Hi um

How are radio waves transmitted

kotegsom [21]

Attractive antennas that pull the waves tword it.

7 0

3 years ago

What is the major organic product obtained from the following reaction?

trapecia [35]

The correct option is (b)

NaNH2 is an effective base. It can be a good nucleophile in the few situations where its strong basicity does not have negative side effects. It is employed in elimination reactions as well as the deprotonation of weak acids.Alkynes, alcohols, and a variety of other functional groups with acidic protons, such as esters and ketones, will all be deprotonated by NaNH2, a powerful base.Alkynes are deprotonated with NaNH2 to produce what are known as "acetylide" ions. These ions are powerful nucleophiles that can react with alkyl halides to create carbon-carbon bonds and add to carbonyls in an addition reaction.Acid/base and nucleophilic substitution are the two types of reactions.Using the right base, terminal alkynes can be deprotonated to produce a carbanion.A good C is the acetylide carbanion.The acetylide carbanion can undergo nucleophilic substitution reactions because it is a potent C nucleophile. (often SN2) with 1 or 2 alkyl halides with electrophilic C to create an internal alkyne (Cl, Br, or I).Elimination is more likely to occur with 3-alkyl halides.It is possible to swap either one or both of the terminal H atoms in ethylene (acetylene) to create monosubstituted (R-C-C-H) and symmetrical (R = R') or unsymmetrical (R not equal to R') disubstituted alkynes (R-C-C-R').

Learn more about NANH2 here :-

brainly.com/question/12601787

#SPJ4

7 0

2 years ago

calculate the difference in slope of the chemical potential against temperature on either side of the normal freezing point of w

kipiarov [429]

Answer:

(a) The normal freezing point of water (J·K−1·mol−1) is $-22Jmole^-^1k^-^1$

(b) The normal boiling point of water (J·K−1·mol−1) is $-109Jmole^-^1K^-^1$

(c) the chemical potential of water supercooled to −5.0°C exceed that of ice at that temperature is 109J/mole

Explanation:

Lets calculate

(a) - General equation -

$(\frac{d\mu(\beta )}{dt})p-(\frac{d\mu(\alpha) }{dt})_p$ = $-5_m(\beta )+5_m(\alpha )$ = $-\frac{\Delta H}{T}$

$\alpha ,\beta$ → phases

ΔH → enthalpy of transition

T → temperature transition

$(\frac{d\mu(l)}{dT})_p -(\frac{d\mu(s)}{dT})_p$ = $= -\frac{\Delta_fH}{T_f}$

= $\frac{-6.008kJ/mole}{273.15K}$ ( $\Delta_fH$ is the enthalpy of fusion of water)

= $-22Jmole^-^1k^-^1$

(b) $(\frac{d\mu(g)}{dT})_p-(\frac{d\mu(l)}{dT})_p= -\frac{\Delta_v_a_p_o_u_rH}{T_v_a_p_o_u_r}$

= $\frac{40.656kJ/mole}{373.15K}$ ( $\Delta_v_a_p_o_u_rH$ is the enthalpy of vaporization)

= $-109Jmole^-^1K^-^1$

(c) $\Delta\mu =\Delta\mu(l)-\Delta\mu(s)$ = $-S_m\DeltaT$

$[\mu(l-5$ ° $C)-\mu(l,0$ ° $C)]$ = $[\mu(s-5$ ° $C)-\mu(s,0$ ° $C)]$ $=-S_m$ ΔT

$\mu(l,-5$ ° $C)-\mu(s,-5$ ° $C)=-Sm\DeltaT [\mu(l,0$

$\Delta\mu=(21.995Jmole^-^1K^-^1)\times (-5K)$

= 109J/mole

6 0

2 years ago

Equilibrium is established in a reversible reaction when

avanturin [10]

The correct answer among the options given is B. . When the forward and reverse paths of a change occur at the same rate, the system is in equilibrium specifically in dynamic equilibrium. Dynamic equilibrium is the balance in a process that is continuing.

6 0

4 years ago

Please Help!! How does alpha decay change the nucleus of radioactive atom.

cluponka [151]

Answer:

When atom undergoes the alpha emission the original atom convert into the atom having mass number less than 4 and atomic number 2 less than the starting atom.

Explanation:

Alpha radiations are emitted as a result of radioactive decay. The atom emit the alpha particles consist of two proton and two neutrons. Which is also called helium nuclei. When atom undergoes the alpha emission the original atom convert into the atom having mass number less than 4 and atomic number 2 less than the starting atom.

Properties of alpha radiation:

Alpha radiations can travel in a short distance.

These radiations can not penetrate into the skin or clothes.

These radiations can be harmful for the human if these are inhaled.

These radiations can be stopped by a piece of paper.

₉₂U²³⁸ → ₉₀Th²³⁴ + ₂He⁴ + energy

5 0

3 years ago