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bogdanovich [222]
2 years ago
5

If you were to perform the reaction KCl (s)→K+ (g)+ Cl−(g), would energy be absorbed or released? Lattice Energies for Some Ioni

c Compounds Compound Lattice Energy (kJ/mol) Compound Lattice Energy (kJ/mol) LiF 1030 MgCl2 2326 LiCl 834 SrCl2 2127 LiI 730 NaF 910 MgO 3795 NaCl 788 CaO 3414 NaBr 732 SrO 3217 NaI 682 KF 808 ScN 7547 KCl 701 KBr 671 CsCl 657 CsI 600 Drag the appropriate labels to their respective targets.
Chemistry
1 answer:
lina2011 [118]2 years ago
6 0

Answer:

Energy would be absorbed.

Explanation:

Lattice energy is defined as the energy required to break apart an ionic solid and convert its component atoms into gaseous ions. That is what you're doing in:

KCl (s) → K⁺(g) + Cl⁻(g)

The energy you require to obtain this reaction is 701 kJ/mol. As the value is positive, <em>energy would be absorbed.</em>

<em />

I hope it helps!

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Drag the terms to the appropriate blanks to complete the following sentences summarizing the redox reactions of photosynthesis.
quester [9]
<h3><u>Answer;</u></h3>

1. In the light reactions, light energy is used to oxidize H2O to O2.

2. The electrons derived from this oxidation reaction in the light reactions are used to reduce NADP+ to NADPH.

3. The Calvin cycle oxidizes the light-reactions product NADPH to NADP+.

4. The electrons derived from this oxidation reaction in the Calvin cycle are used to reduce CO2 to G3P.

<h3><u>Explanation;</u></h3>
  • <em><u>In the light reactions, light energy is used to remove electrons from (oxidize) water, producing O2 gas. These electrons are ultimately used to reduce NADP+ to NADPH. </u></em>
  • In the Calvin cycle, NADPH is oxidized back to NADP+ (which returns to the light reactions). The electrons released by the oxidation of NADPH are used to reduce three molecules of CO2 to sugar (G3P), which then exits the Calvin cycle.
  • As ATP and NADPH are used in the Clavin cycle, they produce ADP and NADP+, respectively, which are returned to the light reactions so that more ATP and NADPH can be formed.
5 0
2 years ago
Explain and illustrate the notation for distinguishing between the different p orbitals in a sublevel.
masha68 [24]

Here, the three different notation of the p-orbital in different sub-level have to generate

The value of azimuthal quantum number (l) for -p orbital is 1. We know that the magnetic quantum number m_{l} depends upon the value of l, which are -l to +l.

Thus for p-orbital the possible magnetic quantum numbers are- -1, 0, +1. So there will be three orbitals for p orbitals, which are designated as p_{x}, p_{y} and p_{z} in space.

The three p-orbital can be distinguish by the quantum numbers as-

For 2p orbitals (principal quantum number is 2)

1) n = 2, l = 1, m = -1

2) n = 2, l = 1, m = 0

3) n = 2, l = 1, m = +1

Thus the notation of different p-orbitals in the sub level are determined.  

6 0
2 years ago
A 0.500 g sample of C7H5N2O6 is burned in a calorimeter containing 600. g of water at 20.0∘C. If the heat capacity of the bomb c
Nata [24]

Answer:

22.7

Explanation:

First, find the energy released by the mass of the sample. The heat of combustion is the heat per mole of the fuel:

ΔHC=qrxnn

We can rearrange the equation to solve for qrxn, remembering to convert the mass of sample into moles:

qrxn=ΔHrxn×n=−3374 kJ/mol×(0.500 g×1 mol213.125 g)=−7.916 kJ=−7916 J

The heat released by the reaction must be equal to the sum of the heat absorbed by the water and the calorimeter itself:

qrxn=−(qwater+qbomb)

The heat absorbed by the water can be calculated using the specific heat of water:

qwater=mcΔT

The heat absorbed by the calorimeter can be calculated from the heat capacity of the calorimeter:

qbomb=CΔT

Combine both equations into the first equation and substitute the known values, with ΔT=Tfinal−20.0∘C:

−7916 J=−[(4.184 Jg ∘C)(600. g)(Tfinal–20.0∘C)+(420. J∘C)(Tfinal–20.0∘C)]

Distribute the terms of each multiplication and simplify:

−7916 J=−[(2510.4 J∘C×Tfinal)–(2510.4 J∘C×20.0∘C)+(420. J∘C×Tfinal)–(420. J∘C×20.0∘C)]=−[(2510.4 J∘C×Tfinal)–50208 J+(420. J∘C×Tfinal)–8400 J]

Add the like terms and simplify:

−7916 J=−2930.4 J∘C×Tfinal+58608 J

Finally, solve for Tfinal:

−66524 J=−2930.4 J∘C×Tfinal

Tfinal=22.701∘C

The answer should have three significant figures, so round to 22.7∘C.

8 0
2 years ago
Marie mixed 5g of carbon with 5g of lead oxide. she heated the mixture strongly for 15 minutes in a fume cupboard.
olga2289 [7]
The problem talks about two questions and these are:

1. Metals are very good conductors of electricity and heat. Directing heat is easier. So let Marie heat the beads and also have heat another substance, for instance, water. If the beads heat quicker, then they are metals. Another test to conduct is called flame test. This test should give you a colored flame (blue/white for lead) the metal is lead if the reaction is: 2PbO+C ==> 2Pb +CO2

2. The beads are possibly to be lead since Ferrous(lead) oxide + carbon = carbon dioxide + lead 
7 0
2 years ago
Item 5 A solution of methanol, CH3OH, in water is prepared by mixing together 128 g of methanol and 108 g of water. The mole fra
Basile [38]

Answer:

Mole fraction of methanol will be closest to 4.

Explanation:

Given, Mass of methanol = 128 g

Molar mass of methanol = 32.04 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{128\ g}{32.04\ g/mol}

Moles\ of\ methanol = 3.995\ mol

Given, Mass of water = 108 g

Molar mass of water = 18.0153 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{108\ g}{18.0153\ g/mol}

Moles\ of\ water= 5.995\ mol

So, according to definition of mole fraction:

Mole\ fraction\ of\ methanol=\frac {n_{methanol}}{n_{methanol}+n_{water}}

Mole\ fraction\ of\ methanol=\frac{3.995}{3.995+5.995}=0.39989

<u>Mole fraction of methanol will be closest to 4.</u>

5 0
2 years ago
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