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steposvetlana [31]
3 years ago
8

Which would be a better base to use if you wanted to remove a proton from ethyne?

Chemistry
1 answer:
myrzilka [38]3 years ago
5 0
-NH2 is the most favorable for the reaction
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I need a real life example when you need to know about viscosity. <br> Thanks!!
NemiM [27]
Viscosity refers to a fuilds resistance to flow.

EX: Honey have higher viscosity than water.
7 0
3 years ago
Why doesn’t neon readily form an ionic bond?
Goryan [66]

By definition of noble gases, neon does not easily form an ionic bond because it belongs to the group of noble or inert gases, so its reactivity is practically nil.

<h3>Noble gases</h3>

Noble gases are not very reactive, that is, they practically do not form chemical compounds. This means that they do not react with other substances, nor do they even react between atoms of the same gas, as is the case with diatomic gases such as oxygen (O₂).

The chemical stability of the noble gases and therefore the absence of spontaneous evolution towards any other chemical form, implies that they are already in a state of maximum stability.

All chemical transformations involve valence electrons, they are involved in the process of covalent bond formation and the formation of ions. Therefore, the practically null reactivity of the noble gases is due to the fact that they have a complete valence shell, which gives them a low tendency to capture or release electrons.

Since the noble gases do not react with the other elements, they are also called inert gases.

<h3>Neon</h3>

Neon does not easily form an ionic bond because it belongs to the group of noble or inert gases, so its reactivity is practically nil.

Learn more about noble gases:

brainly.com/question/8361108

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brainly.com/question/19024000

6 0
3 years ago
Read 2 more answers
The table shows the amount of radioactive element remaining in a sample over a period of time.
Assoli18 [71]

It would take 147 hours for 320 g of the sample to decay to 2.5 grams from the information provided.

Radioactivity refers to the decay of a nucleus leading to the spontaneous emission of radiation. The half life of a radioactive nucleus refers to the time required for the nucleus to decay to half of its initial amount.

Looking at the table, we can see that the initial mass of radioactive material present is 186 grams, within 21 hours, the radioactive substance decayed to half of its initial mass (93 g). Hence, the half life is 21 hours.

Using the formula;

k = 0.693/t1/2

k = 0.693/21 hours = 0.033 hr-1

Using;

N=Noe^-kt

N = mass of radioactive sample at time t

No = mass of radioactive sample initially present

k = decay constant

t = time taken

Substituting values;

2.5/320= e^- 0.033 t

0.0078 = e^- 0.033 t

ln (0.0078) = 0.033 t

t = ln (0.0078)/-0.033

t = 147 hours

Learn more: brainly.com/question/6111443

7 0
2 years ago
Boyles Law P1V1 = P2V2
arsen [322]

Answer:

A. The balloons will increase to twice their original volume.

Explanation:

Boyle's law states that the pressure exerted on a gas is inversely proportional to the volume occupied by the gas at constant temperature. That is:

P ∝ 1/V

P = k/V

PV = k (constant)

P = pressure, V = volume.

P_1V_1=P_2V_2

Let the initial pressure of the balloon be P, i.e. P_1=P, initial volume be V, i.e. V_1=V. The pressure is then halved, i.e. P_2=\frac{P}{2}

P_1V_1=P_2V_2\\\\P*V=\frac{P}{2} *V_2\\\\V_2=\frac{2*P*V}{P}\\\\V_2=2V

Therefore the balloon volume will increase to twice their original volume.

3 0
3 years ago
The half-life of nitrogen-13 is 10.0 minutes. if you begin with 53.3 mg of this isotope, what mass remains after 25.9 minutes ha
zimovet [89]

Hello!

The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.

We have the following data:

mo (initial mass) = 53.3 mg

m (final mass after time T) = ? (in mg)

x (number of periods elapsed) = ?

P (Half-life) = 10.0 minutes

T (Elapsed time for sample reduction) = 25.9 minutes

Let's find the number of periods elapsed (x), let us see:

T = x*P

25.9 = x*10.0

25.9 = 10.0\:x

10.0\:x = 25.9

x = \dfrac{25.9}{10.0}

\boxed{x = 2.59}

Now, let's find the final mass (m) of this isotope after the elapsed time, let's see:

m =  \dfrac{m_o}{2^x}

m =  \dfrac{53.3}{2^{2.59}}

m \approx \dfrac{53.3}{6.021}

\boxed{\boxed{m \approx 8.85\:mg}}\end{array}}\qquad\checkmark

I Hope this helps, greetings ... DexteR! =)

3 0
3 years ago
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