(g) + B(g) ⇄ C(g)                 K₂= 3.61                 (2)

2 A(g) + D(g)  ⇄ C(g)             K₃= 7.19                 (3)

all at 45 ºC.

What we need to do to solve this question is to manipulate equations (2) and (3)  algebraically  to get our desired equilibrium (1).

We are allowed to reverse  reactions, in that case we take the reciprocal of K as our new K' ; we can also  add two equilibria together, and the new equilibrium constant will be the product of their respective Ks .

Finally if we multiply by a number then we raise the old constant to that factor to get the new equilibrium constant.

With all this  in mind, lets try to solve our question.

Notice A is not in our goal equilibrium (3)  and we want D as a reactant . That  suggests we should reverse the first equilibria and multiply it by two since we have 2 moles of B  as product in our  equilibrium (1) . Finally we would add (2) and (3) to get  (1) which is our final  goal.

2C(g)             ⇄  2A(g) + 2B(g)  K₂´= ( 1/ 3.61 )²  

                                   ₊

2 A(g) + D(g)  ⇄     C(g)               K₃ = 7.19  

<u>                                                                                    </u>

C(g) + D(g)     ⇄    2B(g)       K₁ = ( 1/ 3.61 )²   x  7.19

                                             K₁ = 0.55

Kp is the same as K = 0.55 since the equilibrium constant expression only involves  gases.

To compute the last part lets setup the following mnemonic  ICE table to determine the quantities at equilibrium:

pressure (atm)        C             D           B

initial                     1.64          1.64         0

change                    -x             -x        +2x

equilibrium          1.64-x         1.64-       2x

Thus since

Kp =0.55 = pB²/ (pC x pD) = (2x)²/ (1.64 -x)²  where p= partial pressure

Taking square root to both sides of the equation we have

√0.55 = 2x/(1.64 - x)

solving for x  we obtain a value of 0.44 atm.

Thus at equilibrium we have:

(1.64 - 0.44) atm = 1.20 atm = pC = p D

2(0.44) = 0.88 = pB

mole fraction of B = partial pressure of B divided into the total gas pressure:

X(B) = 0.88 / ( 1.20 + 1.20 + 0.88 ) = 0.27

8 0
3 years ago
In the reaction MnO₂ + 4HCI MnCl₂ + 2H₂O + Cl₂, which<br> species is reduced?
Triss [41]

Answer:

4 HCl (aq) + MnO2 (s) → MnCl2 (aq) + Cl2 (g) + 2 H2O (l)

This is an oxidation-reduction (redox) reaction:

MnIV + 2 e- → MnII (reduction)

2 Cl-I - 2 e- → 2 Cl0 (oxidation)

MnO2 is an oxidizing agent, HCl is a reducing agent.

Reactants:

HCl – Chlorane, Hydrogen chloride

Other names: Hydrochloric acid
Appearance: Colorless, transparent liquid, fumes in air if concentrated; Colorless gas; Colourless compressed liquefied gas with pungent odour; Colorless to slightly yellow gas with a pungent, irritating odor. [Note: Shipped as a liquefied compressed gas.]

MnO2 – Manganese oxide, Manganese(IV) oxide

Other names: Manganese dioxide, Pyrolusite, Hyperoxide of manganese

Appearance: Brown-black solid Black-to-brown powder

Products:

MnCl2 – Manganese(II) chloride, Manganese dichloride

Other names: Manganous chloride, Manganous dichloride Manganese(II) chloride (1:2)

Appearance: Pink solid (tetrahydrate)

Cl2

Names: Chlorine, Molecular chlorine

Appearance: Greenish-yellow compressed liquefied gas with pungent odour ; Greenish-yellow gas with a pungent, irritating odor. [Note: Shipped as a liquefied compressed gas.]

H2O – Water, oxidane

Other names: Water (H2O) Hydrogen hydroxide (HH or HOH) Hydrogen oxide

Appearance: White crystalline solid, almost colorless liquid with a hint of blue, colorless gas

Explanation:

6 0
2 years ago
Read 2 more answers
In aqueous solution amino acids are rarely found in the neutral, unionized form.
True [87]

a. True.

There is always an equilibrium of the type

NH₃⁺CHRCOOH ⇌ NH₃⁺CHRCOO⁻ ⇌ NH₂CHRCOO⁻

The compound is <em>always in an ionized form</em>.

There are no unionized NH₂CHRCOOH molecules in the solution.

3 0
3 years ago
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