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2 years ago

Name two methods that can be used to break down compounds into simpler substances!

2 answers:

8 0

The two methods that can be used do break down compounds into simpler substances or smaller units, are known as Hydrolysis and Photolysis.

Hydrolysis uses water to split a molecule. Water is used to break the bonds of a molecule. There are three main types of hydrolysis:1. acid hydrolysis2. base hydrolysis 3. salt hydrolysis

Photolysis uses energy from light to split molecules. This energy is called photons. As in hydrolysis, photons are used to break the bonds of a molecule. Photolysis is a chemical reaction.

kotykmax [81]2 years ago

7 0

Photolysis and hydrolysis. These are two methods that can be used to break down a compound into simpler substances and smaller units.

Water which is used to break the bonds of molecules and split molecules is used from hydrolysis. Hydrolysis is made of three types which include;
1. salt hydrolysis.
2. acid hydrolysis.
3. Base hydrolysis.
Photolysis is well known to use energy from light to split the molecule and the same energy is referred to as photons which are used to break builds of molecules.

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choose the reaction that illustrates delta H *f for Ca(NO3)2.(A) Ca (s) + N2 (g) + 3O2 ---> Ca(NO3)2 (s)(B) Ca2 (aq) + 2 NO3-

kompoz [17]

Answer: The correct answer is Option A.

Explanation:

Standard enthalpy of formation is the change in enthalpy of one mole of a substance present at the standard state that is 1 atm of pressure and 298 K of temperature. The substance is formed from its pure elements under the same conditions.

We are given a chemical compound having chemical formula $Ca(NO_3)_2$

This compound is formed by the combination of calcium, nitrogen and oxygen elements.

The chemical equation for the formation of $Ca(NO_3)_2$ from the components in their standard states follows:

$Ca+N_2+3O_2\rightarrow Ca(NO_3)_3$

Hence, the correct answer is Option A.

3 0

2 years ago

From the following enthalpy of reaction data and data in Appendix C, calculate ΔH∘f for CaC2(s): CaC2(s)+2H2O(l)→Ca(OH)2(s)+C2H2

sashaice [31]

Answer:

From the following enthalpy of reaction data and data in Appendix C, calculate ΔH∘f for CaC2(s): CaC2(s)+2H2O(l)→Ca(OH)2(s)+C2H2(g)ΔH∘=−127.2kJ

ΔHf°(C2H2) = 227.4 kJ/mol

ΔHf°(H2O) = -285.8 kJ/mol and

ΔHf°(Ca(OH)2) = -985.2 kJ/mol

(Ans)

ΔHf° of CaC2 = -59.0 kJ/mol

Explanation:

CaC2(s) + 2 H2O(l) → Ca(OH)2(s) + C2H2 (g) = −127.2kJ

ΔHrxn = −127.2kJ

ΔHrxn = ΔHf°(C2H2) + ΔHf°(Ca(OH)2) - ΔHf°(CaC2)- 2ΔHf°(H2O);

ΔHf°(CaC2) = ΔHf°(C2H2) + ΔHf°(Ca(OH)2) - 2ΔHf°(H2O) – ΔHrxn

Where

ΔHf°(C2H2) = 227.4 kJ/mol

ΔHf°(H2O) = -285.8 kJ/mol and

ΔHf°(Ca(OH)2) = -985.2 kJ/mol

ΔHf°(CaC2) =227.4 - 985.2 + 2x285.8 + 127.2 = -59.0 kJ/mol

ΔHf°(CaC2) = -59.0 kJ/mol

7 0

2 years ago

When an electron in a 2p orbital of a particular atom makes a transition to the 2s orbital, a photon of approximate wavelength 6

Mariulka [41]

Answer:

The energy difference between these 2p and 2s orbitals is $3.07\times 10^{-19} J$

Explanation:

Wavelength of the photon emitted = $\lambda =646.3 nm =646.3\times 10^{-9} m$

Energy of the photon will corresponds to the energy difference between 2p and 2s orbital = E

Energy of the photon is given by Planck's equation:

$E=\frac{hc}{\lambda }$

h = Planck's constant = $6.626\tiomes 10^{-34} Js$

c = Speed of the light = $3\times 10^8 m/s$

$E=\frac{6.626\tiomes 10^{-34} Js\times 3\times 10^8 m/s}{646.3\times 10^{-9} m}$

$E=3.07\times 10^{-19} J$

The energy difference between these 2p and 2s orbitals is $3.07\times 10^{-19} J$

3 0

2 years ago

Assume the weight of an average adult is 70. kg, and that 420. kJ of heat are evolved per mole of oxygen consumed as a result of

seraphim [82]

Answer:

The temperature difference of the body after 3 hours = 5.16 K

Explanation:

we know that the number of moles of O₂ inhaled are 0.02 mole/min⁻¹

or, 1.2 mole.h⁻¹

The average heat evolved by the oxidation of foodstuffs is then:

⇒ Q avg = $\frac{1.2 X 420 X 10^{3} }{70}$ = 7.2 kj.h⁻¹.Kg⁻¹

the heat produced after 3 h would be:

= 7.2 kj. h⁻¹.Kg⁻¹ x 3 h

= 21.6 kj. kg⁻¹

= 21.6 x 10³ j kg⁻¹

We know Qp = Cp x ΔT

Assume the heat capacity of the body is 4.18 J g⁻¹K⁻¹

⇒ ΔT = $\frac{Qp}{Cp}$

⇒ ΔT = $\frac{(21.6 X 10^{3} j.kg^{-1} ) }{(4.18 j k^{-1}g^{-1}) X (1000g.kg^{-1} )}$

⇒ ΔT = 5.16 K

6 0

2 years ago

100 POINTS!!!!Questions/Results Heat energy, Q, is calculated by multiplying the mass of the object in grams, m, by the change o

zepelin [54]

Q = mΔT(Cp)
where Q = heat energy in J (joules),
m = mass in g, ΔT = change in temper. (°C),
Cp = heat capacity in J/(g°C)

Water has a higher heat capacity, meaning that once heat energy is absorbed, it holds that heat longer than bread. Also though, a higher heat capacity of water means that it takes more energy to heat it up.

I don't see any specific data listed for this lab??

5 0

2 years ago

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