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3 years ago

When 3.0 kg of water is warmed from 10 °C to 80 °C, how much heat energy is needed?

Chemistry

1 answer:

Naddik [55]3 years ago

6 0

Answer:

THE HEAT NEEDED TO CHANGE 3KG OF WATER FROM 10 C TO 80 C IS 877.8kJ OR 877,800 J.

Explanation:

Mass = 3.0 kg = 3 * 1000 = 3000 g

Initial temperature = 10 C

Final temperature = 80 C

Change in temperature = 80 - 10 = 70 C

Specific heat of water = 4.18 J/g C

Heat needed = unknown

Heat is the amount of energy in joules needed to change a gram of water by 1 C.

Heat = mass * specific heat * change in temperature

Heat = 3000 g * 4.18 J/g C * 70 C

Heat = 877 800 Joules

Heat = 877.8 kJ.

The heat needed to change 3 kg mass of water from 10 C to 80 C is 877,800 J or 877.8 kJ.

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Which of the following is unchanged at the end of the CNO cycle?

Musya8 [376]

Answer:

The correct option is: A. carbon-12

Explanation:

The CNO cycle, the abbreviation for the carbon-nitrogen-oxygen cycle, is a catalytic cycle by which the stars produce helium from elemental hydrogen, via a series of nuclear fusion reactions.

This cycle involves the fusion of four protons with carbon ( $_{6}^{12}\textrm{C}$ ), nitrogen isotope ( $_{7}^{13}\textrm{N}$ ), and oxygen isotope ( $_{8}^{15}\textrm{O}$ ), to give an alpha particle and two electron neutrinos and positrons.

The reaction involves the regeneration of carbon ( $_{6}^{12}\textrm{C}$ ) nucleus in the last step.

$_{6}^{12}\textrm{C}$ → $_{7}^{13}\textrm{N}$ → $_{6}^{13}\textrm{C}$ → $_{7}^{14}\textrm{N}$ → $_{8}^{15}\textrm{O}$ → $_{7}^{15}\textrm{N}$ → $_{6}^{12}\textrm{C}$

3 0

3 years ago

Commercial grade HCl solutions are typically 39.0% (by mass) HCl in water. Determine the molality of the HCl, if the solution ha

marta [7]

Molality of the solution is defined as the number of moles of a substance dissolved divided by the mass of the solvent:

Molality = number of moles / solvent mass

From the concentration of 39% (by mass) of HCl in water, we construct the following reasoning:

in 100 g solution we have 39 g hydrochloric acid (HCl)

number of moles = mass / molecular weight

number of moles of HCl = 39 / 36.5 = 1.07 moles

solvent (water) mass = solution mass - hydrochloric acid mass

solvent (water) mass = 100 - 39 = 61 g

Now we can determine the molality:

molality = 1.07 moles / 61 g = 0.018

8 0

3 years ago

The difference in an area with high concentration and an area with low concentration is called.

Varvara68 [4.7K]

The difference in an area with high concentration and an area with low concentration is called the concentration gradient.

<h3>What is Concentration Gradient ?</h3>

A concentration gradient occurs when the concentration of particles is higher in one area than another.

In passive transport, particles will diffuse down a concentration gradient, from areas of higher concentration to areas of lower concentration, until they are evenly spaced.

This difference in an area with high concentration and an area with low concentration is called the concentration gradient.

Learn more about diffusion here ;

brainly.com/question/24746577

#SPJ1

5 0

2 years ago

Determine the free energy(ΔG) from the standard cell potential (Ecell0 ) for the reaction:2ClO2-(aq)+Cl2(g)→2ClO2(g)+ 2Cl-(aq)wh

Dima020 [189]

<u>Answer:</u> The $\Delta G^o$ for the given reaction is $-7.84\times 10^4J$

<u>Explanation:</u>

For the given chemical reaction:

$2ClO_2^-(aq.)+Cl_2(g)\rightarrow 2ClO_2(g)+2Cl^-(aq.)$

Half reactions for the given cell follows:

<u>Oxidation half reaction:</u> $ClO_2^-\rightarrow ClO_2+e^-;E^o_{ClO_2^-/ClO_2}=0.954V$ ( × 2)

<u>Reduction half reaction:</u> $Cl_2+2e^-\rightarrow 2Cl(g);E^o_{Cl_2/2Cl^-}=1.36V$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=1.36-(0.954)=0.406V$

To calculate standard Gibbs free energy, we use the equation:

$\Delta G^o=-nFE^o_{cell}$

Where,

n = number of electrons transferred = 2

F = Faradays constant = 96500 C

$E^o_{cell}$ = standard cell potential = 0.406 V

Putting values in above equation, we get:

$\Delta G^o=-2\times 96500\times 0.406=-78358J=-7.84\times 10^4J$

Hence, the $\Delta G^o$ for the given reaction is $-7.84\times 10^4J$

4 0

3 years ago

In the decomposition reaction, 1 mole of water (mw = 18.015 g/mol) was produced for every mole of cuo (mw = 79.545 g/mol) produc

natita [175]

Reactives -> Products

CuO and water are products.

I found this reaction which has CuO and water as products: decomposition of Cu(OH)2.

Cu(OH)2 -> CuO + H2O

Stoichiometry calculus involve the mole proportions you can see in the reaction: When 1 mole of Cu(OH)2 reacts, 1 mole of CuO and 1 mole of H2O are formed.

Considering the molar masses:

Cu(OH)2 = 83.56 g/mol

CuO = 79.545 g/mol

H2O = 18.015 g/mol

Then: When 83.56 g of Cu(OH)2 react, 79.545 g of CuO and 18.015 g H2O are formed.

You should use that numbers in the rule of three:

79.545 g CuO __________18.015 g water

3.327 g CuO__________ x =3.327*18.015 /79.545 g water

x= 0.7535 g water

3 0

3 years ago