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3 years ago

How many electrons does it take to fill the outer energy level of most atoms

Chemistry

1 answer:

Orlov [11]3 years ago

5 0

Depends on the element it can by up to 3, 8, or maybe 16.

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Can someone please help me come up with a hypothesis for an experiment?

m_a_m_a [10]

I believe that sugar will desolve the fastest because of the room temperature

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2 years ago

What is the maximum amount of kcl that can dissolve in 200 g of water? (the solubility of kcl is 34 g/100 g h2o at 20°c.) (1 poi

Solnce55 [7]

The answer is 68g
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5 0

3 years ago

An unknown compound with a molar mass of 223.94 g/mol consists of 32.18% c, 4.50% h, and 63.32% cl. find the molecular formula f

Dima020 [189]

The actual number of atoms of each element present in the molecule of the compound is represented by the formula known as molecular formula.

Molar mass of the unknown compound = 223.94 g/mol (given)

Mass of each element present in the unknown compound is determined as:

Mass of carbon, $C$ :

$\frac{32.18}{100}\times 223.94 = 72.06 g$

Mass of hydrogen, $H$ :

$\frac{4.5}{100}\times 223.94 = 10.08 g$

Mass of chlorine, $Cl$ :

$\frac{63.32}{100}\times 223.94 = 141.79 g$

Now, the number of each element in the unknown compound is determined by the formula:

$number of moles = \frac{given mass}{molar mass}$

Number of moles of $C$ :

$number of moles = \frac{72.06}{12} = 6.005 mole\simeq 6 mole$

Number of moles of $H$ :

$number of moles = \frac{10.08}{1} = 10.08 mole\simeq 10 mole$

Number of moles of $Cl$

$number of moles = \frac{141.79}{35.5} = 3.99 mole\simeq 4 mole$

Dividing each mole with the smallest number of mole, to determine the empirical formula:

$C_{\frac{6}{4}}H_{\frac{10}{4}}Cl_{\frac{10}{4}}$

$C_{1.5}H_{2.5}Cl_{1}$

Multiplying with 2 to convert the numbers in formula into a whole number:

So, the empirical formula is $C_{3}H_{5}Cl_{2}$ .

Empirical mass = $12\times 3+1\times 5+2\times 35.5 = 112 g/mol$

In order to determine the molecular formula:

n = $\frac{molar mass}{empirical mass}$

n = $\frac{223.94}{112} = 1.99 \simeq 2$

So, the molecular formula is:

$2\times C_{3}H_{5}Cl_{2} = C_{6}H_{10}Cl_{4}$

6 0

3 years ago

How much work (in JJ) is required to expand the volume of a pump from 0.0 LL to 2.5 LL against an external pressure of 1.1 atmat

stira [4]

Answer:

- 278.85 J

Explanation:

Given that:

Pressure = 1.1 atm

The initial volume V₁ = 0.0 L

The final volume V₂ = 2.5 L

The work that takes place in a reaction at constant pressure can be expressed by using the equation:

W = P(V₂ - V₁ )

Since the volume of the gas is expanded from 0 to 2.5 L when 1.1 atm pressure is applied. Then, the work can be given by the expression:

W = - P(V₂ - V₁ )

W = -1.1 atm ( 2.5 - 0.0) L

W = -1.1 atm (2.5 L)

W = -2.75 atm L

Recall that:

1 atm L = 101.4 J

Therefore;

-2.75 atm L = ( -2.75 × 101.4 )J

= -278.85 J

Thus, the work required at the chemical reaction when the pressure applied is 1.1 atm = - 278.85 J

8 0

3 years ago

Where are electrons found in the atom?

RoseWind [281]

In the electron shell

5 0

2 years ago