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2 years ago

What is the formal charge on the hydrogen atom in hf?

Chemistry

2 answers:

stepladder [879]2 years ago

4 0

The formal charge of H = +1, because F in compounds can have only -1.

REY [17]2 years ago

4 0

Answer:

The formal charge on H atom in HF = 0

Explanation:

Formal charge is the charge on an atom when all the chemical bonds that are formed by that atom are formed by equal sharing of electrons.

The formal charge of an atom is given by the equation:

Formal charge = [Total number of valence electrons in the ground state (n)] - [Number of non-bonding valence electrons on atom in a molecule (N)] - [Number of shared or bonding electrons (B) ÷ 2]

To find out the formal charge on hydrogen (H) atom in H-F molecule:

For H atom in ground state: n = 1

For H atom in H-F molecule: N = 0 and B = 2

Since, Formal charge = n - N - (B ÷ 2)

So, The formal charge on H atom in HF = 1 - 0 - (2÷ 2) = 1 - 1 = 0

Therefore, the formal charge on H atom in HF = 0

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3k+aici3 3kci+ai is what type of reaction

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3K + AlCl3 --------> 3 KCl + Al
element compound compound element

This is single replacement (displacement) reaction.

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2 years ago

2.15 Liters of helium gas at a pressure of 58 kPa and a temperature of 25 degrees celsius, what is the new volume at stp?

Nastasia [14]

The ideal gas law is P1V1/T1 = P2V2/T2. STP means the temperature is 273 K and pressure is 101.3 kPa. According to this formula, the new volume V2=2.15*58*273/(298*101.3) = 1.13 L.

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3 years ago

A balloon is filled with 0.250 mole of air at 35Â°C. If the volume of the balloon is 6.23 liters, what is the absolute pressure

lord [1]

Answer is: the absolute pressure of the air in the balloon is 1.015 atm (102.84 kPa).

n = 0.250 mol; amount of substance.

V = 6.23 L; volume of the balloon.

T = 35°C = 308.15 K; temperature.

R = 0.08206 L·atm/mol·K, universal gas constant.

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p = n·R·T / V.

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p = 1.015 atm; presure of the air.

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2 years ago

A 125g metal block at a temperature of 93.2 degrees Celsius was immersed in 100g of water at 18.3 degrees Celsius. Given the spe

nikitadnepr [17]

Answer:

$\large \boxed{34.2\, ^{\circ}\text{C}}$

Explanation:

There are two heat transfers involved: the heat lost by the metal block and the heat gained by the water.

According to the Law of Conservation of Energy, energy can neither be destroyed nor created, so the sum of these terms must be zero.

Let the metal be Component 1 and the water be Component 2.

Data:

For the metal:

$m_{1} =\text{125 g; }T_{i} = 93.2 ^{\circ}\text{C; }\\C_{1} = 0.900 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}$

For the water:

$m_{2} =\text{100 g; }T_{i} = 18.3 ^{\circ}\text{C; }\\C_{2} = 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}$

$\begin{array}{rcl}\text{Heat lost by metal + heat gained by water} & = & 0\\q_{1} + q_{2} & = & 0\\m_{1}C_{1}\Delta T_{1} + m_{2}C_{2}\Delta T_{2} & = & 0\\\text{125 g}\times 0.900 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times\Delta T_{1} + \text{100 g} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}\Delta \times T_{2} & = & 0\\112.5\Delta T_{1} + 418.4\Delta T_{2} & = & 0\\112.5\Delta T_{1} & = & -418.4\Delta T_{2}\\\Delta T_{1} & = & -3.719\Delta T_{2}\\\end{array}$

$\Delta T_{1} = T_{\text{f}} - 93.2 ^{\circ}\text{C}\\\Delta T_{2} = T_{\text{f}} - 18.3 ^{\circ}\text{C}$

$\begin{array}{rcl}\Delta T_{1} & = & -3.719\Delta T_{2}\\T_{\text{f}} - 93.2 ^{\circ}\text{C} & = & -3.719 (T_{\text{f}} - 18.3 ^{\circ}\text{C})\\T_{\text{f}} - 93.2 ^{\circ}\text{C} & = & -3.719T_{\text{f}} + 68.06 ^{\circ}\text{C}\\4.719T_{\text{f}} & = & 161.3 ^{\circ}\text{C}\\T_{\text{f}} & = & \mathbf{34.2 ^{\circ}}\textbf{C}\\\end{array}\\\text{The final temperature of the block and the water is $\large \boxed{\mathbf{34.2\, ^{\circ}}\textbf{C}}$}$

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3 years ago

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GREYUIT [131]

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2 years ago

Read 2 more answers