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4 years ago

When it acts as a base, which atom in hydroxylamine accepts a proton?

1 answer:

6 0

<span>The oxygen atom accepts the proton. The oxidation number of O is -2, meaning that there are two unshared electrons in the valence shell; In the ClO- ion, one of these is shared with the Cl- ion, leaving an unshared electron on the oxygen atom, which is what the hydrogen atom shares its electron with, becoming the proton accepted by the O atom.</span>

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Calculate the value of the diffusion coefficient D (in m2/s) at 547°C for the diffusion of some species in a metal; assume that

svetlana [45]

Answer : The value of diffusion coefficient is, $2.97\times 10^{-16}m^2/s$

Explanation :

Formula used :

$D=D_o\times \exp \left(-\frac{Q_d}{RT}\right )$

where,

D = diffusion coefficient = ?

$D_o$ = $5.6\times 10^{-5}m^2/s$

$Q_d$ = $177kJ/mol$

R = gas constant = 8.314 J/mol.K

T = temperature = $547^oC=273+547=820K$

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

$D=5.6\times 10^{-5}m^2/s\times \exp \left(-\frac{177kJ/mol}{(8.314J/mol.K)\times (820K)}\right )$

$D=2.97\times 10^{-16}m^2/s$

Thus, the value of diffusion coefficient is, $2.97\times 10^{-16}m^2/s$

6 0

3 years ago

How does the velocity of a falling object change over time?

valentinak56 [21]

I’m pretty sure the velocity increases as the time of an object falling increases :)

7 0

3 years ago

A 85.2 g copper bar was heated to 221.32 degrees Celsius and placed in a coffee cup calorimeter containing 4250 mL of water at 2

Assoli18 [71]

Answer:

The specific heat of copper is 0.385 J/g°C

Explanation:

A 85.2 g copper bar was heated to 221.32 degrees Celsius and placed in a coffee cup calorimeter containing 425 mL of water at 22.55 degrees Celsius. The final temperature of the water was recorded to be 26.15 degrees Celsius. What is the specific heat of the copper?

Step 1: Data given

Mass of copper = 85.2 grams

Temperature of copper = 221.32 °C

Volume of water = 425 mL

Temperature of water = 22.55 °C

Final temperature = 26.15 °C

Specific heat of water = 4.184 J/g°C

Step 2: Calculat the specific heat of copper

Heat lost = heat gained

Q = m*c*ΔT

Qcopper = -Qwater

m(copper)*c(copper)*ΔT(copper) = - m(water) * c(water) * ΔT(water)

⇒ m(copper) = 85.2 grams

⇒ c(copper) = TO BE DETERMINED

⇒ ΔT(copper) = the change in temeprature = T2 -T1 = 26.15 -221.32 = -195.17 °C

⇒ m(water) = The mass of water = 425 mL * 1g/mL = 425 grams

⇒ c(water) = The specific heat of water = 4.184 J/g°C

⇒ ΔT(water) = The change of temperature of water = 26.15 - 22.55 = 3.6

85.2 * c(copper) * (-195.17) = -425 * 4.184 * 3.6

c(copper) = 0.385 J/g°C

The specific heat of copper is 0.385 J/g°C

(Note, The original question says the volume of the water is 4250 mL. IF this is not an error, the specific heat of copper is 3.85 J/g°C (10x higher than the normal value).

8 0

3 years ago

For the following reaction, 42.2 grams of potassium hydrogen sulfate are allowed to react with 21.4 grams of potassium hydroxide

ASHA 777 [7]

Answer:

53.99g

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

KHSO4(aq) + KOH(aq) —> K2SO4(aq) + H2O(l)

Step 2:

Determination of the masses of KHSO4 and KOH that reacted and the mass of K2SO4 produced from the balanced equation.

This is illustrated below:

Molar mass of KHSO4 = 39 + 1 + 32 + (16x4) = 136g/mol

Mass of KHSO4 from the balanced equation = 1 x 136 = 136g

Molar mass of KOH = 39 + 16 + 1 = 56g/mol

Mass of KOH from the balanced equation = 1 x 56 = 56g

Molar mass of K2SO4 = (39x2) + 32 + (16x4) = 174g/mol

Mass of K2SO4 from the balanced equation = 1 x 174 = 174g.

From the balanced equation above, 136g of KHSO4 reacted with 56g of KOH to produce 174g of K2SO4

Step 3:

Determination of the limiting reactant. This is illustrated below:

From the balanced equation above, 136g of KHSO4 reacted with 56g of KOH.

Therefore, 42.2g of KHSO4 will react with = (42.2 x 56)/136 = 17.38g of KOH.

From the above calculations, we can see that only 17.38g out of 21.4g of KOH given was needed to react completely with 42.2g of KHSO4.

Therefore, KHSO4 is the limiting reactant and KOH is the excess reactant.

Step 4:

Determination of the maximum mass of K2SO4 produced from the reaction.

In this case, the limiting reactant will be used as all of it is used up in the reaction. The limiting reactant is KHSO4 and the maximum amount of K2SO4 produced can be obtained as follow:

From the balanced equation above, 136g of KHSO4 reacted to produce 174g of K2SO4.

Therefore, 42.2g of KHSO4 will react to produce = (42.2 x 174)/136 = 53.99g of K2SO4.

Therefore, the maximum amount of K2SO4 produced is 53.99g.

8 0

3 years ago

For a given mass of reactants, the energy released is greatest for a reaction involving

kari74 [83]

Answer: Option (D) is the correct answer.

Explanation:

In fusion reaction, two small nuclei combine together to result in the formation of a large nuclei.

Whereas in fission reaction, large nuclei dissociates into two or more small nuclei.

Nuclear fusion releases more energy as compared to nuclear fission.

For example, two hydrogen nuceli fuse together and result in the formation of helium atom along with release of energy.

$^{2}_{1}H + ^{2}_{1}H \rightarrow ^{4}_{2}He + Energy$

3 0

3 years ago

Read 2 more answers