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3 years ago

Which element will attract electrons in a chemical bond?

Chemistry

1 answer:

Irina-Kira [14]3 years ago

4 0

Answer:

O Bromine

Explanation:

The element that will attract electrons in a chemical bond is Bromine.

An element that attracts electrons in chemical bonds is said to be an electronegative element.

Such an element have a relative tendency to attract valence electrons in a chemical bond.

The most electronegative group on the periodic table is that of the halogens.
Halogens are a special class of electronegative elements.
They need just one electron to complete their octet, hence the reason for their electronegativity.
Metals are electropositive and are not ready to attract electrons. They prefer reactions where they can lose electrons.

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2) Show the calculation of Kc for the following reaction if an initial reaction mixture of 0.800 mole of CO and 2.40 mole of H2

nadezda [96]

Answer:

Kc = 3.90

Explanation:

CO reacts with $H_2$ to form $CH_4$ and $H_2O$ . balanced reaction is:

$CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g) + H_2O(g)$

No. of moles of CO = 0.800 mol

No. of moles of $H_2$ = 2.40 mol

Volume = 8.00 L

Concentration = $\frac{Moles}{Volume\ in\ L}$

Concentration of CO = $\frac{0.800}{8.00} = 0.100\ mol/L$

Concentration of $H_2$ = $\frac{2.40}{8.00} = 0.300\ mol/L$

$CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g) + H_2O(g)$

Initial 0.100 0.300 0 0

equi. 0.100 -x 0.300 - 3x x x

It is given that,

at equilibrium $H_2O (x)$ = 0.309/8.00 = 0.0386 M

So, at equilibrium CO = 0.100 - 0.0386 = 0.0614 M

At equilibrium $H_2$ = 0.300 - 0.0386 × 3 = 0.184 M

At equilibrium $CH_4$ = 0.0386 M

$Kc=\frac{[H_2O][CH_4]}{[CO][H_2]^3}$

$Kc=\frac{0.0386 \times 0.0386}{(0.184)^3 \times 0.0614} =3.90$

8 0

4 years ago

A reaction between liquid reactants takes place at −10.0°C in a sealed, evacuated vessel with a measured volume of 45.0L. Measur

maks197457 [2]

Answer: 0.0624 atm

Explanation:-

According to the ideal gas equation:

$PV=nRT$

P = Pressure of the gas = ?

V= Volume of the gas = 45.0 L

T= Temperature of the gas = -10.0°C = 263 K $0^00C=273K$

R= Gas constant = 0.0821 atmL/K mol

n= moles of gas

Moles of gas= $\frac{\text{ given mass}}{\text{ molar mass}}= \frac{19.0g}{146g/mole}=0.130moles$

$P=\frac{nRT}{V}=\frac{0.130\times 0.0821\times 263}{45.0}=0.0624atm$

The pressure of sulfur hexafluoride gas in the reaction vessel after the reaction is 0.0624 atm

3 0

3 years ago

What's the atomic mass unit of helium?

Alex17521 [72]

4.002602 u ± 0.000002 u

4 0

3 years ago

Write a half-reaction for the oxidation of the manganese in MnCO3(s) to MnO2(s) in neutral groundwater where the carbonate-conta

Sergeu [11.5K]

Answer:

The half-reaction for the oxidation of the manganese in $MnCO_3(s)$ to $MnO_2(s)$ .:

$MnCO_3+2H_2O\rightarrow MnO_2(s)+HCO_3^{-}+3H^+$

Explanation:

$MnCO_3+H_2O\rightarrow MnO_2(s)+HCO_3^{-}$

Let us say the medium in which reaction is taking place is an acidic medium. And balancing in an acidic mediums done as;

Step 1: Balance all the atom beside oxygen and hydrogen atom;

$MnCO_3+H_2O\rightarrow MnO_2(s)+HCO_3^{-}$

Manganese and carbon are balanced.

Step 2: Balance oxygen atom adding water on the required side:

$MnCO_3+H_2O+H_2O\rightarrow MnO_2(s)+HCO_3^{-}$

$MnCO_3+2H_2O\rightarrow MnO_2(s)+HCO_3^{-}$

Step 3: Now balance hydrogen atom by adding hydrogen ion on the required side:

$MnCO_3+2H_2O\rightarrow MnO_2(s)+HCO_3^{-}+3H^+$

6 0

3 years ago

Which of the following best describes what happens when hydrogen and oxygen to form water?

Sever21 [200]

Atoms are combining to form molecules

6 0

3 years ago