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3 years ago

What is the hybridization of carbon in nco-?

Chemistry

2 answers:

never [62]3 years ago

5 0

Answer : The hybridization of carbon in $NCO^-$ is, $sp$ hybridized.

Explanation :

As we know that when a carbon attached to the four single sigma bonds then the hybridization will be, $sp^3$ .

When a carbon attached to the three single sigma bonds and one- $\pi$ bond that means a carbon bonded to another atom by the double bond then the hybridization will be, $sp^2$ .

When a carbon attached to the two single sigma bonds and two- $\pi$ bonds that means a carbon bonded to another atom by the triple bond then the hybridization will be, $sp$ .

As per question, the given molecule is, $NCO^-$ . In this molecule, a triple bond is present between the nitrogen and carbon atom and a single bond is present between the carbon and oxygen atom. That means, a carbon bonded to another atom by the triple bond then the hybridization will be, $sp$ .

Hence, the hybridization of carbon in $NCO^-$ is, $sp$ hybridized.

Svet_ta [14]3 years ago

4 0

First let us see what kind of bonds are formed in the compound. By drawing the structure, we see that the kind of bonds are:

N =- triple bond -= C – O

So there is only single bond between C and O therefore the hybridization of C is sp.

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A containing vessel holds a gaseous mixture of nitrogen and butane. Thepressure in the vessel at 126.9 Cis 3.0 atm. At 0 C, the

Viefleur [7K]

A vessel that contains a mixture of nitrogen and butane has a pressure of 3.0 atm at 126.9 °C and a pressure of 1.0 atm at 0 °C. The mole fraction of nitrogen in the mixture is 0.33.

A vessel contains a gaseous mixture of nitrogen and butane. At 126.9 °C (400.1 K) the pressure is due to the mixture is 3.0 atm.

We can calculate the total number of moles using the ideal gas equation.

$P \times V = n \times R \times T\\\\n = \frac{P \times V}{R \times T} = \frac{3.0 atm \times V}{0.082 atm.L/mol.K \times 400.1 K} = 0.091 mol/L \times V$

At 0 °C (273.15 K), the pressure due to the gaseous nitrogen is 1.0 atm.

We can calculate the moles of nitrogen using the ideal gas equation.

$P \times V = n \times R \times T\\\\n = \frac{P \times V}{R \times T} = \frac{1.0 atm \times V}{0.082 atm.L/mol.K \times 400.1 K} = 0.030 mol/L \times V$

The mole fraction of nitrogen in the mixture is:

$X(N_2) = \frac{0.030 mol/L \times V}{0.091 mol/L \times V} = 0.33$

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2 years ago

What is the charge of this atom? 65 protons 60 electrons A)-5 B)+5 C)0

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Answer:

B. +5

Explanation:

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2 years ago

Read 2 more answers

A 2.50 g sample of solid sodium hydroxide is added to 55.0 mL of 25 °C water in a foam cup (insulated from the environment) and

zlopas [31]

Answer:

37.1°C.

Explanation:

Firstly, we need to calculate the amount of heat (Q) released through this reaction:

∵ ΔHsoln = Q/n

no. of moles (n) of NaOH = mass/molar mass = (2.5 g)/(40 g/mol) = 0.0625 mol.

The negative sign of ΔHsoln indicates that the reaction is exothermic.

∴ Q = (n)(ΔHsoln) = (0.0625 mol)(44.51 kJ/mol) = 2.78 kJ.

We can use the relation:

Q = m.c.ΔT,

where, Q is the amount of heat released to water (Q = 2781.87 J).

m is the mass of water (m = 55.0 g, suppose density of water = 1.0 g/mL).

c is the specific heat capacity of water (c = 4.18 J/g.°C).

ΔT is the difference in T (ΔT = final temperature - initial temperature = final temperature - 25°C).

∴ (2781.87 J) = (55.0 g)(4.18 J/g.°C)(final temperature - 25°C)

∴ (final temperature - 25°C) = (2781.87 J)/(55.0 g)(4.18 J/g.°C) = 12.1.

∴ final temperature = 25°C + 12.1 = 37.1°C.

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2 years ago

You run an experiment in which two substances chemically react in a closed system. you run the same experiment in an open system

m_a_m_a [10]

If we run an experiment in which two substances chemically react in a closed system and then we run the same experiment in an open system, then the masses of the products in each experiment will be different because the open system allows the interchange of matter and energy with the media.

<h3>What is an open system?</h3>

An open system is an interrelated group pf parts that work together to interchange matter and energy with the surrounding environment, while a closed system does not generate an exchange of matter and energy.

Therefore, with this data, we can see that an open system interchange energy and matter with the environment, while closes systems do it.

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10 months ago

How did uranium become present in Earth's upper crust? Choose one: A. It was delivered by comets that crashed into Earth's surfa

Luda [366]

Answer:

Option A. It was delivered by comets that crashed into Earth's surface.

Explanation:

Uranium (U) is a chemical element with atomic number 92.

For many years, a large number of scientists have been studying the abundance and origin of the isotopes of uranium in Earth. According to some theories, the Earth's uranium was produced in one or more supernovae (an explosive brightening of a star), in wich, the main process consists in the rapid capture of neutrons by seed nuclei at great rates. Another theory proposes that uranium is created during the merger of two neutron stars (neutron stars are very dense), because, when such dense bodies come closer together the gravitational force cause them to merge, producing huge amounts of hevy metals like uranium.

Many analyses have been made of the uranium in rocks of the Earth. These measurements shows that the abundance of uranium is bigger in the crust and upper mantle of the Earth.

So, knowing that Earth's uranium was produced through one of these processes, the best answer is option A, the uranium was delivered by comets that crashed into Earth's surface.

Have a nice day!

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3 years ago