All categories

3 years ago

Element in the noble ga family are considered stabled because their outer energy levels are

Chemistry

1 answer:

blsea [12.9K]3 years ago

3 0

Answer: In octet state.

Explanation: For noble gases they are stable in state since their outer shell contain fully occupied having 8 electrons.

You might be interested in

What is the molarity of a solution that contains 1000.0 mg of AgNO3 that has been dissolved in 500 mL of water

Vaselesa [24]

0.012moldm⁻³

Explanation:

Given parameters:

Mass of AgNO₃ = 1000mg

Volume of water = 500mL

Unknown:

Molarity of solution = ?

Solution:

The molarity of a solution is the number of moles of a solute dissolved in volume of solvent.

Molarity = $\frac{xnumber of moles}{Volume}$

Number of moles of AgNO₃ = ?

Number of moles = $\frac{mass}{molar mass}$

Molar mass of AgNO₃ = 108 + 14 + 3(16) = 170g/mol

convert mass to g;

1000mg = 1g

Number of moles = $\frac{1}{170}$ = 0.00588moles

convert the given volume to dm³;

1000mL = 1dm³;

500mL = 0.5dm³

Now solve;

Molarity = $\frac{0.00588}{0.5}$ = 0.012moldm⁻³

learn more:

Molarity brainly.com/question/9324116

#learnwithBrainly

4 0

3 years ago

Mara and ivy are asked to identify two minerals in science class. To do this, they decided to study the physical properties of t

scZoUnD [109]

Answer:

2 only is a pretty accurate answer the others dont make sense to me.

Explanation:

8 0

3 years ago

Read 2 more answers

he heat of fusion of tetrahydrofuran is . Calculate the change in entropy when of tetrahydrofuran melts at . Be sure your answer

lana [24]

Answer:

$\Delta S=1.8x10^{-3}\frac{kJ}{K}=1.8\frac{J}{K}$

Explanation:

Hello.

In this case, given the heat of fusion of THF to be 8.5 kJ/mol and freezing at -108.5 °C, for the required mass of 5.9 g, we can compute the entropy as:

$\Delta S=\frac{n*\Delta H}{T}$

Whereas n accounts for the moles which are computed below:

$n=5.9g*\frac{1mol}{72g} =0.082mol$

Thus, the entropy turns out:

$\Delta S=\frac{0.0819mol*8.5 kJ/mol}{(-108.5+273.15)K}\\\\\Delta S=1.8x10^{-3}\frac{kJ}{K}=1.8\frac{J}{K}$

Best regards.

3 0

3 years ago

A sample compound with a molar mass of 34.00g/mol is found to consist of 0.44g H and 6.92g O. Calculate both empirical and molec

balu736 [363]

Answer:

E.F= OH

M.F= $O_{2} H_{2}$

Explanation:

Empirical Formula

Step 1: Calculate mols of each element

0.44gH(1 mol H/ 1.008g H)= 0.4365 mol H

*note leave extra sig figs for calculations

6.92gO(1 mol O/ 16 g O)=0.4325 mol O

Step 2: Identify which is the smallest mol

*in this case 0.4365 mol H> 0.4325 mol O so we will use 0.4325 mol O

Step 3: Divide above calculations by the smallest mol

0.4325 mol O/0.4325= 1

0.4365 mol H/0.4325= 1.009 *rounds to 1

Step 4: use calculations as subscripts

Oxygen = 1 so the subscript will 1 (O)

Hydrogen = 1 so the subscript will 1 (H)

making E.F= OH

Molecular Formula:

Step 1: identify molecular mass and mass from the E.F

the molecular mass is given 34.00g/mol

the mass of the E.F is

(oxygen mass from periodic table)+(hydrogen mass from periodic table)

16+1.008= 17.008

Step 2:Divide the molecular mass by the mass given by the emipirical formula.

$\frac{34.00}{17.008}= 1.999$ round to 2

Step 3:Multiply the empirical formula (the subscripts) by this number to get the molecular formula. ANSWER: M.F=2(OH)- $O_{2} H_{2}$

3 0

3 years ago

Which term describes this reaction?

Semmy [17]

Answer:

Condensation is the term used to describe the reaction

5 0

3 years ago

Read 2 more answers