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2 years ago

The location of an element in the periodic table can be used to predict which elements combine with others to form compounds.

Chemistry

1 answer:

skelet666 [1.2K]2 years ago

7 0

False. I rlly hope this helps you!

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Which process would produce a similar pathway? condensation dehydration decomposition melting

babymother [125]

It is A) Condensation

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2 years ago

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Lithium, sodium and potassium are all in group 1 of the periodic table. Which of these has the lowest boiling point?

kvv77 [185]

Answer:

Potassium

Explanation:

In group one of the periodic table both the melting and the boiling points usually decrease down the group.

Now, down the group, it's lithium that comes first, then sodium, then potassium.

Thus, among the 3, potassium is furthermost down the group by virtue of the factor it has the highest atomic number.

Therefore, we can say that potassium has the lowest boiling point among the 3.

5 0

2 years ago

Elements with atomic numbers 7 and 83 have how many valence electrons? *

IRINA_888 [86]

Explanation:

element with atomic numbers 7 and 83 have 5 and 5 valence electrons

5 0

2 years ago

What is the h+ concentration for an aqueous solution with poh = 3.35 at 25 ∘c? express your answer to two significant figures an

RSB [31]

POH value was calculated by the negative logarithm of hydroxide ion concentration.
To know the hydrogen ion concentration, we need to know the pH value, that can be found out if pOH is known
pH + pOH = 14
pH = 14 - pOH
pH = 10.65
once the pH is known we have to find the antilog.
[H⁺] = antilog (-pH)
antilog can be found by
[H⁺] = 10^(-10.65)
[H⁺] = 2.2 x 10⁻¹¹ M

3 0

2 years ago

Given that the initial rate constant is 0.0110s−1 at an initial temperature of 21 ∘C , what would the rate constant be at a temp

gulaghasi [49]

The rate constant is mathematically given as

K2=2.67sec^{-1}

<h3>What is the Arrhenius equation?</h3>

The rate constant for a particular reaction may be calculated with the use of the Arrhenius equation. This constant can be stated in terms of two distinct temperatures, T1 and T2, as follows:

$ln(\frac{K2}{K1})= (\frac{Ea}{R})*(\frac{1}{T1}-\frac{1}{T2})$

Therefore

KT1= 0.0110^{-1}

T1= 21+273.15

T1= 294.15K

T2= 200

T2=200+273.15

T2= 473.15K

Ea= 35.5 Kj/Mol

Hence, in j/mol R Ea is

Ea=35.5*1000 j/mol R

$ln(\frac{K2}{0.0110})= (\frac{35.5*1000}{8.314})*(\frac{1}{294.15}-\frac{1}{473.15}\\\\ln(\frac{K2}{0.0110})=5.492$

K2/0.0110 =e^(5.492)

K2/0.0110 =242.74

K2= 242.74*0.0110

K2=2.67sec^{-1}

In conclusion, rate constant

K2=2.67sec^{-1}