Physics, 24.11.2019 06:31 laura2school

Astudent tested a sample of carbon (c) by combining it with water. there was no reaction between the carbon and water. the student concluded that carbon must be a group 18 element, since it was unreactive. state whether you agree with the student's reasoning, and explain why or why not.

Answers: 2

Show answers

Answers

I wouldn't agree with the student's reasoning. Not every element that doesn't react with water is unreactive. If Carbon were exposed to Hydrogen, it would form Hydrocarbons. In order to conclude it to be a group 18 element, it would have to be unreactive with any compound, not just water.

Answer from: CamarianMcNeal1855

Student's reasoning is wrong.

Explanation:

Carbon is a second period element and its electronic configuration is $1s^{2}2s^{2}2p^{4}$ .

Since, there is no d-orbital present in a carbon atom. Hence, when it is dissolve in water then it becomes unable to accommodate the lone pair donated by donor oxygen atom of water.

As a result, there occurs no reaction when carbon is dissolved in water.

Hence, we can conclude that we do not agree with the student's reasoning.

Answer from: Quest

Answer: a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same sign as the -4.0 µc charge.======electricstatic force between two point charges is given by $f = \dfrac{kq_1q_2}{r^2}$ where k is coulomb's constant, 8.99×10⁹ n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ = 6.0 µc, q₂ = -4.0 µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1 µc = 10⁻⁶ c) $f_{01} = \dfrac{kq_0q_1}{(\text{6.0 m})^2}$ the force from q₂ $f_{02} = \dfrac{kq_0q_2}{(\text{15 m})^2}$ the force from q₃ $f_{03} = \dfrac{kq_0q_1}{(\text{26 m})^2}$ the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃ $\begin{aligned} f_{net} & = 0\\ f_{01} + f_{02} + f_{03} & = 0 \\ \dfrac{kq_0q_1}{(\text{6.0 m})^2} + \dfrac{kq_0q_2}{(\text{15 m})^2} + \dfrac{kq_0q_3}{(\text{26 m})^2} & = 0 \\ kq_0 \left( \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2}\right) & = 0 \\ \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2} & = 0 \end{aligned}$ isolating: $\begin{aligned} \dfrac{q_3}{(\text{26 m})^2} & = -\dfrac{q_1}{(\text{6.0 m})^2} - \dfrac{q_2}{(\text{15 m})^2} \\ q_3 & = -(\text{26 m})^2\left(\dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2}\right) \\ q_3 & = -(\text{26 m})^2\left(\dfrac{6.0\times10^{-6} \text{ c}}{(\text{6.0 m})^2} + \dfrac{-4.0\times10^{-6} \text{ c}}{(\text{15 m})^2}\right) \\ q_3 & = -1.0064888 \times 10^{-4}\text{ c} \\ q_3 & = -1.0\times 10^{-4}\text{ c} \end{aligned}$ a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same as the -4.0 µc charge

Answer from: Quest

Answerlosing electrons;

Another question on Physics

Physics, 22.06.2019 10:00

Atomic nuclei of almost all elements consist of

Answers: 1

Answer

Physics, 22.06.2019 11:30

In a good thermo flask the main cause of heat loss is

Answers: 1

Answer

Physics, 22.06.2019 20:30

We will use a video to study faraday’s law of electromagnetic induction by analyzing the induction that occurs due a coil of copper wire sliding along an air track that passes through an external magnetic field. specifically, we will: examine some of the main principles behind faraday’s law. apply lenz’s law to determine the direction of the induced emf. take measurements of the induced emf from the voltmeter as the coil passes into and out of the external magnetic field. use the horizontal scale provided to determine the speed of the coil. calculate the magnetic field induced in the coil as it passes into and out of the external magnetic field.

Answers: 2

Answer

Physics, 22.06.2019 22:00

If anyonee has done the momentum lab practical in physics can you give me the answers i am s lostt rnn : (

Answers: 2

Answer

You know the right answer?

Astudent tested a sample of carbon (c) by combining it with water. there was no reaction between the...

Questions