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3 years ago

What is Alkali metals

2 answers:

6 0

Any of the elements lithium, sodium, potassium, rubidium , cesium, and francium, occupying group 1A of the periodic table.

seropon [69]3 years ago

3 0

Answer:

The alkali metals consist of the chemical elements lithium, sodium, potassium, rubidium, caesium, and francium. Together with hydrogen, they constitute group 1, which lies in the s-block of the periodic table

The alkali metals are all shiny, soft, highly reactive metals at standard temperature and pressure and readily lose their outermost electron to form cations with charge +1.

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Arrange the molecules in order from strongest intermolecular force to weakest.

motikmotik

Answer:- $C_4H_1_0>C_3H_8>C_2H_6>CH_4$

Explanations:- Alkanes are non polar molecules as these only have carbons and hydrogens. Electron negativity difference of C and H is very low and it makes them non polar. These have weaker London dispersion forces.

The forces of attraction becomes stronger in alkanes as the number of carbon increases because the surface area as well as molecular weight of the alkanes increases with an increase in number of carbons.

Butane has four carbons, propane has three carbons, ethane has two and methane has only one carbon, So, the strongest to weakest order of inter molecular forces is butane > propane > ethane > methane .

3 0

4 years ago

How is the process of mitosis essential to the survival of an organism?

ELEN [110]

Answer:

Explanation:

Mitosis is crucial to this process. Mitosis is the reason we can grow, heal wounds, and replace damaged cells. Mitosis is also important in organisms which reproduce asexually: this is the only way that these cells can reproduce. This is the one key process that sustains populations of asexual organisms.Jul 22, 2020

3 0

3 years ago

Which of these half-reactions represents reduction?

gogolik [260]

Answer: The half-reactions represents reduction are as follows.

$Cr_{2}O^{2-}_{7} \rightarrow Cr^{3+}$
$MnO^{-}_{4} \rightarrow Mn^{2+}$

Explanation:

A half-reaction where addition of electrons take place or a reaction where decrease in oxidation state of an element takes place is called reduction-half reaction.

For example, the oxidation state of Cr in $Cr_{2}O^{2-}_{7}$ is +6 which is getting converted into +3, that is, decrease in oxidation state is taking place as follows.

$Cr_{2}O^{2-}_{7} + 3 e^{-} \rightarrow Cr^{3+}$

Similarly, oxidation state of Mn in $MnO^{-}_{4}$ is +7 which is getting converted into +2, that is, decrease in oxidation state is taking place as follows.

$MnO^{2-}_{4} + 5 e^{-} \rightarrow Mn^{2+}$

Thus, we can conclude that half-reactions represents reduction are as follows.

$Cr_{2}O^{2-}_{7} \rightarrow Cr^{3+}$
$MnO^{-}_{4} \rightarrow Mn^{2+}$

3 0

3 years ago

Which change occurs when an atom in excited state returns to ground state? 1 energy is emitted 2energy is absorbed 3 the number

kozerog [31]

The answer is number 1 - energy is emitted

Energy is released when an atom in an excited state returns to the ground state.

7 0

3 years ago

Read 2 more answers

First-order reaction that results in the destruction of a pollutant has a rate constant of 0.l/day. (a) how many days will it ta

Klio2033 [76]

Rate equation for first order reaction is as follows:

$t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}$

Here, k is rate constant of the reaction, t is time of the reaction, $A_{0}$ is initial concentration and $A_{t}$ is concentration at time t.

The rate constant of the reaction is $0.1 day^{-1}$ .

(a) Let the initial concentration be 100, If 90% of the chemical is destroyed, the chemical present at time t will be 100-90=10, on putting the values,

$t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{10}=23.03 days$

Thus, time required to destroy 90% of the chemical is 23.03 days.

(b) Let the initial concentration be 100, If 99% of the chemical is destroyed, the chemical present at time t will be 100-99=1, on putting the values,

$t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{1}=46.06 days$

Thus, time required to destroy 99% of the chemical is 46.06 days.

(c) Let the initial concentration be 100, If 99.9% of the chemical is destroyed, the chemical present at time t will be 100-99.9=0.1, on putting the values,

$t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{0.1}=69.09 days$

Thus, time required to destroy 99.9% of the chemical is 69.09 days.

5 0

3 years ago