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3 years ago

Ce cantitate de O se gaseste in 5,8 g hidroxid de magneziu

Chemistry

1 answer:

coldgirl [10]3 years ago

4 0

Answer:

separare pâlnii de picurare balon cotat. 1. 2. 3. 4 5 6 7. 8. 9. 1 ... Mod de lucru: 25 g Na2SO4∙7H2O se dizolvă în cantitatea minim ... Exemple: NaOH – hidroxid de sodiu.

Explanation:

make me as brain liest

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3. If the dartboard below is used to model an atom, which subatomic particles would be located at Z?

yarga [219]

Answer : The protons and neutrons subatomic particles will be located at Z.

Explanation :

In the model of an atom, there are three subatomic particles. Protons, neutrons and electrons are the subatomic particles.

The protons and the neutrons subatomic particles are located inside the nucleus and the electrons subatomic particle are located around or outside the nucleus.

The protons are positively charged, electrons are negatively charged and neutrons are neutral that means it has no charge.

In the given dartboard, Z is the nucleus in which the protons and neutrons subatomic particles are present and x, w & y are the electrons because they are located around the nucleus.

Hence, the protons and neutrons subatomic particles will be located at Z.

3 0

3 years ago

A student makes the following two claims about chemical reactions:

Svet_ta [14]

I would agree with the second one, not the first. You can't always see the chemical reaction, and it isn't always sudden. But the second claim is true.

7 0

3 years ago

Aqueous solutions of Na2CO3 and Ca(NO3)2, 0.10 M each, are combined. A white precipitate is observed in the container after mixi

anzhelika [568]

The question is incomplete. Here is the complete question.

Aqueous solutions of $Na_{2}CO_{3}$ and $Ca(NO_{3})_{2}$ , 0.10 M each, are combined. A white precipitate is observed in the container after mixing. he precipitate is filtered andcarefully rinsed with distilled water to remove other ions. A sample of the precipitate is added to 100 mL of 0.1 M NaCl. A second sample of the precipitate is then added to 100 mL of 0.1 M HCl. What would be observed in each case?

Observation upon Observation upon

addition of precipitate addition of precipitate

to NaCl(aq) to HCl(aq)

(A) additional precipitates forms no visible reaction occurs

(B) no visible reaction occurs gas is produced and some precipitate dissolves

(D) additional precipitates forms gas is produced and some

precipitate dissolves

Answer: (B) No visible reaction occurs; Gas is produced and some precipitate dissolves

Explanation: When aqueous solutions of $Na_{2}CO_{3}$ and $Ca(NO_{3})_{2}$ are combined, it reacts according to the following balanced equation:

$Na_{2}CO_{3}+Ca(NO_{3})_{2}$ → $CaCO_{3}_{(s)}+2NaNO_{3}_{(aq)}$

forming calcium carbonate ( $CaCO_{3}$ ), which, as it is insoluble in water, precipitates as a solid of the color white. This process is Precipitation and this reaction is a Precipitation Reaction.

When calcium carbonate reacts with NaCl it produces:

$CaCO_{3}+2NaCl$ → $CaCl_{2}+Na_{2}CO_{3}$

Now, calcium chloride is an inorganic compound very soluble in water, so, in this reaction, there are no precipitate and no visible reaction occurs.

When $CaCO_{3}$ reacts with hydrochloridric acid, the balanced reaction is

$CaCO_{3}+2HCl$ → $CaCl_{2}+H_{2}CO_{3}$

which, also produces calcium chloride and carbonic acid.

Both are soluble in water but, when carbonic acid is in an "aqueous state", carbonic acid, it dissociates, forming carbon dioxide and water. Therefore, gas is produced and some precipitate dissolves.

In conclusion, sentence B is the correct alternative.

5 0

3 years ago

A rock contains 0.623 mg of 206Pb for every 1.000 mg of 238U present. Assuming that no lead was originally present, that all the

maxonik [38]

Answer:

t = 3,496x10⁹ years

Explanation:

The decay of ²³⁸U is:

²³⁸U → ²⁰⁶Pb + 8He + 6e⁻

Moles of ²⁰⁶Pb presents in 0,623mg are:

0,623x10⁻³g×(1mol / 206g) = 3,02x10⁻⁶ moles of ²⁰⁶Pb.

These moles are equals to moles of ²³⁸U before decay, that means, 3,02x10⁻⁶ moles²³⁸U

In grams:

3,02x10⁻⁶ moles²³⁸U× (238g / 1mol) = 7,20x10⁻⁴ g ²³⁸U = 0,720 mg²³⁸U

That means initial ²³⁸U was 1,000mg + 0,720mg = 1,720mg

Applying the formula:

ln (N₀/N) t₁₂ = t ln2

Where N₀ is initial amount of uranium (1,720mg), N is concentration of uranium (1,000mg), half-life time is a constant (t₁₂= 4,468x10⁹ years) and t is the time transcurred for the reaction. Replacing:

ln(1,720/1)*4,468x10⁹ years = t ln2

t = 3,496x10⁹ years

I hope it helps!

6 0

3 years ago

Metals react by what electrons?

fomenos

Answer:

valence electrons

Explanation:

5 0

3 years ago