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2 years ago

What is the maximum number of moles of nickel carbonate (NiCO3) that can form during the precipitation reaction

Chemistry

1 answer:

ZanzabumX [31]2 years ago

3 0

Answer:

The correct answer is : 0.025.

Explanation:

The precipitation reaction is as follows in this procedure:

NiSO4.6H20 (aq) + Na2CO3.10H2O (aq)⇒ NiCO3 (s) + Na2SO4 (aq) + 16H2O

So, the number of moles of reactants can be found by

number of moles : mass used/ molar mass

n (NiSO4×6H2O) = (6.57 g)/(262.84 g/mol) = 0.025 mol (for Ni)

n (Na2CO3×10H2O) = (7.15 g)/(286.14 g/mol) = 0.0250 mol (for CO3).

Thus, The maximum number of moles of NiCO3 that can form is 0.025 mol.

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A major problem caused by humans is the contamination and depletion of _____ resources. A. solar B. water C. wind

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Water is the only one of these that would work by process of elimination.

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2 years ago

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A 63.5 g sample of an unidentified metal absorbs 355 ) of heat when its temperature changes

insens350 [35]

0.208 is the specific heat capacity of the metal.

Explanation:

Given:

mass (m) = 63.5 grams 0R 0.0635 kg

Heat absorbed (q) = 355 Joules

Δ T (change in temperature) = 4.56 degrees or 273.15+4.56 = 268.59 K

cp (specific heat capacity) = ?

the formula used for heat absorbed and to calculate specific heat capacity of a substance will be calculated by using the equation:

q = mc Δ T

c = $\frac{q}{mΔ T}$

c = $\frac{355}{63.5X 268.59}$

= 0.208 J/gm K

specific heat capacity of 0.208 J/gm K

The specific heat capacity is defined as the heat required to raise the temperature of a substance which is 1 gram. The temperature is in Kelvin and energy required is in joules.

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2 years ago

In order for fission reactions to be successful, they must be self-perpetuating, meaning they must be able to keep themselves go

aleksandrvk [35]

Answer:

Option C is correct.

The minimum amount of material that is needed for a fission reaction to keep going is called the critical mass.

Explanation:

Nuclear fission is the term used to describe the breakdown of the nucleus of a parent isotope into daughter nuclei.

Normally, the initial energy supplied for nuclear fission is the energy to initiate the first breakdown of the first set of radioactive isotopes that breakdown. Once that happens, the energy released from the first breakdown is enough to drive further breakdown of numerous isotopas in a manner that leads to more energy generation.

But, for this to be able to be sustained and not fizzle out, a particular amount of radioactive material to undergo nuclear fission must be present. This particular amount is termed 'critical mass'

Hope this Helps!!!

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2 years ago

The reaction SO2(g)+2H2S(g)←→3S(s)+2H2O(g) is the basis of a suggested method for removal of SO2 from power-plant stack gases. T

kifflom [539]

Answer : The equilibrium $SO_2$ pressure is, $3.93\times 10^{-5}torr$

Explanation :

The given balanced chemical reaction is,

$SO_2(g)+2H_2S(g)\rightleftharpoons 3S(s)+2H_2O(g)$

First we have to calculate the standard free energy of reaction $(\Delta G^o)$ .

$\Delta G^o=G_f_{product}-G_f_{reactant}$

$\Delta G^o=[n_{S(s)}\times \Delta G_f^0_{(S(s))}+n_{H_2O(g)}\times \Delta G_f^0_{(H_2O(g))}]-[n_{SO_2(g)}\times \Delta G_f^0_{(SO_2(g))}+n_{H_2S(g)}\times \Delta G_f^0_{(H_2S(g))}]$

where,

$\Delta G^o$ = standard free energy of reaction = ?

n = number of moles

Now put all the given values in this expression, we get:

$\Delta G^o=[3mole\times (0kJ/mol)+2mole\times (-228.57kJ/mol)]-[1mole\times (-300.4kJ/mol)+2mole\times (-33.01kJ/mol)]$

$\Delta G^o=-90.72kJ/mol$

Now we have to calculate the value of $K_p$

$\Delta G^o=-RT\ln K_p$

where,

$\Delta G_^o$ = standard Gibbs free energy = -90.72 kJ/mol

R = gas constant = 8.314 J/mole.K

T = temperature = 298 K

$K_p$ = equilibrium constant = ?

Now put all the given values in this expression, we get:

$-90.72kJ/mol=-(8.314J/mol.K)\times (298K) \ln K_p$

$K_p=7.98\times 10^{15}$

Now we have to calculate the value of $K_p$ .

The given balanced chemical reaction is,

$SO_2(g)+2H_2S(g)\rightleftharpoons 3S(s)+2H_2O(g)$

The expression for equilibrium constant will be :

$K_p=\frac{(p_{H_2O})^2}{(p_{H_2S})^2\times (p_{SO_2})}$

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Let the equilibrium $SO_2$ pressure be, x

Pressure of $SO_2$ = Pressure of $H_2S$ = x

Now put all the given values in this expression, we get

$7.98\times 10^{15}=\frac{(22)^2}{(x)^2\times (x)}$

$x=3.93\times 10^{-5}torr$

Thus, the equilibrium $SO_2$ pressure is, $3.93\times 10^{-5}torr$

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2 years ago

Are the strengths of the interactions between the particles in the solute and between the particles in the solvent before the so

Colt1911 [192]

Answer:

Less than

Explanation:

The process of dissolution occurs as a kind of "tug of war". On one side are the solute-solute and solvent-solvent interaction forces, while on the other side are the solute-solvent forces.

Only when the solute-solvent forces are strong enough to overcome the pre-mixing forces do they overcome the "tug of war", and thus dissolution occurs.

Thus, it is concluded that the interaction forces between solute particles and solvent particles before they are combined are less than the interaction forces after dissolution.

5 0

2 years ago