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3 years ago

Choose the more reactive metal Te or O

Chemistry

2 answers:

kvasek [131]3 years ago

8 0

Te is the answer between the two

Allushta [10]3 years ago

3 0

Tellurium is a more reactive metal.

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A 45.0-gram sample of copper metal was heated from 20.0°C to 100.0°C. Calculate the heat absorbed, in kJ, by the metal.

s2008m [1.1K]

Answer:

1.386 KJ

Explanation:

From the question given above, the following data were obtained:

Mass (M) of copper = 45 g

Initial temperature (T1) = 20.0°C

Final temperature (T2) = 100.0°C

Heat absorbed (Q) =..?

Next, we shall determine the change in temperature. This can be obtained as follow:

Initial temperature (T1) = 20.0°C

Final temperature (T2) = 100.0°C

Change in temperature (ΔT) =?

ΔT = T2 – T1

ΔT = 100 – 20

ΔT = 80 °C

Next, we shall determine the heat absorbed by the sample of copper as follow:

Mass (M) of copper = 45 g

Change in temperature (ΔT) = 80 °C

Specific heat capacity (C) of copper = 0.385 J/gºC

Heat absorbed (Q) =..?

Q = MCΔT

Q = 45 × 0.385 × 80

Q = 1386 J

Finally, we shall convert 1386 J to KJ. This can be obtained as follow:

1000 J = 1 KJ

Therefore,

1386 J = 1386 J × 1 KJ /1000 J

1386 J = 1.386 KJ

Thus, the heat absorbed by the sample of the sample of copper is 1.386 KJ.

5 0

2 years ago

what is the inferred pressure, in million of atmospheres, in Earth's interior at a depth of 2900 kilometers?

NemiM [27]

Answer:

1.4 millions atmospheres

Explanation:

8 0

3 years ago

According to reference table adv-10, which reaction will take place spontaneously?

olga_2 [115]

Missing table!! write the elements with the first letter of the symbol with Upper Caps letters!!!

http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Tables/EStandardTable.htm

Ni2+ +Pb(s) → Ni(s) + Pb2+
The potential of the oxidation of Pb(s) --> Pb2+(aq) is 0.126 V
The potential of the reduction go Ni2+(aq) --> Ni(s) is -0.25 V

Add the two together and the potential for the reaction is -0.124 V (NO SPONTANEOUS THE SIGN IS NEGATIVE)

au3+ + al(s) → au(s) + al3+Au3+(aq) -> Au(s) +1.5 VAl -> Al3+ +1.66VV= 3.16 (SPONTANEOUS THE SIGN OF THE PONTENTIAL IS POSITIVE)Sr2+ + Sn(s) → Sr(s) + Sn2+

Sr2+(aq) + 2 e– Sr(s) V= -2.89V
Sn -> Sn2+ V= 0.14 V
V= -2.75 V (no spontaneous)

Fe2+ + Cu(s) → Fe(s) + Cu2+
Fe2+(aq) + 2 e– Fe(s) V= -0.44 V
Cu -> C2+ V = - 0.337V

V= - 0.777V (no spontaneous)

5 0

3 years ago

Calculate the activity coefficients for the following conditions:

uysha [10]

Answer:

For a: The activity coefficient of copper ions is 0.676

For b: The activity coefficient of potassium ions is 0.851

For c: The activity coefficient of potassium ions is 0.794

Explanation:

To calculate the activity coefficient of an ion, we use the equation given by Debye and Huckel, which is:

$-\log\gamma_i=\frac{0.51\times Z_i^2\times \sqrt{\mu}}{1+(3.3\times \alpha _i\times \sqrt{\mu})}$ ........(1)

where,

$\gamma_i$ = activity coefficient of ion

$Z_i$ = charge of the ion

$\mu$ = ionic strength of solution

$\alpha _i$ = diameter of the ion in nm

To calculate the ionic strength, we use the equation:

