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3 years ago

What information about elements can be collected from the periodic table?

Chemistry

1 answer:

Lostsunrise [7]3 years ago

8 0

Answer:

A lot

Explanation:

- As you go down a group, the elements get larger in size

- All of the elements of the same group have the same # of valence electrons

- As you go to the right of a period, the # of electrons in the outermost shell increases

- The elements at the top right of the table have the highest electronegativity

- Groups 13 through 17 are nonmetals

- The last group contains noble gases

etc.

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Identify the reagents necessary to achieve each of the following transformations: Na2Cr2O7, H2SO4, H2O EtOH, H+ H2O Fe2+, NaOH N

Gala2k [10]

There’s not an answer in there.

8 0

3 years ago

Dawn is comparing how different animals move and the structures they use to move. She made the table shown below.

Dovator [93]

Answer:

tail and fins

Explatanation:

6 0

3 years ago

A reaction is followed and found to have a rate constant of 3.36 × 104 m-1s-1 at 344 k and a rate constant of 7.69 m-1s-1 at 219

g100num [7]

The relation between rate constants at different temperatures, temperature and activation energy is known as Arrhenius equation

<u>Arrhenius equation</u>

$K=Ae^-{\frac{Ea}{RT} }$

For two temperatures

$Ea=R(\frac{ln\frac{k1}{k1} }{(\frac{1}{T1})-(\frac{1}{T2})})$

Where

Ea = ? = activation energy

k1 = 3.36 × 10⁴

T1=344 k

k2=7.69

T2=219K

R= gas constant = 8.314 J /molK

Putting values

$Ea= (8.314)(\frac{ln(\frac{7.69}{33600})}{(\frac{1}{344})(\frac{1}{219})}$

Ea = (-69.69)/(-0.00166) = 41981.93 J/mol

Activation energy is 42.0 kJ /mol

3 0

3 years ago

. 125g of water has an initial temperature of 25.6°C, and is heated by 50.0g of a metal

Eduardwww [97]

Answer : The specific heat of the metal is, $0.451J/g^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of metal = ?

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of metal = 50.0 g

$m_2$ = mass of water = 125 g

$T_f$ = final temperature of mixture = $29.3^oC$

$T_1$ = initial temperature of metal = $115.0^oC$

$T_2$ = initial temperature of water = $25.6^oC$

Now put all the given values in the above formula, we get

$(50.0g)\times c_1\times (29.3-115.0)^oC=-[(125g)\times 4.18J/g^oC\times (29.3-25.6)^oC]$

$c_1=0.451J/g^oC$

Therefore, the specific heat of the metal is, $0.451J/g^oC$

4 0

3 years ago

How to Prevent Onions from Making You Cry

sergejj [24]

Answer:

Deleting everything except the first sentences of each numbered step.

Explanation:

The first sentences actually tell you what to do while the rest just explains why they're telling you to do it. If you read that article, all you would want to know are the steps to prevent crying while cutting onions, not necessarily the science behind those reasons.

6 0

3 years ago