All categories

2 years ago

A beaker contains nickel(ii sulfate, zinc(ii sulfate and solid iron. what species in this beaker will be reduced?

2 answers:

6 0

If a beaker is contained with the substances nickel sulfate, zinc sulfate and solid iron, there is a substance that will be reduced in the following substances that is in it. The substance that will be reduced as it is in the beaker is the Ni^2 + or also known as the Nickel (II)

attashe74 [19]2 years ago

4 0

Answer:

Nickel (II) to solid nickel.

Explanation:

Hello,

In this case, the placed species are nickel (II) sulfate $NiSO_4$ , zinc sulfate $ZnSO_4$ and solid iron $Fe$ . Now, as the nickel sulfate is more reactive than the zinc sulfate, it will experience the following reduction half-reaction from nickel (II) to solid nickel:

$Ni^{2+}+2e^--->Ni^0$

Nevertheless, it is possible that the zinc might experience the same reduction reaction from zinc (II) to solid zinc as well.

Best regards.

You might be interested in

By mistake, you added salt instead of sugar to the oil. How can you remove the salt?

shutvik [7]

Answer,

Add water to dissolve the salt solution.Shake the solution,oil and water are immiscible liquids therefore oil will not dissolve.Put the solution in a separating funnel and leave the solution undisturbed for a few minutes.Water is denser than oil there oil will come to the top.Power the lower layer of water in to a beaker.Boil the water until water turns into vapour leaving the salt behind.

8 0

2 years ago

Read 2 more answers

Acetaldehyde decomposes at 750 K: CH3CHO → CO + CH4. The reaction is first order in acetaldehyde and the half-life of the reacti

Degger [83]

Answer:

k = 1.3 x 10⁻³ s⁻¹

Explanation:

For a first order reaction the integrated rate law is

Ln [A]t/[A]₀ = - kt

where [A] are the concentrations of acetaldehyde in this case, t is the time and k is the rate constant.

We are given the half life for the concentration of acetaldehyde to fall to one half its original value, thus

Ln [A]t/[A]₀ = Ln 1/2[A]₀/[A]₀= Ln 1/2 = - kt

- 0.693 = - k(530s) ⇒ k = 1.3 x 10⁻³ s⁻¹

4 0

2 years ago

Estimate the increase in the molar entropy of O2(g) when the temperature is increased at constant pressure from 298 K to 348 K,

Tamiku [17]

Explanation:

It is known that relation between entropy, heat energy and temperature is as follows.

dS = $\frac{Q}{T}$

$\int dS = \int \frac{Q}{T}$

Also we know that at constant pressure, Q = $\Delta H = C_{p} - dT$

$\int_{S_{1}}^{S_{2}} dS = \int_{T_{1}}^{T_{2}} C_{p} \frac{dT}{T}$

$\Delta S = C_{p} \int_{T_{1}}^{T_{2}} \frac{dT}{T}$

As the given data is as follows.

$T_{1}$ = 298 K, $T_{2}$ = 348 K

$C_{p}$ = 29.355 J/K mol

Now, putting the given values into the above formula as follows.

$\Delta S = C_{p} ln {T_{1}}^{T_{2}} \frac{dT}{T}$

= $29.355 [ln (348) - ln (298)]$

= $29.355 [5.85 - 5.69]$

= 4.48 J/k mol

Thus, we can conclude that the increase in the molar entropy of given oxygen gas is 4.48 J/k mol.

6 0

2 years ago

50 kg of N2 gas and 10kg of H2 gas are mixed to produce NH3 gas calculate the NH3gas formed. Identify the limiting reagent in th

statuscvo [17]

Answer:

1. H2 is the limiting reactant.

2. 56666.67g ( i.e 56.67kg) of NH3 is produced.

Explanation:

Step 1:

The equation for the reaction. This is given below:

N2 + H2 —> NH3

Step 2:

Balancing the equation.

N2 + H2 —> NH3

The above equation can be balanced as follow :

There are 2 atoms of N on the left side and 1 atom on the right side. It can be balance by putting 2 in front of NH3 as shown below:

N2 + H2 —> 2NH3

There are 6 atoms of H on the right side and 2 atoms on the left side. It can be balance by putting 3 in front of H2 as shown below

N2 + 3H2 —> 2NH3

Now the equation is balanced.

Step 3:

Determination of the masses of N2 and H2 that reacted and the mass of NH3 produced from the balanced equation. This is illustrated below:

N2 + 3H2 —> 2NH3

Molar Mass of N2 = 2x14 = 28g/mol

Molar Mass of H2 = 2x1 = 2g/mol

Mass of H2 from the balanced equation = 3 x 2 = 6g

Molar Mass of NH3 = 14 + (3x1) = 14 + 3 = 17g/mol

Mass of NH3 from the balanced equation = 2 x 17 = 34g

From the balanced equation above,

28g of N2 reacted with 6g of H2 to produce 34g of NH3

Step 4:

Determination of the limiting reactant. This is illustrated below:

N2 + 3H2 —> 2NH3

Let us consider using all the 10kg (i.e 10000g) of H2 to see if there will be any left of for N2.

From the balanced equation above,

28g of N2 reacted with 6g of H2.

Therefore, Xg of N2 will react with 10000g of H2 i.e

Xg of N2 = (28 x 10000)/6

Xg of N2 = 46666.67g

We can see from the calculations above that there are leftover for N2 as only 46666.67g reacted out of 50kg ( i.e 50000g) that was given. Therefore, H2 is the limiting reactant.

Step 5:

Determination of the mass of NH3 produced during the reaction. This is illustrated below:

N2 + 3H2 —> 2NH3

From the balanced equation above,

6g of H2 reacted to produce 34g of NH3.

Therefore, 10000g of H2 will react to produce = ( 10000 x 34)/6 = 6g of 56666.67g of NH3.

Therefore, 56666.67g ( i.e 56.67kg) of NH3 is produced.

3 0

2 years ago

Read 2 more answers

Will is learning about a new kind of plastic used to make models. He learns that infrared light is absorbed by the plastic, X-ra

tamaranim1 [39]

Answer:

no it dosen't matter

Explanation:

because it absorbed the light so it had to damage it.

8 0

2 years ago

Read 2 more answers