How many miles of MgCI2 are there in 345 g of the compound?

An air mass usually forms in an area of _____. low pressure because it's stable low pressure because it's unstable high pressure

horsena [70]

Answer: high pressure because it's stable

Explanation:

Air mass is volume of air which has stable temperature, humidity and pressure horizontally. Over time, each air mass acquires properties of the region by residing over same part of a surface.

Areas of low pressure and high pressure occur where there is warm air and cold air respectively. An air mass usually forms over an area of high pressure. Warm air rises up and cold air takes its place. Warm air has low density and low pressure where as cold air has high density and pressure and therefore, sinks to the bottom. This is a stable condition. The movement of air mass is responsible for maintenance of temperature conditions on Earth.

6 0

3 years ago

Standard reduction half-cell potentials at 25∘c half-reaction e∘ (v) half-reaction e∘ (v) au3 (aq) 3e−→au(s) 1.50 fe2 (aq) 2e−→f

Zanzabum

K = 5.0 × 10^25

<h2>Part (a). Calculate E° for the reaction </h2>

Step 1. Write the equations for the two half-reactions

2H^(+)(aq) + 2e^(-) → H2(g); _0.00 V

Zn^(2+)(aq) + 2e^(-) → Zn(s); -0.76 V

Step 2. Identify the cathode and the anode

The half-cell with the more negative E° (Zn) is the anode.

Step 3. Calculate E°

Zn(s) → Zn^(2+)(aq) + 2e^(-); _________+0.76 V

2H^(+)(aq) + 2e^(-) → H2(g); __________0.00 V

Zn(s) + 2H^(+)(aq) → Zn^(2+)(aq) + H2(g); +0.76 V

E° = +0.76 V

<h2>Part (b). Calculate K for the reaction </h2>

The relation between E° and K is

E° = (RT)/(nF)lnK

where

R = the universal gas constant: 8.314 J·K^(-1)mol^(-1)

T = the Kelvin temperature

n = the moles of electrons transferred

F = the Faraday constant: 96 485 J·V^(-1)mol^(-1)

Then

0.76 V = [8.314 J·K^(-1)mol^(-1) × 298.15 K]/[2 × 96 485 J·V^(-1)mol^(-1)]lnK

0.76 = 0.012 85 lnK

lnK = 0.76/0.012 85 = 59.16

K =e^59.16 = 5.0 × 10^25

4 0

3 years ago

Which substance contains only 1 kind of atom

Nataliya [291]

Elements are substances that contain only 1 kind of atom.

8 0

3 years ago

Read 2 more answers

1. Name the compound present in a sample of dry air.

aev [14]

Answer:

not air

Explanation:

I think so

hahaha

7 0

2 years ago

Acetic acid and ethanol react to form ethyl acetate and water, like this:

ladessa [460]

Answer:

1.) Option C is correct.

The rate of reverse reaction is greater than zero, but equal to the rate of the forward reaction.

2) Option B is correct.

The rate of reverse reaction is Greater than zero, but less than the rate of the forward reaction.

3) Option C is correct.

The rate of reverse reaction is Greater than zero, and equal to the rate of the forward reaction.

4) Option A is correct.

How much less C2H5CO2CH3 is in the flask when the system has again reached equilibrium? Zero.

Explanation:

HCH,CO2(aq) + C2H5OH(aq) ⇌ C2H,CO2CH3(aq) + H2O

1) Before the main product is removed from the reaction setup, the chemical reaction is at equilibrium.

Chemical equilibrium is a state of dynamic equilibrium such that the concentration of the reactants and the products do not always remain the same but the rate of forward reaction always matches the rate of backward reaction.

2) When 246. mmol of C2HCO2CH3 are removed from the reaction mixture....

And when one of the factors involved in chemical equilibrium changes, Le Chatellier's principle explains that the system then adjusts to remedy this change and takes time to go back to equilibrium again.

When one of the species involved in the chemical reaction at equilibrium, is removed from the reaction mixture, the rate of reaction begins to favour that side of the reaction until equilibrium is re-established.

So, when 246 mmol of one of the products is removed, the response is to cause the rate of forward reaction to be favoured to produce more of products as there are fewer, and the rate of reverse reaction at this moment becomes less than the rate of forward reaction.

3) The rate of the reverse reaction when the system has again reached equilibrium

Like I said in (2) above, the reaction remedies this change in concentration of one of the products until equilibrium is re-established and when chemical equilibrium is re-established the rate of forward reaction once again matches the rate of backward reaction.

4) How much less C2H5CO2CH3 is in the flask when the system has again reached equilibrium?

By the time equilibrium is re-established, the system goes back to how it all was and the concentration of C2H5CO2CH3 goes back to the same as it was at the start of the reaction.

Hope this Helps!!!

3 0

3 years ago