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3 years ago

What is accretion?????

Chemistry

2 answers:

Monica [59]3 years ago

8 0

Answer:

the process of growth or increase, typically by the gradual accumulation of additional layers or matter.

Explanation:

nasty-shy [4]3 years ago

4 0

Answer:

The process of growth or increase, typically by the gradual accumulation of additional layers or matter.

Explanation:

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Identify and label the Brønsted-Lowry acid, its conjugate base, the Brønsted-Lowry base, and its conjugate acid in each of the f

julia-pushkina [17]

Explanation:

As per Brønsted-Lowry concept of acids and bases, chemical species which donate proton are called Brønsted-Lowry acids.

The chemical species which accept proton are called Brønsted-Lowry base.

(a) $HNO_3 + H_2O \rightarrow H_3O^+ + NO_3^-$

$HNO_3$ is Bronsted lowry acid and $NO_3^-$ is its conjugate base.

$H_2O$ is Bronsted lowry base and $H_3O^+$ is its conjugate acid.

(b)

$CN^- + H_2O \rightarrow HCN + OH^-$

$CN^-$ is Bronsted lowry base and HCN is its conjugate acid.

$H_2O$ is Bronsted lowry acid and $OH^-$ is its conjugate base.

(c)

$H_2SO_4 + Cl^- \rightarrow HCl + HSO_4^-$

$H_2SO_4$ is Bronsted lowry acid and $HSO_4^-$ is its conjugate base.

Cl^- is Bronsted lowry base and HCl is its conjugate acid.

(d)

$HSO_4^-+OH^- \rightarrow SO_4^{2-}+H_2O$

$HSO_4^-$ is Bronsted lowry acid and $SO_4^{2-}$ is its conjugate base.

OH^- is Bronsted lowry base and $H_2O$ is its conjugate acid.

(e)

$O_{2-}+H_2O \rightarrow 2OH^-$

$O_{2-}$ is Bronsted lowry base and OH- is its conjugate acid.

$H_2O$ is Bronsted lowry acid and OH- is its conjugate base.

6 0

3 years ago

When the concentrations of CH 3 Br and NaOH are both 0.100 M, the rate of the reaction is 0.0030 M/s. What is the rate of the re

mash [69]

Answer : The rate of the reaction if the concentration of $CH_3Br$ is doubled is, 0.006 M/s

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

The balanced equations will be:

$CH_3Br+NaOH\rightarrow CH_3OH+NaBr$

In this reaction, $CH_3Br$ and $NaOH$ are the reactants.

The rate law expression for the reaction is:

$\text{Rate}=k[CH_3Br][NaOH]$

As we are given that:

$[CH_3Br]$ = concentration of $CH_3Br$ = 0.100 M

$[NaOH]$ = concentration of $NaOH$ = 0.100 M

Rate = 0.0030 M/s

Now put all the given values in the above expression, we get:

$0.0030M/s=k\times (0.100M)\times (0.100M)$

$k=0.3M^{-1}s^{-1}$

Now we have to calculate the rate of the reaction if the concentration of $CH_3Br$ is doubled.

$\text{Rate}=k[CH_3Br][NaOH]$

$\text{Rate}=(0.3M^{-1}s^{-1})\times (2\times 0.100M)\times (0.100M)$

$\text{Rate}=0.006M/s$

Thus, the rate of the reaction if the concentration of $CH_3Br$ is doubled is, 0.006 M/s

3 0

4 years ago

6th grade work east to some people! (:

blagie [28]

b. is correct

wind can cool the land

7 0

3 years ago

Find the density of the object below,

Kitty [74]

Answer:

19.32g/cm³

Explanation:

Density = m/w

Density = 10g/0.5176cm³

Density = 19.32 g/cm³

4 0

3 years ago

Which of the three has the largest ei1? which of the three has the largest ? 1s22s22p63s23p64s1 1s22s22p63s23p5 1s22s22p63s23p1?

coldgirl [10]

$E_i1$ will be largest for $1s^22s^22p^63s^23p^5$ .

Explanation: Ionization energy is the energy to knock off an electron from a gaseous atom of ion. First ionization energy or $E_i1$ is the energy required to remove 1 loosely held electron from 1 mole of gaseous atoms to produce 1 mole of gaseous ion carrying (+)1 charge.

$M(g)\rightarrow M^+(g)+e^-$

The electrons are filled according to Aufbau's rule and the orbitals which are strongly held to the nucleus follows the order $s>p>d>f$ .

Electron is released from the outermost shell that is from the electrons which are loosely held to the nucleus, this follows the pattern $s$ .

In configurations, $1s^22s^22p^63s^23p^64s^1,1s^22s^22p^63s^23p^5\text{ and } 1s^22s^22p^63s^23p^1$

The loosely held orbital is 4s, therefore electron will be lost from that easily.

Now, in 3p orbital, one configuration has 5 electrons and one has 1 electron.

The configuration having 5 electrons will be more tightly held by the nucleus because it has more electrons that the one having only 1 electron. Hence, the electron will be lost easily from the configuration having $3p^1$ as the valence shell.

Therefore, the configuration $1s^22s^22p^63s^23p^5$ will the largest $E_i1$ .

8 0

3 years ago