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Vinvika [58]
3 years ago
6

What is the empirical formula for a compound if a 2.50g sample contains 0.900g of calcium and 1.60g

Chemistry
1 answer:
Sergeeva-Olga [200]3 years ago
4 0

Answer:

CaCl₂

Explanation:

First we <u>convert the given masses of elements into moles</u>, using <em>their respective molar masses</em>:

  • 0.900 g Ca ÷ 40 g/mol = 0.0225 mol Ca
  • 1.60 g Cl ÷ 35.45 g/mol = 0.045 mol Cl

Now we divide those numbers of moles by the lowest value among them:

  • 0.0225 mol Ca / 0.0225 mol = 1
  • 0.045 mol Cl / 0.0225 mol = 2

Meaning the empirical formula for the compound is CaCl₂.

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(WILL MARK AS BRAINLIEST!)
nekit [7.7K]

Answer:

Explanation:

As one moves down the vertical groupings of elements on the periodic table, it is evident that new shells are being added from top to down.

An atomic orbital is the region of space surrounding the nucleus where there is a high probability of finding an electron.

Down a group, the atomic radius increases as more shells are added to an atom.

3 0
3 years ago
What is the name of the solid that will form from the following reaction: a
Reptile [31]

Answer:

yeild solid iron

Explanation:

When two solutions of ionic compounds are mixed, a solid may form. This type of reaction is called a precipitation reaction, and the solid produced in the reaction is known as the precipitate.

4 0
3 years ago
Help!!!!!!!!!!!!!!!!!!
skad [1K]

The density of the sample : 0.827 g/L

<h3>Further explanation</h3>

In general, the gas equation can be written  

\large {\boxed {\bold {PV = nRT}}}

where  

P = pressure, atm , N/m²

V = volume, liter  

n = number of moles  

R = gas constant = 0.082 l.atm / mol K (P= atm, v= liter),or 8,314 J/mol K (P=Pa or N/m2, v= m³)

T = temperature, Kelvin  

n= 1 mol

MW Neon = 20,1797 g/mol

mass of Neon :

\tt mass=mol\times MW\\\\mass=1\times 20,1797 =20.1797~g

The density of the sample :

\tt \rho=\dfrac{m}{V}\\\\\rho=\dfrac{20,1797}{24.4}=0.827~g/L

or We can use the ideal gas formula ta find density :

\tt \rho=\dfrac{P\times MW}{RT}\\\\\rho=\dfrac{1\times 20.1797}{0.082\times 298}\\\\\rho=0.826~g/L

8 0
3 years ago
What volume of 15.7 M H2SO4 is required to prepare 12.0 L of 0.156 M sulfuric acid?
kirza4 [7]

The volume of 15.7 M H2SO4 is required to prepare 12.0 L of 0.156 M sulfuric acid is 0.12 L

<h3>Data obtained from the question</h3>

From the question given above, the following data were obtained:

  • Molarity of stock solution (M₁) = 15.7 M
  • Volume of diluted solution (V₂) = 12 L
  • Molarity of diluted solution (M₂) = 0.156 M
  • Volume of stock solution needed (V₁) = ?

<h3>How to determine the volume of the stock solution needed</h3>

The volume of the stock solution needed can be obtained by using the dilution formula as shown below:

M₁V₁ = M₂V₂

15.7 × V₁ = 0.156 × 12

15.7 × V₁ = 1.872

Divide both side by 15.7

V₁ = 1.872 / 15.7

V₁ = 0.12 L

Thus, the volume of the stock solution needed to prepare the solution is 0.12 L

Learn more about dilution:

brainly.com/question/15022582

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4 0
1 year ago
Interpretacao gerais das praticas <br>sobre densidade de amostra solido e liquido<br>​
Savatey [412]

Answer:

O sólido tem densidade mais alta em comparação ao líquido.

Explicação:

A densidade da amostra sólida é maior do que a densidade do líquido porque há pouco espaço entre as partículas do sólido. A densidade tem relação inversa com o volume de uma substância, se uma substância ocupa mais espaço então sua densidade é menor, enquanto se a substância ocupa menos espaço então tem maior densidade. As substâncias sólidas ocupam menos espaço em comparação com as substâncias líquidas, então podemos dizer que a densidade do sólido é maior do que as substâncias líquidas.

6 0
3 years ago
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