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vladimir2022 [97]

1 year ago

8

What is the temperature change on a 75.0 gram sample of mercury if 480.0 cal of heat are added to it? The specific heat of mercu

ry = 0.033 cal/g C

1 answer:

Olegator [25]1 year ago

6 0

Answer:

$\Delta T=194^oC$

Explanation:

Hello,

In this case, in terms of the heat, mass, heat capacity and change in temperature, we can analyze thermal changes as:

$Q=mCp\Delta T\\$

In such a way, we compute the required change in temperature as shown below:

$\Delta T=\frac{Q}{mCp}=\frac{480.0cal}{75.0g*0.033\frac{cal}{g^oC} } \\\\\Delta T=194^oC$

Such change in temperature is positive indicating an increase in the temperature as the involved heat is positive, in means that heat was added to increase the temperature.

Best regards.

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The reaction 2NO + O2 → 2NO2 is third order. Assuming that a small amount of NO3 exists in rapid reversible equilibrium with NO

Rina8888 [55]

Answer:

The equation for the rate of this reaction is R = [NO] + {O2}

Explanation:

The rate-determining step of a reaction is the slowest step of a chemical reaction which determines the rate (speed) at which the overall reaction would take place.

Reaction mechanism:

The slow and fast reactions both have NO3 which is cancelled out on both sides, in order to get the overall reaction.

The rate law for this reaction would be that for the rate determining step:

R = [NO] + {O2}

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2 years ago

Calculate the heat of reaction, ΔH°rxn, for overall reaction for the production of methane, CH4.

Lesechka [4]

<u>Answer:</u> The enthalpy of the reaction for the production of $CH_4$ is coming out to be -74.9 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H^o$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]$

For the given chemical reaction:

$C(s)+2H_2(g)\rightarrow CH_4(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(CH_4(g))})]-[(1\times \Delta H^o_f_{(C(s))})+(2\times \Delta H^o_f_{(H_2(g))})]$

We are given:

$\Delta H^o_f_{(C(s))}=0kJ/mol\\\Delta H^o_f_{(H_2)}=0kJ/mol\\\Delta H^o_f_{CH_4}=-74.9kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-74.9))]-[1\times 0)+(2\times 0)]\\\\\Delta H^o_{rxn}=-74.9kJ$

Hence, the enthalpy of the reaction for the production of $CH_4$ is coming out to be -74.9 kJ

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2 years ago

Who helps recycle nutrients through an ecosystem? A. producers and decomposers B. producers and consumers C. consumers and decom

s2008m [1.1K]

D. Producers im sure of it hope im right

8 0

2 years ago

Read 2 more answers

Carbonic acid, H2CO3, has two acidic hydrogens. A solution containing an unknown concentration of carbonic acid is titrated with

stepan [7]

Answer:

1) Net ionic equation :

$2H^+(aq)+2OH^-(aq)\rightarrow 2H_2O(l)$

2) 0.765 M is the molarity of the carbonic acid solution.

Explanation:

1) In aqueous carbonic acid , carbonate ions and hydrogen ion is present.:

$H_2CO_3(aq)\rightarrow 2H^+(aq)+CO_3^{2-}(aq)$ ..[1]

In aqueous potassium hydroxide , potassium ions and hydroxide ion is present.:

$KOH(aq)\rightarrow K^+(aq)+OH^{-}(aq)$ ..[2]

In aqueous potassium carbonate , potassium ions and carbonate ion is present.:

$K_2CO_3(aq)\rightarrow 2K^+(aq)+CO_3^{2-}(aq)$ ..[3]

$H_2CO_3(aq)+2KOH(aq)\rightarrow K_2CO_3(aq)+2H_2O(l)$

From one:[1] ,[2] and [3]:

$2H^+(aq)+CO_3^{2-}(aq)+2K^+(aq)+2OH^{-}(aq)\rightarrow 2K^+(aq)+CO_3^{2-}(aq)+H_2O(l)$

Cancelling common ions on both sides to get net ionic equation :

$2H^+(aq)+2OH^-(aq)\rightarrow 2H_2O(l)$

2)

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2CO_3$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is KOH.

We are given:

$n_1=2\\M_1=?\\V_1=50.0 mL\\n_2=1\\M_2=3.840 M\\V_2=20.0 mL$

Putting values in above equation, we get:

$M_1=\frac{1\times 3.840 M\times 20.0 mL}{2\times 50.2 mL}=0.765 M$

0.765 M is the molarity of the carbonic acid solution.

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2 years ago

An atom of lead has a radius of and the average orbital speed of the electrons in it is about . Calculate the least possible unc

victus00 [196]

The question is incomplete. Here is the complete question.

An atom of lead has a radius of 154 pm and the average orbitalspeed of the electron in it is about 1.8x $10^{8}$ m/s. Calculate the least possible uncertainty in a measurement of the speed of an electron in an atom of lead. Write your answer as a percentage of the average speed, and round it to significant 2 digits.

Answer: v% = 0.21 m/s

Explanation: To calculate the uncertainty, use <u>Heisenberg's Uncertainty Principle</u>, which states that: ΔpΔx≥ $\frac{h}{4\pi }$

where h is <u>Planck's constant</u> and it is equal to 6.626. $10^{-34}$ m²kg/s.

Since p (momentum) is p = m.v:

mΔv.Δx ≥ $\frac{h}{4\pi }$

Δv = $\frac{h}{4\pi.x.m }$

Given that: r = x = 1.54. $10^{-10}$ m and mass of an electron is m=9.1. $10^{-31}$ kg

Δv = $\frac{6.626.10^{-34} }{4.3.14.1.54.10^{-10}.9.1.10^{-31}}$

Δv = 0.0376. $10^{7}$

As percentage of average speed:

Δv. $\frac{1}{v}$ .100% = $\frac{0.0376.10^{7} }{1.8.10^{8} }$ .10² = 0.021.10 = 0.21%

The least possible uncertainty in a speed of an electron is 0.21%.

5 0

2 years ago