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2 years ago

Describe what the isotopes of an element have in common and how are they different.

Chemistry

2 answers:

Neporo4naja [7]2 years ago

8 0

Answer:

Isotopes of an element are atoms of the same element that have different number of neutrons.

babunello [35]2 years ago

8 0

Yea what she said^^^^^

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What kind of change is sodium hydroxide dissolving in water

Korolek [52]

Physical change because water is not a gas or change . you can see through it.

6 0

3 years ago

How many atoms are in Cu2MgHO

Illusion [34]

To find how many atoms are in the compound, we first have to find how many moles there are. The numbers in the subscript tell you how many moles of each element are present in the compound. Therefore, to find the number of total moles, we can add up each of the numbers in the subscript:

Cu₂Mg₁H₁O₁ *when there is no number under an element, it is implied that there is 1 mole of the element
2+1+1+1=5 moles

We can then use Avogadro's number and dimensional analysis to find how many atoms are in the compound (Avogadro's number is 6.02x10^23):

5 moles (6.02x10^23/1 mole)= 3.01x10^24 atoms

4 0

3 years ago

Lithium has the electron arrangement 2,1. When lithium forms an ion,

iragen [17]

Answer:

when lithium forms an ion it donates 1 electron

which makes its electron configuration to become 2

4 0

2 years ago

23.495 g sample of aqueous waste leaving a fertilizer manufacturer contains ammonia. The sample is diluted with 72.311 g of wate

Otrada [13]

Answer:

1.86% NH₃

Explanation:

The reaction that takes place is:

HCl(aq) + NH₃(aq) → NH₄Cl(aq)

We calculate the moles of HCl that reacted, using the volume used and the concentration:

32.27 mL ⇒ 32.27/1000 = 0.03227 L

0.1080 M * 0.03227 L = 3.4852x10⁻³ mol HCl

The moles of HCl are equal to the moles of NH₃, so now we calculate the mass of NH₃ that was titrated, using its molecular weight:

3.4852x10⁻³ mol NH₃ * 17 g/mol = 0.0592 g NH₃

The weight percent NH₃ in the aliquot (and thus in the diluted sample) is:

0.0592 / 12.949 * 100% = 0.4575%

Now we calculate the total mass of NH₃ in the diluted sample:

Diluted sample total mass = Aqueous waste Mass + Water mass = 23.495 + 72.311 = 95.806 g

0.4575% * 95.806 g = 0.4383 g NH₃

Finally we calculate the weight percent NH₃ in the original sample of aqueous waste:

0.4383 g NH₃ / 23.495 g * 100% = 1.86% NH₃

6 0

2 years ago

The voltage generated by the zinc concentration cell described by the line notation Zn ( s ) ∣ ∣ Zn 2 + ( aq , 0.100 M ) ∥ ∥ Zn

Oxana [17]

Answer:

0.193 M

Explanation:

We need to calculate the Zn²⁺ concentration at the cathode where reduction occurs which is the right side in the expression:

Zn2+(aq,0.100 M) ‖ Zn2+(aq,? M) | Zn(s)

Zn²⁺ (aq,?) + 2 e⁻ ⇒ Zn (s)

and oxidation will occur in the anode

Zn (s) ⇒ Zn²⁺ (aq, 0.100 M ) + 2 e⁻

and the overall reaction is

Zn²⁺ (aq,?) ⇒ Zn²⁺ (0.100 M )

The driving force is the difference in concentration and E the electromotive force will be given by

E = Eº - 0.0592/2 log [Zn²⁺ (0.100 M) / Zn²⁺ (M) ]

Plugging the value for E and knowing Eº is cero because we have the same electrodes, we have

17.0 x 10⁻³ = 0 - 0.0592 log 0.100/ X =

- 17.0 x 10⁻³ / 0.0592 = log 0.100 / X

- 0.287 = log (0.100 / X)

Taking inverse log to both sides of the equation

0.516 = 0.100 / X ⇒ X = 0.100 / 0.516 = 0.193 M

4 0

3 years ago