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3 years ago

The _______ geometry of a water molecule is tetrahedral even though the molecular geometry is bent

Chemistry

2 answers:

IRINA_888 [86]3 years ago

6 0

The electron geometry of a water molecule is tetrahedral even though the molecular geometry is bent.

As water molecule hybridisation is sp³ that provides it a electron geometry tetrahedral but due to presence of 2 lone pairs and 2 bond pairs its molecular geometry is bent.

The hybridisation sp³ makes electron geometry of a water molecule tetrahedral but the presence of 2 lone pairs makes its molecular geometry bent

Stolb23 [73]3 years ago

6 0

Answer:

the electron geometry of a water molecule is tetrahedral.

Explanation:

There are four pair of electrons on oxygen central atom, these electron pairs are arranged in tetrahedral manner and thus electron geometry is tetrahedral.

The hybridization is sp³ and thus there are four hybridized orbitals,

However there are two bonded electron pairs and two lone pair of electrons.

Thus the two hydrogen atoms are arranged in bent shape.

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3 years ago

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How many grams of hydrogen chloride can be produced from 1.00g of hydrogen and 55.0g of chlorine? what is the limiting reactant?

zloy xaker [14]

Answer:

$m_{HCl}=36.1gHCl$

Explanation:

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In this case, according to the given information, it turns out possible for us to calculate the required grams of HCl by firstly identifying the limiting reactant via the moles of each reactant as they are in a 1:1 mole ratio:

$n_{H_2}=1.00gH_2*\frac{1molH_2}{2.02gH_2}=0.500molH_2\\\\ n_{Cl_2}=55.0gCl_2*\frac{1molCl_2}{70.9gCl_2}=0.776molCl_2$

Thus, we infer the hydrogen is the limiting reactant and therefore we use its 1:2 mole ratio with HCl whose molar mass is 36.46 g/mol:

$m_{HCl}=0.500molH_2*\frac{2molHCl}{1molH_2}*\frac{36.46gHCl}{1molHCl}\\\\m_{HCl}=36.1gHCl$

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2 years ago

The question is in the picture below

Rus_ich [418]

Answer:

$\Delta\text{H}_1+2\Delta\text{H}_2-\Delta\text{H}_3$

Explanation:

Hess's Law of Constant Heat Summation states that if a chemical equation can be written as the sum of several other chemical equations, the enthalpy change of the first chemical equation is equal to the sum of the enthalpy changes of the other chemical equations. Thus, the reaction that involves the conversion of reactant A to B, for example, has the same enthalpy change even if you convert A to C, before converting it to B. Regardless of how many steps it takes for the reactant to be converted to the product, the enthalpy change of the overall reaction is constant.

With Hess's Law in mind, let's see how A can be converted to 2C +E.

$\bf{\text{A} \rightarrow 2\text{B}}$ (Δ $\text{H}_1$ ) -----(1)

Since we have 2B, multiply the whole of II. by 2:

$\bf{2\text{B} \rightarrow 2\text{C} +2\text{D}}$ (2Δ $\text{H}_2$ ) -----(2)

This step converts all the B intermediates to 2C +2D. This means that the overall reaction at this stage is $\text{A} \rightarrow 2\text{C} +2\text{D}$ .

Reversing III. gives us a negative enthalpy change as such:

$\bf{2\text{D} \rightarrow \text{E}}$ (-Δ $\text{H}_3$ ) -----(3)

This step converts all the D intermediates formed from step (2) to E. This results in the overall equation of $\text{A} \rightarrow 2\text{C} +\text{E}$ , which is also the equation of interest.