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3 years ago

Kernel structure for manganese

Chemistry

1 answer:

Amanda [17]3 years ago

3 0

Answer:

The nucleus consists of 25 protons (red) and 30 neutrons (blue). 25 electrons (green) bind to the nucleus, successively occupying available electron shells (rings). Manganese is a transition metal in group 7, period 4, and the d-block of the periodic table. It has a melting point of 1246 degrees Celsius.

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All energy originates from the _______.

Phantasy [73]

Answer:

sun

Explanation:

because all plants use energy from the sun to make food and grow

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3 years ago

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ascorbic acid is a diprotic acid (Ka= 8.0x10^-5 and Ka2= 1.6x10^-12). What is the pH of a 0.260 M solution of ascorbic acid

zlopas [31]

The pH of a 0.260 M solution of ascorbic acid is 0.585. Details about pH can be found below.

<h3>How to calculate pH?</h3>

The pH of a solution can be calculated using the following expression:

pH = - log {H+}

According to this question, ascorbic acid is a diprotic acid and posseses a concentration of 0.260M. The pH can be calculated as follows;

pH = - log {0.260}

pH = 0.585

Therefore, the pH of a 0.260 M solution of ascorbic acid is 0.585.

Learn more about pH at: brainly.com/question/15289741

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1 year ago

Which object will have the greatest gravitational pull?

sukhopar [10]

Answer:

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2 years ago

Calculate the standard reaction Gibbs free energy for the following cell reactions: (a) 2 Ce41(aq) 1 3 I2(aq) S 2 Ce31(aq) 1 I32

Law Incorporation [45]

Answer:

For a: The standard Gibbs free energy of the reaction is -347.4 kJ

For b: The standard Gibbs free energy of the reaction is 746.91 kJ

Explanation:

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$ ............(1)

For a:

The given chemical equation follows:

$2Ce^{4+}(aq.)+3I^{-}(aq.)\rightarrow 2Ce^{3+}(aq.)+I_3^-(aq.)$

Oxidation half reaction: $Ce^{4+}(aq.)\rightarrow Ce^{3+}(aq.)+e^-$ ( × 2)

Reduction half reaction: $3I^_(aq.)+2e^-\rightarrow I_3^-(aq.)$

We are given:

$n=2\\E^o_{cell}=+1.08V\\F=96500$

Putting values in equation 1, we get:

$\Delta G^o=-2\times 96500\times (+1.80)=-347,400J=-347.4kJ$

Hence, the standard Gibbs free energy of the reaction is -347.4 kJ

For b:

The given chemical equation follows:

$6Fe^{3+}(aq.)+2Cr^{3+}+7H_2O(l)(aq.)\rightarrow 6Fe^{2+}(aq.)+Cr_2O_7^{2-}(aq.)+14H^+(aq.)$

Oxidation half reaction: $Fe^{3+}(aq.)\rightarrow Fe^{2+}(aq.)+e^-$ ( × 6)

Reduction half reaction: $2Cr^{2+}(aq.)+7H_2O(l)+6e^-\rightarrow Cr_2O_7^{2-}(aq.)+14H^+(aq.)$

We are given:

$n=6\\E^o_{cell}=-1.29V\\F=96500$

Putting values in equation 1, we get:

$\Delta G^o=-6\times 96500\times (-1.29)=746,910J=746.91kJ$

Hence, the standard Gibbs free energy of the reaction is 746.91 kJ

7 0

3 years ago

Enter your answer in the provided box.

Lina20 [59]

Answer:

5.06atm

Explanation:

Using the combined gas law equation;

P1V1/T1 = P2V2/T2

Where;

P1 = initial pressure (atm)

P2 = final pressure (atm)

V1 = initial volume (Litres)

V2 = final volume (Litres)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the information provided in this question;

P1 = 1.34 atm

P2 = ?

V1 = 5.48 L

V2 = 1.32 L

T1 = 61 °C = 61 + 273 = 334K

T2 = 31 °C = 31 + 273 = 304K

Using P1V1/T1 = P2V2/T2

1.34 × 5.48/334 = P2 × 1.32/304

7.34/334 = 1.32P2/304

Cross multiply

334 × 1.32P2 = 304 × 7.34

440.88P2 = 2231.36

P2 = 2231.36/440.88

P2 = 5.06

The final pressure is 5.06atm

6 0

3 years ago