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zavuch27 [327]
3 years ago
11

An element that has the same number of protons and electrons is called a __________ atom. The charges balance out.

Chemistry
1 answer:
Virty [35]3 years ago
8 0
It is called a neutral atom
You might be interested in
Select the correct electron configuration for Boron. (Atomic Number 5) 1s 22s 12p 2 1s 22s 3 1s 32s 12p 1 1s 22s 22p 1
kodGreya [7K]

ans is: 1s2, 2s2 ,2p1,...

7 0
3 years ago
A compound contains 6.0 g of carbon and 1.0 g of hydrogen and has a molar mass of 42.0 g/mol.
makvit [3.9K]

Answer:

%C = 85.71 wt%; %H = 14.29 wt%; Empirical Formula => CH₂; Molecular Formula => C₃H₆

Explanation:

%Composition

Wt C = 6 g

Wt H = 1 g

TTL Wt = 6g + 1g = 7g

%C per 100wt = (6/7)100% = 85.71 wt%

%H per 100wt = (1/7)100% = 14.29 wt % or, %H = 100% - %C = 100% - 85.71% = 14.29 wt% H

What you should know when working empirical formula and molecular formula problems.

Empirical Formula=> <u>smallest</u> whole number ratio of elements in a compound

Molecular Formula => <u>actual</u> whole number ratio of elements in a compound

Empirical Formula Weight x Whole Number Multiple = Molecular Weight

From elemental %composition values given (or, determined as above), the empirical formula type problem follows a very repeatable pattern. This is ...

% => grams => moles => ratio => reduce ratio => empirical ratio

for determination of molecular formula one uses the empirical weight - molecular weight relationship above to determine the whole number multiple for the molecular ratios.

Caution => In some 'textbook' empirical formula problems, the empirical ratio may contain a fraction in the amount of 0.25, 0.50 or 0.75. If such an issue arises, multiply all empirical ratio numbers containing 0.25 and/or 0.75 by '4'  to get the empirical ratio and multiply all empirical ration numbers containing 0.50 by '2' to get the final empirical ratio.

This problem:

Empirical Formula:

Using the % per 100wt values in part 'a' ...

              %     =>         grams                 =>                 moles

%C => 85.71% => 85.71 g* / 100 g Cpd => (85.71 / 12) = 7.14 mol C

%H => 14.29% => 14.29 g / 100 g Cpd => (14.29 / 1) = 14.29 mol H

=> Set up mole Ratio and Reduce to Empirical Ratio:

mole ratio C:H =>  7.14 : 14.29

<u>To reduce mole values to the smallest whole number ratio,  divide all mole values by the smaller mole value of the set.</u>

=> 7.14/7.14 : 14.29/7.14 => Empirical Ration=> 1 : 2

∴ Empirical Formula => CH₂

Molecular Formula:

(Empirical Formula Wt)·N = Molecular Wt => N = Molecular Wt / Empirical Wt

N = 42 / 14 = 3 => multiply subscripts of empirical formula by '3'.

Therefore, the molecular formula is C₃H₆

3 0
3 years ago
Valance shell example..​
Natali5045456 [20]

Answer:

<h2>Oxygen has six valence electrons, two in the 2s subshell and four in the 2p subshell.</h2>

<h3>Valence electrons are the electrons in the outermost shell, or energy level, of an atom. </h3>

<h3>Configuration of oxygen's valence electrons as 2s²2p⁴.</h3>

Explanation:

#Let's Study

#I Hope It's Help

#Keep On Learning

#Carry On Learning

\\ \green{MaggieEve}

3 0
2 years ago
If an object has more positive charges than negative charges, the object has an overall?
djyliett [7]
The object has an overall positive charge.
7 0
3 years ago
Suppose you have a spherical balloon filled with air at room temperature and 1.0 atm pressure; its radius is 17 cm. You take the
Sladkaya [172]
<span>Answer: 17.8 cm
</span>

<span>Explanation:
</span>

<span>1) Since temperature is constant, you use Boyle's law:
</span>

<span>PV = constant => P₁V₁ = P₂V₂


</span><span>=> V₁/V₂ = P₂/P₁</span>
<span>
2) Since the ballon is spherical:


</span><span>V = (4/3)π(r)³</span>
<span>
Therefore, V₁/V₂ = (r₁)³ / (r₂)³
</span>

<span>3) Replacing in the equation V₁/V₂ = P₂/P₁:


</span><span><span>(r₁)³ / (r₂)³ </span>= P₂/P₁</span>
<span>
And you can solve for r₂: (r₂)³ = (P₁/P₂) x (r₁)³


</span>(r₂)³ = (1.0 atm / 0.87 atm) x (17 cm)³ = 5,647.13 cm³
<span>
r₂ = 17.8 cm</span>

4 0
3 years ago
Read 2 more answers
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