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3 years ago

What is the trend for atomic radii across the periods and down the group?

Chemistry

1 answer:

jok3333 [9.3K]3 years ago

5 0

Answer:

So, in general, atomic radius decreases across a period and increases down a group. Across a period, effective nuclear charge increases as electron shielding remains constant.

Explanation:

Moving from left to right across a period, the atomic radius decreases. The nucleus of the atom gains protons moving from left to right, increasing the positive charge of the nucleus and increasing the attractive force of the nucleus upon the electrons.

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Acid precipitation is an agent of_______ weathering

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Chemical weathering

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4 years ago

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Most chemical reactions can be classified as one of four types. Using the chemical equations in the Gizmo as a guide, match the

Lostsunrise [7]

Answer:

<h2>The sequence is; b, e, a, d, c </h2>

Explanation:

1. In a decomposition reaction; One reactant is broken down into two or more than two products is called decomposition.

2. A combustion reaction; A fuel is combined with oxygen to produce carbon dioxide and water, this reaction is called combustion reaction.

3. A synthesis reaction; it occurs when two or more reactants combine to form one product is known as synthesis reaction.

4. Double Replacement Reaction; Two compounds react to form two different compounds is known as double Replacement Reaction.

5. A single replacement reaction; occurs when a compound reacts with an element to form a new compound , this reaction is called as single replacement reaction.

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3 years ago

A compound is 42.9% C, 2.4% H, 16.7% N, and 38.1% O, by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, lowers the

Romashka [77]

This is an incomplete question, here is a complete question.

A compound is 42.9% C, 2.4% H, 16.7% N and 38.1% O by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, C₆H₆ (d= 0.879 g/mL; Kf= 5.12 degrees Celsius/m), lowers the freezing point from 5.53 to 1.37 degrees Celsius. What is the molecular formula of this compound?

Answer : The molecular of the compound is, $C_6H_4N_2O_4$

Explanation :

First we have to calculate the mass of benzene.

$\text{Mass of benzene}=\text{Density of benzene}\times \text{Volume of benzene}$

$\text{Mass of benzene}=0.879g/mL\times 50.0mL=43.95g$

Now we have to calculate the molar mass of unknown compound.

Given:

Mass of unknown compound (solute) = 6.45 g

Mass of benzene (solvent) = 43.95 g = 0.04395 kg

Formula used :

$\Delta T_f=K_f\times m\\\\\Delta T_f=K_f\times\frac{\text{Mass of unknown compound}}{\text{Molar mass of unknown compound}\times \text{Mass of benzene in Kg}}$

where,

$\Delta T_f$ = change in freezing point = $5.53-1.37=4.16^oC$

$\Delta T_s$ = freezing point of solution

$\Delta T^o$ = freezing point of benzene

Molal-freezing-point-depression constant $(K_f)$ for benzene = $5.12^oC/m$

m = molality

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

$4.16^oC=(5.12^oC/m)\times \frac{6.45g}{\text{Molar mass of unknown compound}\times 0.04395kg}$

$\text{Molar mass of unknown compound}=180.6g/mol$

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 42.9 g

Mass of H = 2.4 g

Mass of N = 16.7 g

Mass of O = 38.1 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{42.9g}{12g/mole}=3.575moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{2.4g}{1g/mole}=2.4moles$

Moles of N = $\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.7g}{14g/mole}=1.193moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{38.1g}{16g/mole}=2.381moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{3.575}{1.193}=2.99\approx 3$

For H = $\frac{2.4}{1.193}=2.01\approx 2$

For N = $\frac{1.193}{1.193}=1$

For O = $\frac{2.381}{1.193}=1.99\approx 2$

The ratio of C : H : N : O = 3 : 2 : 1 : 2

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_3H_2N_1O_2$

The empirical formula weight = 3(12) + 2(1) + 1(14) + 2(16) = 84 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{180.6}{84}=2$

Molecular formula = $(C_3H_2N_1O_2)_n=(C_3H_2N_1O_2)_2=C_6H_4N_2O_4$

Therefore, the molecular of the compound is, $C_6H_4N_2O_4$

3 0

3 years ago

Kipish [7]

Answer:

11 proton and 12 neutron

11 electrons

Explanation:

3 0

3 years ago

Adi loves the smell of lavender, so she decides to press a few petals of the flower until they release a few drops of oil. She w

soldier1979 [14.2K]

Answer:

See Explanation

Explanation:

What Adi failed to realize is that the oily substance that was obtained from lavender consists of a mixture of substances. It is not only the required fragrance that is present in the extract.

This experiment will not work because those other components in the mixture may be erroneously identified when they show up in the mass spectrum of the extract and may be mistaken for the fragrance in question.

Hence the experiment will not work because; if some kind of separation method is not used to identify other impurities in the oil, many other substances may be mistaken for the actual fragrance.

4 0

3 years ago