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3 years ago

Given the balanced equation representing a reaction:

Chemistry

2 answers:

gtnhenbr [62]3 years ago

8 0

Answer is: (1) Energy is absorbed and a bond is broken.

This energy is called bond-dissociation energy.

Bond-dissociation energy is the measure of bond strength in a chemical bond.

Bond energy of chlorine molecule (sigma bond) is energy needed for breaking up one molecule of chlorine into two chlorine atoms.

When two chlorine atoms form molecule of chlorne, than energy is released and bond is formed.

sleet_krkn [62]3 years ago

6 0

Energy is absorbed and a bond is Broken.
You Have the bond of Cl2 which has been broken into 2 Cl.
And in order to break, energy need to be absorbed.

Hope this Helps :)

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What are some similarities between sublimation and deposition?

Korolek [52]

Sublimation: the substance changes directly from a solid to a gas without going through the liquid phase. Deposition: the substance changes directly from a gas to a solid without going through the liquid phase.

7 0

3 years ago

Help fast !

jok3333 [9.3K]

Answer:

1 mole represents 6.023×1023 particles.

1 mole of iodine atom= 6.023×1023

Given 127.0g of iodine.

no. of iodine atom = 1 mole of iodine

1mole of magnesium = 24g of Mg = 6.023×1023no.of Mg

Given 48g of Mg = 2×6.023×1023

no. of Mg = 2 moles of Mg

1 mole of chlorine atom= 6.023× 1023

no. of chlorine atom = 35.5g of chlorine atom

Given 71g of chlorine atom=2× 6.023× 1023

no. of chlorine atom = 6.023×1023

2 moles of chlorine atom.

Given that 4g of hydrogen atom.

will be equal to 4 × 6.023 × 1023

no. of atoms of hydrogen= 4 moles of hydrogen atom.

7 0

3 years ago

BRAINLIESTTT ASAP!! PLEASE HELP ME :)

AlladinOne [14]

Answer : False, there will be two lithium and one oxygen atoms in a unit molecular structure of lithium oxide.

Explanation:

Electronic configuration of lithium is :

$Li=1s^2s^1$

In order to attain stable electronic configuration it will loose an electron and form positively charge cation.

$Li^+=1s^22s^0$

The electronic configuration of oxygen is:

$O=1s^22s^22p^4$

Oxygen being second most electronegative atom requires two electrons to attain noble gas configuration stability and form negatively charge ion with 2- charge.:

$O^{2-}=1s^22s^22p^6$

When two atom of lithium and oxygen comes together , one electron from each lithium atom get transferred to an oxygen atom which results in formation of lithium oxide.

$2Li^++O^{2-}\rightarrow Li_2O$

4 0

3 years ago

Read 2 more answers

En un balneario necesitan calentarse 1 millón de litros de agua anuales, subiendo la temperatura desde 15 ºC a 50 ºC y para ello

MAXImum [283]

Answer:

a) $m_{CH_4}=2630kg$

b) 1657 €

Explanation:

Hola,

a) En este problema, vamos a considerar el millón de litros de agua anuales, ya que con ellos podemos calcular el calor requerido para dicho calentamiento, sabiendo que la densidad del agua es de 1 kg/L:

$Q_{H_2O}=m_{H_2O}Cp(T_2-T_1)=1x10^6LH_2O*\frac{1kgH_2O}{1LH_2O}*4.18\frac{kJ}{kg\°C}(50-15) \°C\\Q_{H_2O}=146.3x10^6kJ$

Luego, usamos la entalpía de combustión del metano para calcular su requerimiento en kilogramos, sabiendo que la energía ganada por el agua, es perdida por el metano:

$Q_{H_2O}=-Q_{CH_4}=-146.3x10^5kJ=m_{CH_4}\Delta _cH_{CH_4}$

$m_{CH_4}= \frac{Q_{CH_4}}{\Delta _cH_{CH_4}} =\frac{-146.3x10^5kJ}{-890kJ/molCH_4} *\frac{16gCH_4}{1molCH_4} \\\\m_{CH_4}=2630112.36g=2630kg$

b) En este caso, consideramos que a condiciones normales de 1 bar y 273 K, 1 metro cúbico de metano cuesta 0,45 €, con esto, calculamos las moles de metano a dichas condiciones:

$n_{CH_4}=\frac{PV}{RT}=\frac{1atm*1000L}{0.082\frac{atm*L}{mol*K}*273K} =44.67mol$

Con ello, los kilogramos de metano que cuestan 0,45 €:

$44.67molCH_4*\frac{16gCH_4}{1molCH_4}*\frac{1kg}{1000g} =0.715kgCH_4$

Luego, aplicamos la regla de tres:

0.715 kg ⇒ 0.45 €

2630 kg ⇒ X

X = (2630 kg x 0.45 €) / 0.715 kg

X = 1657 €

Regards.

3 0

4 years ago

If you have a 1.0 L buffer containing 0.208 M NaHSO3 and 0.134 M Na2SO3, what is the pH of the solution after addition of 50.0 m

Vlada [557]

Answer:

pH = 7.233

Explanation:

Initially, the buffer contains 0.208 moles of NaHSO₃ and 0.134 moles of Na₂SO₃.

NaHSO₃ reacts with NaOH thus:

NaHSO₃ + NaOH → Na₂SO₃ + H₂O

50.0 mL of 1.00 M NaOH are:

0.0500L × (1mol / 1L) = 0.0500moles of NaOH added. That means after the addition are produced 0.0500moles of Na₂SO₃ and consumed 0.0500moles of NaHSO₃. That means final moles of the buffer are:

NaHSO₃: 0.208 mol - 0.050 mol = 0.158 mol

Na₂SO₃: 0.134 mol + 0.050 mol = 0.184 mol

As pKa of this buffer is 7.167, it is possible to use H-H equation to find pH, thus:

pH = pKa + log₁₀ [Na₂SO₃] / [NaHSO₃]

pH = 7.167 + log₁₀ [0.184] / [0.158]

pH = 7.233

6 0

3 years ago