Can someone help me please?

Select all that apply.

Shkiper50 [21]

it is also

-many OH- ions

-few H+ ions

-large Kb and small Ka

6 0

3 years ago

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The mineral magnesite contains magnesium carbonate, MgCO3 (molar mass= 84 g/mol), and other impurities. When a 1.26-g sample of

Leviafan [203]

The answer is 33.33 %

The explanation:

According to the reaction equation:

MgCO3(s) + 2HCl (aq) --> MgCl2(aq) + H20(l) + CO2(g)

we can see that 1 mole of MCO3 will produce → 1 mole of CO2

-Now we need o get number of mole of CO2:

and when we have 0.22 g of CO2, so number of mole = mass / molar mass

moles = 0.22 g / 44 g/mol = 0.005 mole

∴ moles of Mg = moles of CO2 = 0.005 mole

∴ mass of Mg = moles * molar mass

= 0.005 * 84 /mol = 0.42 g

∴ Percent of MgCO3 by mass of Mg = 0.42 g / 1.26 * 100

= 33.33 %

6 0

3 years ago

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I need some information about greenhouse effect, please. Could you help me?

LenKa [72]

Answer: green house affect is the sun on plants it is transfering energy to the plants to help them grow same with every plant, tree, fruit tree. everything needs sun.

Explanation: look up there.

4 0

2 years ago

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In thermodynamics, we determine the spontaneity of a reaction by the sign of ΔG. In electrochemistry, spontaneity is determined

faltersainse [42]

Answer:

For A: The standard cell potential of the reaction is 4.4 V

For B: The standard Gibbs free energy of the reaction is $-8.50\times 10^5J$

For C: The reaction is spontaneous as written.

Explanation:

For A:

The given chemical reaction follows:

$2Li(s)+Cl_2(g)\rightarrow 2Li^+(aq.)+2Cl^-(aq.)$

The given half reaction follows:

Oxidation half reaction: $Li(s)\rightarrow Li^+(aq.)+e^-;E^o_{Li^+/Li}=-3.04V$ ( × 2)

Reduction half reaction: $Cl_2(g)+2e^-\rightarrow 2Cl^-(aq.);E^o_{Cl_2/2Cl^-}=+1.36V$

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

Here, chlorine will undergo reduction reaction will get reduced.

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

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

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

$E^o_{cell}=1.36-(-3.04)=4.4V$

Hence, the standard cell potential of the reaction is 4.4 V

For B:

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$

where,

n = number of electrons transferred = $2mol\text{ e}^-$

F = Faradays constant = $96500J/V.mol\text{ e}^-$

$E^o_{cell}$ = standard cell potential = 4.4 V

Putting values in above equation, we get:

$\Delta G^o=-2\times 96500\times 4.4=-849200J=-8.50\times 10^5J$

Hence, the standard Gibbs free energy of the reaction is $-8.50\times 10^5J$

For C:

For a reaction to be spontaneous, the standard Gibbs free energy change of the reaction must be negative.

From above, the standard Gibbs free energy change of the reaction is coming out to be negative.

Hence, the reaction is spontaneous as written.

7 0

3 years ago

The electronic configuration of an element is given below.

Shalnov [3]

Answer:

It is reactive because it has to gain an electron to have a full outermost energy level.

Explanation:

The electron configuration of oxygen is 1s2,2s2 2p4.

Oxygen is in group six in the periodic table so it has six electrons in its valence shell. This means that it needs to gain two electrons to obey the octet rule and have a full outer shell of electrons (eight).

3 0

2 years ago