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kap26 [50]
2 years ago
10

What is the value of Avogadro's constant?​

Chemistry
1 answer:
diamong [38]2 years ago
6 0

Answer:

6.02214154 x 1023

Explanation:

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Using Le Chatelier’s principle, predict which equilibrium system below decrease the concentration of the reactant (H2CO3) as a r
seropon [69]

Answer:

The correct answer is c add more HCO3- by adding NaHCO3.

Explanation:

The reaction mentioned in the question is carried out by bicarbonate buffer system of our body to maintain the normal acid  base balance.

         Now concentration of the reactant (H2C03) is decreased then NaHCO3 should be added which undergo breakdown to release bicarbonate ions(HCO3-).

        The released bicarbonate ions then reacts with H+ to form Carbonic acid(H2CO3).Thus homeostasis of H2CO3 is maintained.

4 0
3 years ago
Read 2 more answers
You use 50.00 mL of a 12.0 M solution of HCl solution to make a 500.00 mL solution. What is the concentration of the new solutio
Alexandra [31]

Answer:

1.2 M

Explanation:

If you use the dilution equation (M1V1=M2V2), you end up with (50)(12)=(500)(M2), and when you solve for M2 you get 1.2 M.

6 0
3 years ago
Consider the reaction mg(oh)2(s)→mgo(s)+h2o(l) with enthalpy of reaction δhrxn∘=37.5kj/mol what is the enthalpy of formation of
Anni [7]
Answer is: -601,2 kJ/mol
Chemical reaction: Mg(OH)₂ → MgO + H₂O.
ΔHrxn = 37,5 kJ/mol.
ΔHf(Mg(OH)₂) = <span>−924,5 kJ/mol.
</span>ΔHf(H₂O) = <span>−285,8 kJ/mol.
</span>ΔHrxn -enthalpy of reaction.
ΔHf - enthalpy of formation.
<span>ΔHrxn=∑productsΔHf−∑reactantsΔHf. 
</span>ΔHf(MgO) = -924,5 kJ/mol - (-285,8 kJ/mol) + 37,5 kj/mol.
ΔHf(MgO) = -601,2 kJ/mol.
7 0
3 years ago
Read 2 more answers
What is the volume of stank of nitrogen that contains 17 moles of nitrogen at 34 C under 12,000Pa?
Otrada [13]

Answer:

3626.76dm³

Explanation:

Given parameters:

Number of moles of Nitrogen in tank = 17moles

Temperature of the gas = 34°C

Pressure on the gas = 12000Pa

Unkown:

Volume of the tank, V =?

Converting the parameters to workable units:

We take the temperature from °C to Kelvin

K = 273 +  °C  = 273 + 34 = 307k

Taking the pressure in Pa to atm:

101325Pa = 1atm

12000Pa = 0.118atm

Solution:

To solve this problem, we employ the use of the ideal gas equation. The ideal gas law combines three gas laws which are the Boyle's law, Charles's law and the Avogadro's law.

            It is expressed as PV = nRT

The unknown is the Volume and we make it the subject of the formula

             V = \frac{nRT}{P}

Where R is called the gas constant and it is given as 0.082atmdm³mol⁻¹K⁻¹

Therefore  V = \frac{17 x 0.082 x 307 }{0.118} = 3626.76dm³

3 0
3 years ago
Read 2 more answers
How many electrons would be transferred in either a voltaic or electrolytic cell that uses the following half reactions
omeli [17]

Answer:

Numbers of electrons transferred in the electrolytic or voltaic cell is 6 electrons.

Explanation:

Fe^{3+} (aq) + 3 e^-\rightarrow Fe (s) ,E^o = -0.036 V

Mg^{2+}(aq) + 2 e^- \rightarrow Mg (s),E^o = -2.37 V

The substance having highest positive reduction E^o potential will always get reduced and will undergo reduction reaction.

Reduction : cathode

Fe^{3+} (aq) + 3 e^-\rightarrow Fe (s) ,E^o = -0.036 V..[1]

Oxidation: anode

Mg(s)\rightarrow Mg^{2+}(aq) + 2 e^-,E^o = 2.37 V..[2]

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{red,cathode}-E^o_{red,anode}

E^o_{cell}=-0.036V-(-2.37 V)=2.334 V

The overall reaction will be:

2 × [1] + 3 × [2] :

2Fe^{3+} (aq) + 3Mg(s)+6e^-\rightarrow 2Fe (s)+3Mg^{2+}(aq)+6e^-

Electrons on both sides will get cancelled :

2Fe^{3+} (aq) + 3Mg(s)\rightarrow 2Fe (s)+3Mg^{2+}(aq)

Numbers of electrons transferred in the electrolytic or voltaic cell is 6 electrons.

5 0
3 years ago
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