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2 years ago

The number of particles of 0.2 mole

Chemistry

1 answer:

strojnjashka [21]2 years ago

3 0

Answer:

There are 1.205x10²³ particles in 0.2 moles.

Explanation:

In order to <u>convert from moles to number of particles</u> we need to use <em>Avogadro's number</em>, which states the number of particles contained in 1 mol:

0.2 mol * 6.023x10²³ particles /mol = 1.205x10²³ particles

Thus, there are 1.205x10²³ particles in 0.2 moles.

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Which statement supports the claim that the continental crust was once joined together as one supercontinent, Pangaea?

Liula [17]

Answer:

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2 years ago

Clear selection

marin [14]

The molarity of the acid given the data from the question is 0.30 M

<h3>Balanced equation </h3>

2HNO₃ + Ba(OH)₂ —> Ba(NO₃)₂ + 2H₂O

From the balanced equation above,

The mole ratio of the acid, HNO₃ (nA) = 2
The mole ratio of the base, Ba(NO₃)₂ (nB) = 1

<h3>How to determine the molarity of the acid</h3>

From the question given above, the following data were obtained:

Volume of acid, HNO₃ (Va) = 39.7 mL
Volume of base, Ba(NO₃)₂ (Vb) = 24 mL
Molarity of base, Ba(NO₃)₂ (Cb) = 0.250 M
Molarity of acid, HNO₃ (Ma) =?

MaVa / MbVb = nA / nB

(Ma × 39.7) / (0.25 × 24) = 2

(Ma × 39.7) / 6 = 2

Cross multiply

Ma × 39.7 = 6 × 2

Ma × 39.7 = 12

Divide both side by 39.7

Ma = 12 / 39.7

Ma = 0.30 M

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1 year ago

What does anhydrous mean in your own words?

dalvyx [7]

Answer: no water

Explanation:

4 0

2 years ago

Read 2 more answers

A hiker has packed a bag of chips as a snack. At the start of the hike the pressure was 1.5 atm and the temperature was 35*C. At

Mazyrski [523]

According to <span>Gay-Lussac's Law the temperature and Pressure are directly proportional to each other if the amount and volume of given gas are kept constant.
Mathematically for initial and final states it is expressed as,

P</span>₁ / T₁ = P₂ / T₂ ----- (1)
Data Given;
P₁ = 1.5 atm

T₁ = 35 °C + 273 = 308 K

P₂ = ?

T₂ = 0 °C + 273 = 273 K

Solving Eq. 1 for P₂,

P₂ = P₁ T₂ / T₁

Putting values,
P₂ = (1.5 atm × 273 K) ÷ 308 K

P₂ = 1.32 atm
Result:
As the temperature is decreased so the pressure also decreases from 1.5 atm to 1.32 atm. Therefore the bag will contract.

5 0

3 years ago

25.0cm3 of s saturated potassium hydroxide is neutralized by 35.0cm3 of hydrogen chloride acid of concentration 0.75 mol/dm3. Ca

Kamila [148]

Answer:

Concentration of the original $\rm KOH$ solution: approximately $1.05\; \rm mol \cdot dm^{-3}$ .

Explanation:

Notice that the concentration of the $\rm HCl$ solution is in the unit $\rm mol\cdot dm^{-3}$ . However, the unit of the two volumes is $\rm cm^{3}$ . Convert the unit of the two volumes to $\rm dm^{3}$ to match the unit of concentration.

$\begin{aligned} V(\mathrm{NaOH}) &= 25.0\; \rm cm^{3} \\ &= 25.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0250\; \rm dm^{3} \end{aligned}$ .

$\begin{aligned} V(\mathrm{HCl}) &= 35.0\; \rm cm^{3} \\ &= 35.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0350\; \rm dm^{3} \end{aligned}$ .

Calculate the number of moles of $\rm HCl$ molecules in that $0.0350\; \rm dm^{3}$ of $0.75\; \rm mol \cdot dm^{3}$ $\rm HCl\!$ solution:

$\begin{aligned}n(\mathrm{HCl}) &= c(\mathrm{HCl}) \cdot V(\mathrm{HCl}) \\ &= 0.00350\; \rm dm^{3} \times 0.75\; \rm mol \cdot dm^{3} \\ &\approx 0.02625\; \rm mol\end{aligned}$ .

$\rm HCl$ is a monoprotic acid. In other words, each $\rm HCl\!$ would release up to one proton $\rm H^{+}$ .

On the other hand, $\rm KOH$ is a monoprotic base. Each $\rm KOH\!$ formula unit would react with up to one $\rm H^{+}$ .

Hence, $\rm HCl$ molecules and $\rm KOH\!$ formula units would react at a one-to-one ratio.

${\rm HCl}\, (aq) + {\rm NaOH}\, (aq) \to {\rm NaCl}\, (aq) + {\rm H_2O}\, (l)$ .

Therefore, that $0.02625\; \rm mol$ of $\rm HCl$ molecules would neutralize exactly the same number of $\rm NaOH$ formula units. That is: $n(\mathrm{NaOH}) = 0.02625\; \rm mol$ .

Calculate the concentration of a $\rm NaOH$ solution where $V(\mathrm{NaOH}) = 0.0250\; \rm dm^{3}$ and $n(\mathrm{NaOH}) = 0.02625\; \rm mol$ :

$\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} \\ &= \frac{0.02625\; \rm mol}{0.0250\; \rm dm^{3}}\approx 1.05\; \rm mol \cdot dm^{-3}\end{aligned}$ .

7 0

2 years ago

The number of particles of 0.2 mole​

The number of particles of 0.2 mole