Can someone do this for me i wasent there when we learned it

How many particles are in 23 g of H 2 O?

Sedaia [141]

1 mole of any substance contains 6.022 × 1023 particles.

⚛ 6.022 × 1023 is known as the Avogadro Number or Avogadro Constant and is given the symbol NA

N = n × NA

· N = number of particles in the substance

· n = amount of substance in moles (mol)

· NA = Avogardro Number = 6.022 × 10^23 particles mol-1

For H2O we have:

2 H at 1.0 each = 2.0 amu
1 O at 16.0 each = 16.0 amu
Total for H2O = 18.0 amu, or grams/mole

It takes 18 grams of H2O to obtain 1 mole, or 6.02 x 1023 molecules of water. Think about that before we answer the question. We have 25.0 grams of water, so we have more than one mole of water molecules. To find the exact number, divide the available mass (25.0g) by the molar mass (18.0g/mole). Watch how the units work out. The grams cancel and moles moves to the top, leaving moles of water. [g/(g/mole) = moles].

Here we have 25.0 g/(18.0g/mole) = 1.39 moles water (3 sig figs).

Multiply 1.39 moles times the definition of a mole to arrive at the actual number of water molecules:

1.39 (moles water) * 6.02 x 1023 molecules water/(mole water) = 8.36 x 1023 molecules water.

That's slightly above Avogadro's number, which is what we expected. Keeping the units in the calculations is annoying, I know, but it helps guide the operations and if you wind up with the unit desired, there is a good chance you've done the problem correctly.

N = n × (6.022 × 10^23)

1 grams H2O is equal to 0.055508435061792 mol.

Then 23 g of H2O is 1.2767 mol

To calculate the number of particles, N, in a substance:

N = n × NA

N = 1.2767 × (6.022 × 10^23)

N= 176.26

N=

3 0

2 years ago

a 2.7 L of N2 is collected at 121kpa and 288 K . if the pressure increases to 202 kpa and the temperature rises to 303 K , what

jok3333 [9.3K]

Answer:

The gas will occupy a volume of 1.702 liters.

Explanation:

Let suppose that the gas behaves ideally. The equation of state for ideal gas is:

$P\cdot V = n\cdot R_{u}\cdot T$ (1)

Where:

$P$ - Pressure, measured in kilopascals.

$V$ - Volume, measured in liters.

$n$ - Molar quantity, measured in moles.

$T$ - Temperature, measured in Kelvin.

$R_{u}$ - Ideal gas constant, measured in kilopascal-liters per mole-Kelvin.

We can simplify the equation by constructing the following relationship:

$\frac{P_{1}\cdot V_{1}}{T_{1}} = \frac{P_{2}\cdot V_{2}}{T_{2}}$ (2)

Where:

$P_{1}$ , $P_{2}$ - Initial and final pressure, measured in kilopascals.

$V_{1}$ , $V_{2}$ - Initial and final volume, measured in liters.

$T_{1}$ , $T_{2}$ - Initial and final temperature, measured in Kelvin.

If we know that $P_{1} = 121\,kPa$ , $P_{2} = 202\,kPa$ , $V_{1} = 2.7\,L$ , $T_{1} = 288\,K$ and $T_{2} = 303\,K$ , the final volume of the gas is:

$V_{2} = \left(\frac{T_{2}}{T_{1}} \right)\cdot \left(\frac{P_{1}}{P_{2}} \right)\cdot V_{1}$

$V_{2} = 1.702\,L$

The gas will occupy a volume of 1.702 liters.

6 0

3 years ago

The more the energy, the larger the (a,b,c,d)?

Elanso [62]

Answer: D:wavelenght

Explanation: Students will understand that shorter wavelengths have higher frequency and energy.

3 0

3 years ago

Read 2 more answers

Calculate the ph of the solution made by adding 0.50 mol of hobr and 0.30 mol of kobr to 1.00 l of water. the value of ka for ho

nikdorinn [45]

To fully understand the problem, we use the ICE table to identify the concentration of the species. We calculate as follows:

Ka = 2.0 x 10^-9 = [H+][OBr-] / [HOBr]

HOBr = 0.50 M
KOBr = 0.30 M = OBr-

<span> HOBr + H2O <-> H+ + OBr- </span>
<span>I 0.50 - 0 0.30 </span>
<span>C -x x x
</span>---------------------------------------------
<span>E(0.50-x) x (0.30+x) </span>

<span>Assuming that the value of x is small as compared to 0.30 and 0.50 </span>

<span>Ka = 2.0 x 10^-9 = x (0.30) / 0.50) </span>

<span>x = 3.33 x 10^-9 = H+</span>
pH = 8.48

3 0

3 years ago

Aluminum and oxygen react according to the following equation: 4Al + 3O2 -> 2Al2O3 In a certain experiment, 4.6g Al was react

stiv31 [10]

Answer:

Percent yield: 78.2%

Explanation:

Based on the reaction:

4Al + 3O₂ → 2Al₂O₃

<em>4 moles of Al produce 2 moles of Al₂O₃</em>

<em />

To find percent yield we need to find theoretical yield (Assuming a yield of 100%) and using:

(Actual yield (6.8g) / Theoretical yield) × 100

Moles of 4.6g of Al (Molar mass: 26.98g/mol) are:

4.6g Al × (1mol / 26.98g) = 0.1705 moles of Al.

As 4 moles of Al produce 2 moles of Al₂O₃, theoretical moles of Al₂O₃ obtained from 0.1705 moles of Al are:

0.17505 moles Al × (2 moles Al₂O₃ / 4 moles Al) = <em>0.0852 moles of Al₂O₃</em>,

In grams (Molar mass Al₂O₃ = 101.96g/mol):

0.0852 moles of Al₂O₃ × (101.96g / mol) =

<h3>8.7g of Al₂O₃ can be produced (Theoretical yield)</h3>

Thus, Percent yield is:

(6.8g / 8.7g) × 100 =

8 0

3 years ago