Answer this plz I'm desperat

In which or areas on the image dose salinity of the ocean increases

Ronch [10]

Answer and Explanation:

D I believe..

8 0

3 years ago

In terms of intermolecular forces why ethanol has a much higher boiling point than ethene at standard point

leva [86]

Ethanol contains an alcohol group (-OH), which includes hydrogen bonding. Hydrogen bonding is stronger than other IMFs such as Van der Waals forces and Dipole-Dipole.

Ethene only has Van der Waals forces as it is an alkene with only C-H or C-C bonds. These means that the forces are not as strong.

Since ethanol contains hydrogen bonding, it will take a higher temperature in order to break these bonds, and thus results in a higher boiling point than ethene.

6 0

3 years ago

Need help !!!!! ASAP

Nat2105 [25]

<h2>Hello!</h2>

The answer is:

The temperature will be the same, 37°C.

Since from the statemet we know the first temperature, pressure and volumen of a gas, and we need to calculate the new temperature after the pressure and the volume changed, we need to use the Combined Gas Law.

The Combined Gas Law establishes a relationship between the temperature, the pressure and the volume of an ideal gas using Boyle's Law, Gay-Lussac's Law and Charles's Law.

The law establishes the following equation:

$\frac{P_{1}V{1}}{T_{1}}=\frac{P_{2}V{2}}{T_{2}}$

Where,

$P_{1}$ is the first pressure.

$V_{1}$ is the first volume.

$T_{1}$ is the first temperature.

$P_{2}$ is the second pressure.

$V_{2}$ is the second volume.

$T_{2}$ is the second temperature.

Then, we are given the following information:

$V_{1}=200mL\\P_{1}=4atm\\T_{1}=37\°C\\V_{2}=400mL\\P_{2}=2atm$

So, isolating the new temperature and substituting the given information, we have:

$\frac{P_{1}V{1}}{T_{1}}=\frac{P_{2}V{2}}{T_{2}}\\\\T_{2}=P_{2}V{2}*\frac{T_{1}}{P_{1}V_{1}} \\\\T_{2}=2atm*400mL*\frac{37\°C}{4atm*200mL}=37\°C$

Hence, we have that the temperature will not change because both pressure and volume decreased and increased proportionally, creating the same relationship that we had before the experiment started.

The temperature will be the same, 37°C

Have a nice day!

4 0

3 years ago

How many grams of calcium will react with 10.0 grams of nitrogen according to the

riadik2000 [5.3K]

Answer:

The answer to your question is 42.9 g of Ca

Explanation:

Data

mass of Ca = ?

mass of N₂ = 10 g

Balanced chemical reaction

3Ca + N₂ ⇒ Ca₃N₂

Process

1.- Calculate the molar mass of the reactants.

Ca = 3(40) = 120 g

N₂ = 2(14) = 28 g

2.- Calculate the mass of Calcium using proportions and the molar mass of the reactants.

120 g of Ca ---------------- 28 g of N₂

x ---------------- 10 g of N₂

x = (10 x 120) / 28

x = 1200 / 28

x = 42.9 g of Ca

3 0

3 years ago

Based on the reactivities of the elements involved, which reactions will form products that are more stable than the reactants?

myrzilka [38]

2 LiI + Cl₂ → 2 LiCl + I₂

2 LiBr + F₂ → 2 LiF + Br₂

<h3>Explanation</h3>

Each of the five reactions involve one halogen molecule (F₂, Cl₂, Br₂, and I₂) substituting the ion of another halogen (F⁻, Cl⁻, Br⁻, and I⁻).

Halogen atoms are found in group 17 of the periodic table. They are all non-metal elements. Each of the halogen atom will gain one electron to form an ion of charge -1. However, the tendency to do so decreases down the group.

F is the first halogen in group 17. It has only two shells of electrons.
Cl is right under F. Its electrons occupy three main energy shells.
Br follows with four main energy shells.
I is under Br and has five main energy shells.

Atoms of all four elements have the same effective nuclear charge of +7. However, F has the smallest radius. As a result, it has the strongest hold on electrons around it. Its ion F⁻ is more stable than ions of Cl, Br, or I. Similarly, its molecule F₂ is more reactive than Cl₂, Br₂, and I₂.

As a result, the stability of halogen molecules increases down the group:

Stability: F₂ < Cl₂ < Br₂ < I₂.

The stability of halogen ions decreases down the group:

Stability: F⁻ > Cl⁻ > Br⁻ > I⁻.

Cl₂ repaces F⁻ (from LiF) in first reaction. F₂ and Cl⁻ are produced. F₂ is less stable than Cl₂. Cl⁻ is less stable than F⁻.

Cl₂ replaces I⁻ (from LiI) in the second reaction. I₂ and Cl⁻ are produced. I₂ is more stable than Cl₂. Cl⁻ is more stable than I⁻.

Br₂ replaces Cl⁻ (from LiCl) in the third reaction. Cl₂ and Br⁻ are produced. Cl₂ is less stable than Br₂. Br⁻ is less stable than Cl⁻.

F₂ replaces Br⁻ (from LiBr) in the fourth reaction. Br₂ and F⁻ are produced. Br₂ is more stable than F₂. F⁻ is more stable than Br⁻.

I₂ replaces Br⁻ (from LiBr) in the fifth reaction. Br₂ and I⁻ are produced. Br₂ is less stable than I₂. I⁻ is less stable than Br⁻.

5 0

3 years ago