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3 years ago

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

Chemistry

2 answers:

stepladder [879]3 years ago

4 0

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

REY [17]3 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.

<u>The formal charge of an atom is given by the equation:</u>

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]

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

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

<em><u>Therefore, the formal charge on H atom in HF = 0</u></em>

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What volume of O2 is needed to combust 425 mL of butane?

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Answer:

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3 years ago

A parachute on a racing dragster opens and changes the speed of the car from 85 m/s to 45 m/s in a period of 4.5 seconds. What i

LiRa [457]

Answer:

- 8.89 $m/s^{2}$

Explanation:

$\frac{(Vf-Vo)}{time}$ = $\frac{(45-85 m/s)}{4.5s}$ = - 8.89 $m/s^{2}$

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3 years ago

Read 2 more answers

Part A

inysia [295]

Pressure of the gas inside the container is 662.59 torr.

<h3>What is ideal gas law?</h3>

The ideal gas law (PV = nRT) connects the macroscopic characteristics of ideal gases. An ideal gas is one in which the particles are both non-repellent and non-attractive to one another (have no volume).

The general law of ideal gas can be applied here: PV is equal to nRT, where P is the gas pressure in atm.

V is the number of moles of the gas in a mole, and n is the volume of the gas in L. R is the universal gas constant. T is the temperature(Kelvin) of the gas.

If P and T are different values and n and V are constants, then

(P₁T₂) = (P₂T₁).

P₁ = 735 torr, T₁ = 29°C + 273 = 302 K,

P₂ = ??? torr, T₂ = 62°C + 273 = 335 K.

∴ P₂ = (P₁T₂)/(P₁) = (735 torr)(302 K)/(335 K) = 662.59 torr.

To know more about ideal gas law visit:

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1 year ago

Arranges the following molecules in order of increasing dipole moment: <br> H2O, H2S, H2Te, H2Se.

erastova [34]

Explanation:

Dipole moment is defined as the measurement of the separation of two opposite electrical charges.

$H_{2}O$ is a bent shaped molecule with a dipole moment of 1.87.

$H_{2}S$ is also a bent shaped molecule with a dipole moment of 1.10.

$H_{2}Te$ is a also a bent shaped molecule and has a negligible dipole moment.

$H_{2}Se$ has a dipole moment of 0.29.

Therefore, given molecules are arranged according to their increasing dipole moment as follows.

$H_{2}Te$ < $H_{2}Se$ < $H_{2}S$ < $H_{2}O$

7 0

3 years ago

Considering the volumes and the concentrations used in mixture 1, what percentage of the moles of H2O2 present have been consume

Kisachek [45]

Answer:

1 mol of H2O2 consumes 1 mol of S2O32, then 0.0005 mol of I- reacts with 0.00025 moles of H2O2

Thus, initial moles of H2O2 present in the mix is equal to 0.0102 moles.

Amount of moles of H2O2 that reacts with I- is equal to 0.00025 moles.

The amount of I ions that is oxidized by H2O2 and the iodine, I2 that is produced, is consumed by the S2O32- ions and are transformed into I ions. The amount of I ions that are consumed by H2O2 is equal to the amount of ion that is produced by S2O32-.

Explanation:

the total volume of the mixture is equal to:

Vmix = 75 + 30 + 25 + 5 + 5 + 10 = 150 mL

the moles of each species in the mix equals:

Na2S2O3 = 0.05 * 0.005 = 0.00025 moles

KI = 0.05 * 0.025 = 0.00125 moles

H2O2 = 1.02 * 0.01 = 0.0102 moles

the following equation shows the reaction between I2 and S2O32:

I2 + S2O32 = 2I- + S4O62-

The same way:

2I- + 2H+ + H2O2 = I2 + 2H2O

1 mol of H2O2 consumes 1 mol of S2O32, then 0.0005 mol of I- reacts with 0.00025 moles of H2O2

Thus, initial moles of H2O2 present in the mix is equal to 0.0102 moles.

Amount of moles of H2O2 that reacts with I- is equal to 0.00025 moles.

3 0

3 years ago