$\mu=\frac{1}{2}\sum_{i=1}^n(C_iZ_i^2)$ ......(2)

where,

$C_i$ = concentration of i-th ions

$Z_i$ = charge of i-th ions

For a:

We are given:

0.01 M NaCl solution:

Calculating the ionic strength by using equation 2:

$C_{Na^+}=0.01M\\Z_{Na^+}=+1\\C_{Cl^-}=0.01M\\Z_{Cl^-}=-1$

Putting values in equation 2, we get:

$\mu=\frac{1}{2}[(0.01\times (+1)^2)+(0.01\times (-1)^2)]\\\\\mu=0.01M$

Now, calculating the activity coefficient of $Cu^{2+}$ ion in the solution by using equation 1:

$Z_{Cu^{2+}}=2+\\\alpha_{Cu^{2+}}=0.6\text{ (known)}\\\mu=0.01M$

Putting values in equation 1, we get:

$-\log\gamma_{Cu^{2+}}=\frac{0.51\times (+2)^2\times \sqrt{0.01}}{1+(3.3\times 0.6\times \sqrt{0.01})}\\\\-\log\gamma_{Cu^{2+}}=0.17\\\\\gamma_{Cu^{2+}}=10^{-0.17}\\\\\gamma_{Cu^{2+}}=0.676$

Hence, the activity coefficient of copper ions is 0.676

For b:

We are given:

0.025 M HCl solution:

Calculating the ionic strength by using equation 2:

$C_{H^+}=0.025M\\Z_{H^+}=+1\\C_{Cl^-}=0.025M\\Z_{Cl^-}=-1$

Putting values in equation 2, we get:

$\mu=\frac{1}{2}[(0.025\times (+1)^2)+(0.025\times (-1)^2)]\\\\\mu=0.025M$

Now, calculating the activity coefficient of $K^{+}$ ion in the solution by using equation 1:

$Z_{K^{+}}=+1\\\alpha_{K^{+}}=0.3\text{ (known)}\\\mu=0.025M$

Putting values in equation 1, we get:

$-\log\gamma_{K^{+}}=\frac{0.51\times (+1)^2\times \sqrt{0.025}}{1+(3.3\times 0.3\times \sqrt{0.025})}\\\\-\log\gamma_{K^{+}}=0.070\\\\\gamma_{K^{+}}=10^{-0.070}\\\\\gamma_{K^{+}}=0.851$

Hence, the activity coefficient of potassium ions is 0.851

For c:

We are given:

0.02 M $K_2SO_4$ solution:

Calculating the ionic strength by using equation 2:

$C_{K^+}=(2\times 0.02)=0.04M\\Z_{K^+}=+1\\C_{SO_4^{2-}}=0.02M\\Z_{SO_4^{2-}}=-2$

Putting values in equation 2, we get:

$\mu=\frac{1}{2}[(0.04\times (+1)^2)+(0.02\times (-2)^2)]\\\\\mu=0.06M$

Now, calculating the activity coefficient of $K^{+}$ ion in the solution by using equation 1:

$Z_{K^{+}}=+1\\\alpha_{K^{+}}=0.3\text{ (known)}\\\mu=0.06M$

Putting values in equation 1, we get:

$-\log\gamma_{K^{+}}=\frac{0.51\times (+1)^2\times \sqrt{0.06}}{1+(3.3\times 0.3\times \sqrt{0.06})}\\\\-\log\gamma_{K^{+}}=0.1\\\\\gamma_{K^{+}}=10^{-0.1}\\\\\gamma_{K^{+}}=0.794$

Hence, the activity coefficient of potassium ions is 0.794

6 0

3 years ago

How can the gravitational potential energy of an object be changed?

Aleksandr [31]

Answer:

Gravitational potential energy may be converted to other forms of energy, such as kinetic energy. If we release the mass, gravitational force will do an amount of work equal to mgh on it, thereby increasing its kinetic energy by that same amount (by the work-energy theorem).

Explanation:

7 0

2 years ago