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Klio2033 [76]
2 years ago
11

When the elements in group 1 are considered in order from top to bottom, each successive element at standard pressure has

Chemistry
2 answers:
KengaRu [80]2 years ago
6 0
When the elements in group 1 are considered in order from top to bottom, each successive element at a standard pressure has a lower melting point and boiling point. This is because melting and boiling points decrease when the increasing atomic number. The correct answer will be D. 
Talja [164]2 years ago
4 0

Answer is: d. a lower melting point and boiling point.

Alkaline metals are lithium, sodium, rubidium, potassium, cesium and francium.

Bond between these elements is metallic bond.

Both the melting and boiling points decrease down the group 1, because the decrease in the strength of each metallic bond.

From top to bottom, atomic radius increase and the distance between the nucleus and delocalized electrons increases, so metallic bond is weaker (attractions fall).

Metallic bond is formed between electrons and positively charged metal ions.

You might be interested in
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
How many neutrons and protons (nucleons) does an atom with
mezya [45]

Answer:

E

Explanation:

None of the above

The atom with the symbol S is called sulphur Sulphur has atomic number 16 which means that it has 16 protons. Sulphur-32 has 32 nucleons - 16 protons and 16 neutrons.Sulfur is a chemical element  It is  nonmetallic. Under normal conditions, sulfur atoms form cyclic octatomic molecules with a chemical formula S8. Elemental sulfur is a bright yellow, crystalline solid at room temperature.

4 0
3 years ago
As part of your job you are asked to make 1 liter of a 0.5 molar sucrose solution. how much sucrose (c12h22o11) do you need? use
kakasveta [241]

Answer:-  171 g

Solution:- It asks to calculate the grams of sucrose required to make 1 L of 0.5 Molar solution of it.

We know that molarity is moles of solute per liter of solution.

If molarity and volume is given then, moles of solute is molarity times volume in liters.

moles of solute = molarity* liters of solution

moles of solute = 0.5*1 = 0.5 moles

To convert the moles to grams we multiply the moles by molar mass.

Molar mass of sucrose = 12(12) + 22(1) + 11(16)  

= 144 + 22 + 176

= 342 grams per mol

grams of sucrose required = moles * molar mass

grams of sucrose required = 0.5*342  = 171 g

So, 171 g of sucrose are required to make 1 L of 0.5 molar solution.




6 0
3 years ago
What are the primary atoms that make up the scents that we smell?
Scorpion4ik [409]

Answer:

Molecules make scents. Aromatic ones (That is, containing rings of carbon atoms with delocalised electrons). Some unpleasant smells are due to hydrogen and sulphur groups.

Explanation:

3 0
2 years ago
Read 2 more answers
How many moles of sulfate ions are in a sample of aluminum sulfate in which the number of formula units is 7.534 × 1023?
valentina_108 [34]

Answer:a unit of grammatical organization next below the sentence in rank and in traditional grammar said to consist of a subject and predicate

Explanation:

7 0
2 years ago
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