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2 years ago

An atom has 11 protons, 10 electrons and 13 neutrons. What is the charge for this atom?

Chemistry

1 answer:

Aliun [14]2 years ago

3 0

Answer:

<h3>the charge is +1 </h3>

Explanation:

<h3>as we know nutral atom have equal number of protons and electrons</h3><h3>from the give this element have 11 protons so if it is nutral it must have 11 electrons,but in the question this atom is charged this means it gains or losts certain amount of electrons , this atom has 11 proton and 10 electron from this we can understand this atom dicreases by 1 from its proton, this means it losts one electron .</h3><h3>when an atom lostes electron it's charge become positive with the number of electrons it lostes .</h3><h3>this atom lost 1 electron there fore it have +1 charge and become ion called cathion</h3>

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Lithium, is used in dry cells and storage batteries and in high temperature lubricants, it has two naturally occurring isotopes,

erastovalidia [21]

<u>Answer:</u> The average atomic mass of lithium is 6.9241 u.

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ ....(1)

<u>For $_3^6\textrm{Li}$ isotope:</u>

Mass of $_3^6\textrm{Li}$ isotope = 6 u

Percentage abundance of $_3^6\textrm{Li}$ isotope = 7.59 %

Fractional abundance of $_3^6\textrm{Li}$ isotope = 0.0759

<u>For $_3^7\textrm{Li}$ isotope:</u>

Mass of $_3^7\textrm{Li}$ isotope = 7 u

Percentage abundance of $_3^7\textrm{Li}$ isotope = 92.41%

Fractional abundance of $_3^7\textrm{Li}$ isotope = 0.9241

Putting values in equation 1, we get:

$\text{Average atomic mass of Lithium}=[(6\times 0.0759)+(7\times 0.9241)]$

$\text{Average atomic mass of Lithium}=6.9241u$

Hence, the average atomic mass of lithium is 6.9241 u.

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2 years ago

An experiment requires that each student use an 8.5 cm length of magnesium ribbon. How many students can do the experiment if th

melisa1 [442]

570/8.5=67.0 58... you only have to take the natural part, si the answer is 67 students

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2 years ago

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What is the potential energy of the reactants from the following energy diagram?

slavikrds [6]

The potential energy of the reactants is 200J.
From the energy diagram, the energy of the product formed is 350J; this means that, this reaction is an endothermic reaction, because it absorbs energy from its environment.<span />

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2 years ago

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If a sample of sodium chloride with a mass of

alex41 [277]

Original molarity was 1.7 moles of NaCl

Final molarity was 0.36 moles of NaCl

Given Information:

Original (concentrated) solution: 25 g NaCl in a 250 mL solution, solve for molarity

Final (diluted) solution: More water is added to make the new total volume 1.2 liters, solve for the new molarity

1. Solve for the molarity of the original (concentrated) solution.

Molarity (M) = moles of solute (mol) / liters of solution (L)

Convert the given information to the appropriate units before plugging in and solving for molarity.

Molarity (M) = 0.43 mol NaCl solute / 0.250 L solution = 1.7 M NaCl (original solution)

2. Solve for the molarity of the final (diluted) solution.

Remember that the amount of solute remains constant in a dilution problem; it is just the total volume of the solution that changes due to the addition of solvent.

Molarity (M) = 0.43 mol NaCl solute / 1.2 L solution

Molarity (M) of the final solution = 0.36 M NaCl

I hope this helped:))

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2 years ago

what volume of a 0.149 m potassium hydroxide solution is required to neutralize 17.0 ml of a 0.112 m hydrobromic acid solution?

IgorLugansk [536]

Answer: 12.78ml

Explanation:

Given that:

Volume of KOH Vb = ?

Concentration of KOH Cb = 0.149 m

Volume of HBr Va = 17.0 ml

Concentration of HBr Ca = 0.112 m

The equation is as follows

HBr(aq) + KOH(aq) --> KBr(aq) + H2O(l)

and the mole ratio of HBr to KOH is 1:1 (Na, Number of moles of HBr is 1; while Nb, number of moles of KOH is 1)

Then, to get the volume of a 0.149 m potassium hydroxide solution Vb, apply the formula (Ca x Va)/(Cb x Vb) = Na/Nb

(0.112 x 17.0)/(0.149 x Vb) = 1/1

(1.904)/(0.149Vb) = 1/1

cross multiply

1.904 x 1 = 0.149Vb x 1

1.904 = 0.149Vb

divide both sides by 0.149

1.904/0.149 = 0.149Vb/0.149

12.78ml = Vb

Thus, 12.78 ml of potassium hydroxide solution is required.

5 0

2 years ago