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Alekssandra [29.7K]
3 years ago
10

5. What is the charge of an atom that has lost four electrons? *

Chemistry
1 answer:
Mashcka [7]3 years ago
7 0

Answer:

+4

Explanation:

An atom is considered to be neutral before losing or gaining any electrons. They are neutral in charge because they initially have the same number of protons (with charge of +1 each) and electrons (with a charge of -1 each).

Consider charge in an electron like a balance, with positively charged protons on one side and negatively charged electrons on the other. When you have the same number of each, there is equal amounts of positive and negative charge so the atom is neutral with a charge of zero.

For example, potassium has 19 protons, and 19 electrons. This means it has

19 positive charges and 19 negative charges. If you take away one electron, it now has 19 positive charges and 18 negative charges. So there is one more positive charge than negative, which means it is now an ion (a charged atom) with a charge of +1. If you take away two electrons the charge becomes +2... 4 electrons the charge becomes +4.

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What is the total distance covered if you walk 5 meters to the north 5 meter east 5 meters south
mihalych1998 [28]

0! because you walked back in forth in diferent direcions

8 0
4 years ago
Given these reactions, where X represents a generic metal or metalloid 1) H2(g)+12O2(g)⟶H2O(g)ΔH1=−241.8 kJ 1) H2(g)+12O2(g)⟶H2O
Bond [772]

Answer:

ΔH = -793,6 kJ

Explanation:

It is possible to obtain ΔH of this reaction using Hess's law that says you can sum the half-reactions ΔH to obtain the ΔH of the global reaction:

If half-reactions are:

1) H₂(g) + ¹/₂O₂(g) ⟶ H₂O(g) ΔH₁ = −241.8 kJ

2) X(s) + 2Cl₂(g) ⟶ XCl₄(s) ΔH₂ = +356.9 kJ  

3) ¹/₂H₂(g) + ¹/₂Cl₂(g) ⟶ HCl(g) ΔH₃ = −92.3 kJ

4) X(s) + O₂(g) ⟶ XO₂(s) ΔH₄ = −639.1 kJ

5) H₂O(g) ⟶ H₂O(l) ΔH₅ = −44.0 kJ

The sum of (4) + 4×(3) - (2) - 2×(1) - 2×(5) is:

(4) X(s) + O₂(g) ⟶ XO₂(s) ΔH = −639.1 kJ

+4×(3) 2H₂(g) + 2Cl₂(g) ⟶ 4HCl(g) ΔH = −369,2 kJ

-(2) XCl₄(s) ⟶ X(s) + 2Cl₂(g) ΔH = -356,9 kJ

-2×(1) 2H₂O(g) ⟶ 2H₂(g) + O₂(g) ΔH = +483,6 kJ

-2×(5) 2H₂O(l) ⟶ 2H₂O(g) ΔH = +88.0 kJ

= <em>XCl₄(s) + 2H₂O(l) ⟶ XO₂(s) + 4HCl(g)</em>

Where ΔH is:

ΔH = -639,1 kJ -369,2 kJ -356,9 kJ +483,6 kJ +88,0 kJ

<em>ΔH = -793,6 kJ</em>

I hope it helps!

5 0
3 years ago
you have accidentally broken a test tube and spilled a chemical on the bench. which of the following best describes what you sho
ANEK [815]

Tell the teacher, do NOT clean it up yourself.

4 0
3 years ago
If your end product is 1.5 moles of KMnO 4 how many moles of manganese oxide were used in the reaction? The equation for the pro
vovikov84 [41]

Answer:

1.5 moles

Explanation:

The equation of the reaction is given as:

2 MnO2 + 4 KOH + O2 --> 2KMnO 4 + 2KOH + H2

From the equation,

2 moles of MnO2 produces 2 moles of KMnO4

x moles of MnO2 would produce 1.5 moles of KMnO4

2 = 2

x = 1.5

Solving for x;

x = 1.5 * 2 / 2

x = 1.5 moles

4 0
3 years ago
All of the following are true of bases except
Oksana_A [137]
All of these are correct except the first option, as Arrhenius bases increase the concentration of hydroxide ions.
4 0
3 years ago
Read 2 more answers
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