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3 years ago

Which of the following takes place during a double displacement reaction?

2 answers:

4 0

Double displacement: parts of compounds switch places to form two new compounds

FYI
decomposition: a complex substance breaks down into two or more simple substances

Single displacement: a single substance replaces another substance in a compound

Synthesis: two simple substances combine to form a new complex substance

Rama09 [41]3 years ago

4 0

Answer: The correct answer is parts of compounds switch places to form two new compounds.

Explanation:

Double displacement reaction is defined as the reaction in which exchange of ions takes place to form two different compounds.

Decomposition reaction is defined as the chemical reaction in which a larger molecule breaks down into two or more smaller molecules.

Single displacement reaction is defined as a reaction in which more reactive metal displaces a less reactive metal from its chemical reaction.

Synthesis reaction is defined as the chemical reaction in which two or more smaller molecules combine to form a single larger molecule.

Hence, the correct answer is parts of compounds switch places to form two new compounds.

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The driver of a 1750 kg car traveling on a horizontal road at 110 km/h suddenly applies the brakes. Due to a slippery pavement,

Alexeev081 [22]

Answer: a=-2.4525 m/s^2

d=s=190.3 m

Explanation:The only force that is stopping the car and causing deceleration is the frictional force Fr

Fr = 25% of weight

W=mg

W=1750*9.81

W=17167.5

Hence

$Fr=\frac{25}{100} * -17167.5\\\\Fr=-4291.875 N$

Frictional force is negative as it acts in opposite direction

According to newton second law of motion

F=ma

hence

$a=Fr/m$

$a=-4291.875/1750\\a=-2.4525$

given

u= 110 km/h

u=110*1000/3600

u=30.55 m/s

to get t we know that final velocity v=0

$v^2=u^2+2as\\0=30.55^2-2*2.4525*s\\s=190.34m$

3 0

3 years ago

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natali 33 [55]

It should be the B
Low frequency and long wavelength

6 0

2 years ago

An electron and a proton are held on an x axis, with the electron at x = + 1.000 m

mixas84 [53]

Answer:

r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

Explanation:

For this exercise we must use conservation of energy

the electric potential energy is

U = $k \frac{q_1q_2}{r_{12}}$

for the proton at x = -1 m

U₁ = $- k \frac{e^2 }{r+1}$

for the electron at x = 1 m

U₂ = $k \frac{e^2 }{r-1}$

starting point.

Em₀ = K + U₁ + U₂

Em₀ = $\frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1}$

final point

Em_f = $k e^2 ( -\frac{1}{r_2 +1} + \frac{1}{r_2 -1})$

energy is conserved

Em₀ = Em_f

\frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e^2 (- \frac{1}{r_2 +1} + \frac{1}{r_2 -1})

\frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e²( $\frac{2}{(r_2+1)(r_2-1)}$ )

we substitute the values

½ 9.1 10⁻³¹ 450 + 9 10⁹ (1.6 10⁻¹⁹)² [ $- \frac{1}{20+1} + \frac{1}{20-1}$ ) = 9 109 (1.6 10-19) ²( $\frac{2}{r_2^2 -1}$ )

2.0475 10⁻²⁸ + 2.304 10⁻³⁷ (5.0125 10⁻³) = 4.608 10⁻³⁷ ( $\frac{1}{r_2^2 -1}$ )

2.0475 10⁻²⁸ + 1.1549 10⁻³⁹ = 4.608 10⁻³⁷ $\frac{1}{r_2^2 -1}$

$\frac{2.0475 \ 10^{-28} }{1.1549 \ 10^{-37} } = \frac{1}{r_2^2 -1}$

r₂² -1 = (4.443 10⁸)⁻¹

r2 = $\sqrt{1 + 2.25 10^{-9}}$

r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

4 0

3 years ago

The number of electrons an atom would gain or lose when forming ionic bonds. It can be positive, negative or zero.

vaieri [72.5K]

Answer:

In an ionic bond , an element will have to lose or gain electrons.

Explanation:

Ionic bond, also called electrovalent bond, type of linkage formed from the electrostatic attraction between oppositely charged ions in a chemical compound.
Such a bond forms when the valence (outermost) electrons of one atom are transferred permanently to another atom.
The atom that loses the electrons becomes a positively charged ion (cation), while the one that gains them becomes a negatively charged ion (anion).

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