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3 years ago

Fifth grade chemistry please help me with this. The questions are related to each other and are attached to the problem.

Chemistry

2 answers:

Juli2301 [7.4K]3 years ago

8 0

What Grant felt was something known as a ‘static current’. Experiencing a light electrical shock when you touch another person, (in this case when Grant touched Olivia) is when electrons move quickly towards the protons.

r-ruslan [8.4K]3 years ago

8 0

Answer:

This might not be right but. It could be the motion of olivia when she was walking through the room she could of made a static or force to make the shock

Explanation:

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The building block of a nucleic acid, consisting of a five-carbon sugar covalently bonded to a nitrogenous base and a phosphate

Romashka-Z-Leto [24]

Answer:

Nucleotides

Explanation:

Nucleotides are the organic molecules which serve as monomer units for the formation of nucleic acid polymers which are the deoxyribonucleic acid and the ribonucleic acid (RNA) and both are the essential biomolecules within the life on the Earth.

Nucleotides are building blocks of the nucleic acids. They are the molecules which are composed of three sub units which are:

Nitrogenous base which is also called as nucleobase
Five-carbon sugar which can be ribose or deoxyribose
At least one phosphate group which is attached to the sugar.

8 0

3 years ago

CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at 25 ∘C under these conditions: PCO2PCCl4PCOCl2===0.140 atm0.185 atm0.7

padilas [110]

<u>Answer:</u> The $\Delta G$ for the reaction is 54.425 kJ/mol

<u>Explanation:</u>

For the given balanced chemical equation:

$CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)$

We are given:

$\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol$

To calculate $\Delta G^o_{rxn}$ for the reaction, we use the equation:

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

For the given equation:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J$

Conversion factor used = 1 kJ = 1000 J

The expression of $K_p$ for the given reaction:

$K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}$

We are given:

$p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm$

Putting values in above equation, we get:

$K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85$

To calculate the gibbs free energy of the reaction, we use the equation:

$\Delta G=\Delta G^o+RT\ln K_p$

where,

$\Delta G$ = Gibbs' free energy of the reaction = ?

$\Delta G^o$ = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = $8.314J/K mol$

T = Temperature = $25^oC=[25+273]K=298K$

$K_p$ = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

$\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol$

Hence, the $\Delta G$ for the reaction is 54.425 kJ/mol

7 0

3 years ago

The half-cell is a chamber in the voltaic cell where one half-cell is the site of the oxidation reaction and the other half-cell

marysya [2.9K]

Answer:

$\boxed{\rm \text{Fe(s) $\rightleftharpoons$ Fe$^{2+}$(aq) + 2e$^{-}$}}$

Explanation:

The half-cell reduction potentials are

Ag⁺(aq) + e⁻ ⇌ Ag(s) E° = 0.7996 V

Fe²⁺(aq) + 2e⁻ ⇌ Fe(s) E° = -0.447 V

To create a spontaneous voltaic cell, we reverse the half-reaction with the more negative half-cell potential.

The anode is the electrode at which oxidation occurs.

The equation for the oxidation half-reaction is

$\boxed{\rm \textbf{Fe(s) $\rightleftharpoons$ Fe$^{2+}$(aq) + 2e$^{-}$}}$

4 0

3 years ago

The pressure of a basketball is 412 mmHg. How many kPa is that equal to?

pychu [463]

Answer:

d. 54.9 kPa

Explanation:

mmHg and Pa are units of pressure used in chemistry principally in the study of gases. 1mmHg is equal to 133.322Pa. 412mmHg are:

412 mmHg * (133.322Pa / 1mmHg) = 54929 Pa

The prefix K (Kilo) represents one thousand of the determined unit.

54929Pa are:

54929Pa * (1KPa / 1000Pa) = 54.9kPa

Right answer is:

3 0

3 years ago

A 50.00 g sample of an unknown metal is heated to 45.00°C. It is then placed in a coffee-cup calorimeter filled with water. The

AysviL [449]

Answer:- Heat lost by the metal is 279.45 cal.

Solution:- This type of problems are solved by using the concept, heat given = - heat taken

Metal temperature is decreasing from 45.00 degree C to 11.08 degree C. It means the heat is lost by the metal and this heat lost by metal is gained by water and the calorimeter to raise their temperature.

the equation we use is, $q=mc\Delta T$ .