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Aneli [31]
2 years ago
11

When a sodium atom bonds with a chlorine atom , it acquires positive charge true or false?

Chemistry
2 answers:
VikaD [51]2 years ago
5 0

true

hope this helps

heh

have a nice day

ArbitrLikvidat [17]2 years ago
3 0
True.

The sodium atom becomes a positive ion and the chlorine atom becomes a negative charge because the sodium atom loss an electron and the chlorine atom gained an electron.
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Which discovery is considered to have led to the greatest advancements in the field of medicine?
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3 years ago
What is the volume of a sample if the density is 2.35 g/mL and the mass is 33.67 g?
lawyer [7]
  • Density=2.35g/mL
  • Mass=33.67g

\\ \star\sf\longmapsto Density=\dfrac{Mass}{Volume}

\\ \star\sf\longmapsto Volume=\dfrac{Mass}{Density}

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\\ \star\sf\longmapsto Volume=14.32mL

6 0
2 years ago
A large cyclotron directs a beam of He++ nuclei onto a target with a beam current of 0.250 mA. (a) How many He++ nuclei per seco
nikitadnepr [17]

Answer:

a. 7.8*10¹⁴ He⁺⁺ nuclei/s

b. 4000s

c. 7.7*10⁸s

Explanation:

I = 0.250mA = 2.5 * 10⁻³A

Q = 1.0C

1 e- contains 1.60 * 10⁻¹⁹C

But He⁺⁺ Carrie's 2 charge = 2 * 1.60*10⁻¹⁹C = 3.20*10⁻¹⁹C

(A).

No. Of charge per second = current passing through / charge

1 He⁺⁺ = 2.50 * 10⁻⁴ / 3.2*10⁻¹⁹C

1 He⁺⁺ = 7.8 * 10¹⁴ He⁺⁺ nuclei

(B).

I = Q / t

From this equation, we can determine the time it takes to transfer 1.0C

I = 1.0 / 2.5*10⁻⁴ = 4000s

(C).

Time it takes for 1 mol of He⁺⁺ to strike the target =?

Using Avogadro's ratio,

1.0 mole of He = (6.02 * 10²³ ions/mol ) * (1 / 7.81*10¹⁴ He ions)

Note : ions cancel out leaving the value of the answer in mols.

1.0 mol of He = 7.7 * 10⁸s

8 0
3 years ago
Which atom would it be most difficult to remove an electron from
Norma-Jean [14]
I would be difficult to remove an electron from a Noble or Inert Gas (also known as the group 8 or 0 elements).  This is because they all have filled outermost shells and as such the outermost shell would be held tightly to the nucleus and as such make it difficult to remove.  Examples Helium, Neon, Argon, Xenon, Krypton and Radon 
6 0
3 years ago
What is the excess reactant in the combustion of 23 g of methane in the open atmosphere?
Marta_Voda [28]

Answer : The excess reactant in the combustion of methane in opem atmosphere is O_{2} molecule.

Solution : Given,

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The Net balanced chemical reaction for combustion of methane is,

CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)

First we have to calculate the moles of methane.

\text{ Moles of methane}=\frac{\text{ Given mass of methane}}{\text{ Molar mass of methane}} = \frac{23g}{16.04g/mole} = 1.434 moles

From the above chemical reaction, we conclude that

1 mole of methane react with the 2 moles of oxygen

and 1.434 moles of methane react to give \frac{2moles\times 1.434moles}{1moles} moles of oxygen

The Moles of oxygen = 2.868 moles

Now we conclude that the moles of oxygen are more than the moles of methane.

Therefore, the excess reactant in the combustion of methane in open atmosphere is O_{2} molecule.


6 0
3 years ago
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