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2 years ago

What is the molecular formula of chalk ?

Chemistry

2 answers:

guapka [62]2 years ago

7 0

Answer:

CaCO3

Explanation:

Hope this helps!

NemiM [27]2 years ago

4 0

Explanation:

Calcium carbonate (Chalk) is a chemical compound, with the chemical formula CaCO3. Calcium carbonate (Chalk) is a chemical compound, with the chemical formula CaCO3.

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8 0

3 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

How to calculate IMA

Cloud [144]

Using IMA is used in Engineering, but there is a simple rhyme I use called IMA Deer. IMA=De/Dr

7 0

3 years ago

3. The doctor notices that Janet has long fingernails, and comments that fingernails grow at a rate (or

Viktor [21]

Answer:

7.93 × 10⁻¹³ km/s

Explanation:

Step 1: Given data

Rate of growth of the fingernails (r): 2.50 cm/year

Step 2: Convert "r" from cm/year to km/year

We will use the following conversion factors.

1 m = 100 cm

1 km = 1000 m

$\frac{2.50cm}{year} \times \frac{1m}{100cm} \times \frac{1km}{1000m} = 2.50 \times 10^{-5} km/year$

Step 3: Convert "r" from km/year to km/s

We will use the following conversion factors.

1 year = 365 day

1 day = 24 h

1 h = 60 min

1 min = 60 s

$\frac{2.50 \times 10^{-5} km}{year} \times \frac{1year}{365day} \times \frac{1day}{24h} \times \frac{1h}{60min} \times \frac{1min}{60s} = 7.93 \times 10^{-13} km/s$

2.50 cm/year is equal to 7.93 × 10⁻¹³ kilometers/second.

6 0

2 years ago

Question 13: Consider the strength of the Hβ absorption line in the spectra of stars of various surface temperatures. This is th

finlep [7]

Answer:

The absorption and strength of the H-beta lines change with the temperature of the stellar surface, and because of this, one can find the temperature of the star from their absorption lines and strength. To better comprehend, let us look into the concept of the atom's atomic structure.

Atoms possess distinct energy levels and these levels of energy are constant, that is, the temperature has no influence on it. However, temperature possesses an influence on the electron numbers found within these levels of energy. Therefore, to generate an absorption line of hydrogen in the electromagnetic spectrum's visible band, the electrons are required to be present in the second energy level, that is when it captivates a photon.

Therefore, after captivating the photons the electrons jump from level 2 to level 4, which shows that there is an increase in the stellar surface temperature and at the same time one can witness a decline in the strength of the H-beta lines. In case, if the temperature of the surface increases too much, then one will witness no attachment of electron with the hydrogen atom and thus no H lines, and if the temperature of the surface becomes too low, then the electrons will stay in the ground state and no formation of H lines will take place in that condition too.

Hence, to generate a very robust H line, after captivating photons the majority of the electrons are required to stay in the second energy level.

5 0

3 years ago