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Serjik [45]
3 years ago
6

Wind and ocean currents do not move in straight lines; instead, they curve as they move across the planet. What is responsible f

or this pattern of movement? A.differences in water temperature B. differences in water salinity C.the Coriolis effect D.differences in water density
please help now
Chemistry
2 answers:
blsea [12.9K]3 years ago
5 0
C............................
klasskru [66]3 years ago
3 0

The correct answer is:    [C]:

____________________________________________

             →  " the Coriolis effect " .

____________________________________________

Hope this helps!

       Best wishes!

____________________________________________

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How does the density of protostars compare to other stars?
lorasvet [3.4K]

Protostars are less dense than other stars.  

Explanation:

Protostars are very young ‘stars’ made from hydrogen clouds that are beginning to coalesce and collapse under their weight. The hydrogen has not even begun fusing. Therefore, they are mainly made of hydrogen which is the lightest element in the universe.

Stars, however, have begun fusing hydrogen to other heavier elements like helium, carbon, oxygen, and iron. The elements are much heavier than hydrogen making other stars much denser than protostars.

Learn More:

For more on protostars vs stars check out;

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#LearnWithBrainly

5 0
3 years ago
Help me pls!!!!!!!!!!!!!!!
tankabanditka [31]
The answer is: 2 NH3 —> N2 + 3 H2
6 0
2 years ago
A 110.0-mL sample of a solution that is 3.0×10−3 M in AgNO3 is mixed with a 230.0- mL sample of a solution that is 0.10 M in NaC
almond37 [142]

Answer:

[Ag⁺] = 0.0666M

Explanation:

For the addition of Ag⁺ and CN⁻, the (Ag(CN)₂⁻ is produced, thus:

Ag⁺ + 2CN⁻  ⇄  Ag(CN)₂⁻

Kf = 1x10²¹ = [Ag(CN)₂⁻] / [CN⁻]² [Ag⁺]

As initial concentrations of Ag⁺ and CN⁻ are:

[Ag⁺] = 0.110L × (3.0x10⁻³mol / L) = 3.3x10⁻⁴mol / (0.110L + 0.230L) = 9.7x10⁻⁴M

[CN⁻] = 0.230L × (0.1mol / L) = 0.023mol / (0.110L + 0.230L) = 0.0676M

The equilibrium concentrations of each compound are:

[CN⁻] = 9.7x10⁻⁴M - x

[Ag⁺] = 0.0676M - x

[Ag(CN)₂⁻] = x

<em>Where x is reaction coordinate</em>

Replacing in Kf formula:

1x10²¹ = [x] / [9.7x10⁻⁴M - x]² [0.0676M - x]

1x10²¹ = [x] / 6.36048×10⁻⁸ - 0.000132085 x + 0.06954 x² - x³

-1x10²¹x³ + 6.954x10¹⁹x² - 1.32085x10¹⁷ x + 6.36x10¹³ = x

-1x10²¹x³ + 6.954x10¹⁹x² - 1.32085x10¹⁷ x + 6.36x10¹³ = 0

Solving for x:

X = 9.614x10⁻⁴M

Thus, equilibrium concentration of Ag⁺ is:

[Ag⁺] = 0.0676M - 9.614x10⁻⁴M = <em>0.0666M</em>

6 0
2 years ago
What is the mass of 0.0150 moles of Na₂SO4 (molar mass = 142.04 g/mol)?
olga2289 [7]

Answer:

2.13 g

Explanation:

(142.04)(0.0150) = 2.13 g

4 0
2 years ago
In the following reaction, which species is reduced? Au(s) + 3NO3-(aq) + 6H+(aq) → Au3+(aq) + NO(g) + 3H2O (l) H+ N+5 O2- H2O Au
weqwewe [10]

Answer:

NO3-

Explanation:

Given the reaction equation;

Au(s) + 3NO3-(aq) + 6H+(aq)→Au3+(aq) + 3NO2(g) + 3H2O (l).

We can consider the oxidation states of species on the left and right hand sides of the reaction equation;

Au is in zero oxidation state on the left hand side and an oxidation state of +3 on the righthand side.

NO3- is in oxidation state of +5 on the righthand side and NO2 is in + 4 oxidation state.

H+ is in + 1 oxidation state on both the left and right hand sides of the reaction equation.

Since reduction has to do with a decrease in oxidation number, it follows that NO3- was reduced in the reaction.

6 0
3 years ago
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