How many valence electrons configuration of oxygen?

What is the direction of the polarity of the indicated bond in h3c−oh marked by δ+ and δ−?

Sedbober [7]

Answer:- The direction of the polarity of the indicated bond is from carbon to oxygen.

Explanations:- There are two types of covalent compounds, polar and non polar. If the bond is between two same atoms for example, H-H, Cl-Cl etc then the bond is non polar. If the bond is between two different atoms then the bond would be polar. The direction of the polarity is from loss electron negative atom to more electron negative atom.

Oxygen is more electron negative than carbon. So, being more electron negative, the bonding electrons are more towards oxygen and it cases partial negative charge on oxygen and partial positive charge on carbon. The direction of the polarity is from less electron negative carbon to more electron negative oxygen.

It is shown in the diagram below:

6 0

3 years ago

1.12g H2 is allowed to react with 9.60 g N2, producing 1.23 g NH3.

andriy [413]

Answer:

A. $m_{NH_3}^{theo} =1.50gNH_3$

B. $Y=82.2\%$

Explanation:

Hello!

In this case, since the undergoing chemical reaction between nitrogen and hydrogen is:

$N_2+3H_2\rightarrow 2NH_3$

Thus we proceed as follows:

A. Here, we first need to compute the moles of ammonia yielded by each reactant, in order to identify the limiting one:

$n_{NH_3}^{by \ H_2}=1.12gH_2*\frac{1molH_2}{2.02gH_2}*\frac{2molNH_3}{3molH_2}=0.370molNH_3\\\\ n_{NH_3}^{by \ N_2}=1.23gN_2*\frac{1molN_2}{28.02gN_2}*\frac{2molNH_3}{1molN_2}=0.0878molNH_3$

Thus, since nitrogen yields the fewest moles of ammonia, we realize it is the limiting reactant, so the theoretical yield, in grams, of ammonia is:

$m_{NH_3}^{theo}=0.0878mol*\frac{17.04gNH_3}{1molNH_3} =1.50gNH_3$

B. Finally, since the actual yield of ammonia is 1.23, the percent yield turns out:

$Y=\frac{1.23gNH_3}{1.50gNH_3} *100\%\\\\Y=82.2\%$

Best regards!

5 0

3 years ago

For the solution resulting from dissolved 0.32 g of naphthalene (C10H8) in 25 g of benzene (C6H6) at temperature of 26.1°C, calc

grandymaker [24]

Answer:

See explanation

Explanation:

Number of moles of naphthalene = 0.32g/128.1705 g/mol = 0.0025 moles

Molality = number of moles/ mass of Solvent in kilograms

Molality = 0.0025/0.025 Kg

Morality = 0.1 m

But

∆T= K × i × m

Where ∆T = boiling point elevation

i= number of particles (this is equal to 1 because naphthalene is molecular and not ionic)

m= molality of naphthalene = 0.1 m

K= boiling point elevation constant = 5.12 °C/m

∆T= 5.12 °C/m ×0.1 = 0.512°C

For freezing point depression

∆T= K× i × m

Where ∆T= freezing point depression

i= number of particles (this is equal to 1 because naphthalene is molecular and not ionic)

m= molality of naphthalene = 0.1 m

K= freezing point depression constant = 2.67 °C/m

∆T= 2.67 °C/m ×0.1 = 0.267°C

From Raoult's law;

∆P = XBPA°

Where;

∆P = vapour pressure lowering

XB = mole fraction of solute

PA° = vapour pressure of pure solvent

Number of moles of solvent = mass/molar mass = 25g/ 78 g/mol= 0.3205 moles

Total number of moles = number of moles of solute + number of moles of solvent = 0.0025 moles + 0.3205 moles = 0.323 moles

Mole fraction of solute = 0.0025 moles/0.323 moles = 0.0077

Vapour pressure of benzene = 100 torr

Therefore;

∆P = 0.0077 × 100torr = 0.77 torr

Hence;

∆P = 0.77 torr

5 0

3 years ago

A certain plant requires moisture oxygen carbon dioxide light and minerals in order to survive.This statement describes a living

TEA [102]

The Answer is C. Abiotic Factors

3 0

3 years ago

Two moles of magnesium (Mg) and five moles of oxygen (O2) are placed in a reaction vessel. When magnesium is ignited, it reacts

SOVA2 [1]

Answer:

$\boxed{\text{Mg is the limiting reactant}}$

Explanation:

We are given the amounts of two reactants, so this is a limiting reactant problem.

We know that we will need moles, so, lets assemble the data in one place.

2Mg + O₂ ⟶ 2MgO

n/mol: 2 5

Calculate the moles of MgO we can obtain from each reactant.

From Mg:

The molar ratio of MgO:Mg is 2:2

$\text{Moles of MgO} = \text{2 mol Mg} \times \dfrac{\text{2 mol MgO}}{\text{2 mol Mg}} = \text{2 mol MgO}$

From O₂:

The molar ratio of MgO:O₂ is 2:1.

$\text{Moles of MgO} = \text{5 mol O}_{2} \times \dfrac{\text{2 mol MgO}}{\text{1 mol O}_{2}} = \text{10 mol MgO}\\\\\boxed{\textbf{Mg is the limiting reactant}} \text{ because it gives the smaller amount of MgO}$

6 0

3 years ago