All categories

3 years ago

How many moles are in 3.00x 10^22 atoms of He?

Chemistry

1 answer:

sp2606 [1]3 years ago

3 0

Answer:

<h2>0.05 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

$n = \frac{N}{L} \\$

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

$n = \frac{3.0 \times {10}^{22} }{6.02 \times {10}^{23} } \\ = 0.049833...$

We have the final answer as

<h3>0.05 moles</h3>

Hope this helps you

You might be interested in

8.00*10^-5 mol was diluted to 500ml what is the concentration in Mol/L?

OLEGan [10]

Make sure all units are consistent. Convert ml to L first. (500ml = 0.5L). Then divide the mole by the volume which would result to 1.6*10^-4mol/L. This can also be expressed as interms of molarity (M). 1 M = 1 mol/L hence the final answer is 1.6*10^-4 M.

8 0

4 years ago

Read 2 more answers

Can someone help me plz!!! Calculate the mass, in grams, of 2 moles of CaCl2.

Luden [163]

Answer:

222.2 grams of CaCl2 in 2 moles

Explanation:

Mole=given mass/gram mass formula

2moles=x/Gram Mass Formula

(Gram Mass Formula of CaCl2):

Ca=40.1

Cl= 35.5 x 2= 71

-----------------------

GFM=111.1

2 moles = x/111.1 g

x=222.2

7 0

3 years ago

Read 2 more answers

How many ml of a 0.50m solution of hno3 solution are needed to make 500 ml of 0.15m hno3

Anna71 [15]

Answer:

150.0 mL.

Explanation:

It is known that the no. of millimoles of HNO₃ before dilution = the no. of millimoles of HNO₃ after dilution.

∵ (MV) before dilution = (MV) after dilution.

∴ V before dilution = (MV) after dilution / M before dilution = (0.15 M)(500.0 mL)/(0.50 M) = 150.0 mL.

7 0

3 years ago

Which Rocky planet's Biosphere is most different from Earth

MakcuM [25]

Answer:

i guess it a right answer for you ✌

3 0

3 years ago

Perform the following for Part C of this lab:

kaheart [24]

Answer:

a. 0.0110 L

b. 0.0020 L

c. 0.011 mol

d. 5.5 M

e. 0.66 g

f. 33%

Explanation:

There is some info missing. I will use some values to show you the procedure and then you can replace them with your values.

Titrant (NaOH) concentration: 1.0 M

Vinegar volume: 2.0 mL

Initial buret reading (initial NaOH volume): 0.1 mL

Final buret reading (final NaOH volume): 11.1 mL

a. Calculate the volume of NaOH that was added to the vinegar. Convert this volume to liters. Show your work.

The volume of NaOH is the difference between the final and the initial buret reading.

11.1 mL - 0.1 mL = 11.0 mL × (1 L/1000 mL) = 0.0110 L

b. Convert the measured volume of vinegar to liters. Show your work.

2.0 mL × (1 L/1000 mL) = 0.0020 L

c. Calculate the moles of NaOH using the volume and molarity of NaOH. Show your work. moles = molarity x volume

moles = molarity × volume

moles = (1.0 mol/L) × 0.0110 L = 0.011 mol

d. Since the reaction ratio is 1:1, the moles of acetic acid in the vinegar is equal to the moles of NaOH reacted during the titration. Calculate the molarity of the acetic acid in the vinegar. Show your work. molarity = moles / volume

molarity = moles / volume

molarity = 0.011 mol/0.0020 L = 5.5 M

e. Calculate the grams of acetic acid in the vinegar. Show your work. mass = moles x molar mass (g/mol)

mass = moles × molar mass

mass = 0.011 mol × 60.05 g/mol = 0.66 g

f. Assuming that the density of vinegar is very close to 1.0 g/mL, the 2.0 mL sample of vinegar used in the titration should weigh 2.0 g. Use this to calculate the mass % of acetic acid in the vinegar sample. mass % = (mass acetic acid / mass vinegar) * 100%

mass % = (mass acetic acid / mass vinegar) * 100%

mass % = (0.66 g /2.0 g) * 100% = 33%

6 0

3 years ago