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igomit [66]
3 years ago
8

Does gravity affect the orbit of the international space station

Chemistry
2 answers:
ivanzaharov [21]3 years ago
7 0

Answer:

Explanation:gbbbynuy

B g gyngybfvtfftdtcdt

Serga [27]3 years ago
6 0
GfsdfCngnmsgagmsgme GM bcsmetmrhmerulrr
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Which value gives the number of particles in 1 mol of a substance?
Alexxx [7]
The number of particles (molecules or atoms) is: 6.022 x 10²³ particles (atoms or molecules).

1 mol of H₂O has 6.022 x 10²³ molecules.
1 mol of Al has 6.022 x 10²³ atoms. 
7 0
3 years ago
In an aqueous chloride solution cobalt(ii) exists in equilibrium with the complex ion cocl42-. co2 (aq) is pink and cocl42-(aq)
liraira [26]

First you should know that the Principle of Le Chatelier states that <u>if a system in equilibrium is subjected to a change of conditions, it will move to a new position in order to counteract the effect that disturbed it and recover the state of equilibrium.</u>

The variation of one or several of the following factors can alter the equilibrium condition in a chemical reaction:

  • Temperature
  • The pressure
  • The volume
  • The concentration of reactants or products

1. This reaction is: <u>a. exothermic</u>

The chemical equilibrium at issue is:

CoCl₄²⁻  ⇄   Co²⁺ + 4Cl⁻ + heat

<em>blue</em>            <em>pink</em>

The reaction of the question is <u>exothermic</u> because when adding heat (at high temperature) the equilibrium moves to the left since the blue color is strong which means that there are more reagent (CoCl₄²⁻) that product (Co²⁺). On the other hand, when extracting heat from the system (at low temperature) the equilibrium moves to the right since the pink color predominates and more product is present in the solution.

2. When the temperature is decreased the equilibrium constant, k: <u>c. remains the same.</u>

As mentioned above, <u>a system in equilibrium that is disturbed will move to a new position in order to counteract the effect that disturbed it and recover the state of equilibrium. </u>In this way, the system will always remain in equilibrium and its equilibrium constant will remain constant. This is why, despite altering the temperature of the system, the equilibrium constant remains constant.

3. When the temperature is decreased the equilibrium concentration of Co²⁺:  <u>a. increases</u>

Again, the chemical equilibrium at issue is:

CoCl₄²⁻  ⇄   Co²⁺ + 4Cl⁻ + heat

As the reaction in question is exothermic when the temperature decreases, heat is extracted from the system. <u>To compensate this decrease in heat, the system will react by shifting the balance to the right, increasing the concentration of the products and, therefore, the concentration of Co²⁺.</u>

5 0
4 years ago
How much<br>much hydrogen gas evolved<br>when 1.5 current is passed through water for 1.5 hours?​
evablogger [386]

0.042 moles of Hydrogen evolved

<h3>Further explanation</h3>

Given

I = 1.5 A

t = 1.5 hr = 5400 s

Required

Number of Hydrogen evolved

Solution

Electrolysis of water ⇒ decomposition reaction of water into Oxygen and Hydrogen gas.

Cathode(reduction-negative pole) : 2H₂O(l)+2e⁻ ⇒ H₂(g)+2OH⁻(aq)

Anode(oxidation-positive pole) : 2H₂O(l)⇒O₂(g)+4H⁻(aq)+4e⁻

Total reaction : 2H₂O(l)⇒2H₂(g)+O₂(g)

So at the cathode H₂ gas is produced

Faraday : 1 mole of electrons (e⁻) contains a charge of 96,500 C

\tt mol~e^-=\dfrac{Q}{96500}

Q = i.t

Q = 1.5 x 5400

Q = 8100 C

mol e⁻ = 8100 : 96500 = 0.084

From equation at cathode , mol ratio e⁻ : H₂ = 2 : 1, so mol H₂ = 0.042

4 0
3 years ago
What is the entropy change for the freezing process of 1 mole of liquid methanol at its freezing temperature (–97.6˚C) and 1 atm
Rudiy27

Answer : The value of change in entropy for freezing process is, -18.07 J/mol.K

Explanation :

Formula used :

\Delta S=\frac{\Delta H_{freezing}}{T_f}

where,

\Delta S = change in entropy

\Delta H_{fus} = change in enthalpy of fusion = 3.17 kJ/mol

As we know that:

\Delta H_{fus}=-\Delta H_{freezing}=-3.17kJ/mol=-3170J/mol

T_f = freezing point temperature = -97.6^oC=273+(-97.6)=175.4K

Now put all the given values in the above formula, we get:

\Delta S=\frac{\Delta H_{freezing}}{T_m}

\Delta S=\frac{-3170J/mol}{175.4K}

\Delta S=-18.07J/mol.K

Therefore, the value of change in entropy for freezing process is, -18.07 J/mol.K

4 0
4 years ago
All objects have mass<br> O True<br> O False
ludmilkaskok [199]
The answerrrrr is truw
7 0
3 years ago
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