All categories

3 years ago

How many atom of cd are in 3.00g of cadmium

2 answers:

8 0

The molar mass of Cadmium is 112.41 grams, so first you need to convert your given value to moles. To do that divide 3 grams by the molar mass of 112.41 grams. That gives you an answer of .0267 moles of cadmium. Then to convert that value to atoms, you must use the constant of 6.022X10^23 atoms in one mole. Multiply the moles by that number, which will give you a final answer of (1.61X10^22) atoms in 3 grams of cadmium.

Hope this helps!

Crazy boy [7]3 years ago

5 0

I wish i was smart enough

You might be interested in

BARSIC [14]

The number of C atoms in 0.524 moles of C is 3.15 atoms.

The number of $SO_2$ molecules in 9.87 moles $SO_2$ is 59.43 molecules.

The moles of Fe in 1.40 x $10^{22}$ atoms of Fe is 0.23 x $10^{-1}$

The moles of $C_2H_6O$ in 2.30x $10^{24}$ molecules of $C_2H_6O$ is 3.81.

<h3>What are moles?</h3>

A mole is defined as 6.02214076 × $10^{23}$ of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.

A. The number of C atoms in 0.524 mole of C:

6.02214076 × $10^{23}$ x 0.524 mole

3.155601758 atoms =3.155 atoms

B. The number of $SO_2$ molecules in 9.87 moles of $SO_2$ :

6.02214076 × $10^{23}$ x 9.87

59.4385293 molecules= 59.43 molecules

C. The moles of Fe in 1.40 x $10^{22}$ atoms of Fe:

1.40 x $10^{22}$ ÷ 6.02214076 × $10^{23}$

0.2324754694 x $10^{-1}$ moles.

0.23 x $10^{-1}$ moles.

D. The moles of $C_2H_6O$ in 2.30x $10^{24}$ molecules of $C_2H_6O$ :

2.30x $10^{24}$ ÷ 6.02214076 × $10^{23}$

3.819239854 moles=3.81 moles

Learn more about moles here:

brainly.com/question/8455949

#SPJ1

6 0

2 years ago

How many atoms of hydrogen are present in 7.63 g of ammonia? please help me ??

GarryVolchara [31]

The atoms of hydrogen that are present in 7.63 g of ammonia(NH3)

find the moles of NH3 =mass/molar mass
7.63 g/ 17 g/mol = 0.449 moles

since there is 3 atoms of H in NH3 the moles of H = 0.449 x 3 = 1.347 moles

by use of 1 mole = 6.02 x10^23 atoms
what about 1.347 moles

= 1.347 moles/1 moles x 6.02 x10^23 atoms = 8.11 x10^23 atoms of Hydrogen

3 0

3 years ago

Butane, C4H10, reacts with oxygen, O2, to form water, H2O, and carbon dioxide, CO2, as shown in the following chemical equation:

Wittaler [7]

The balanced chemical reaction is:

2C4H10(g)+13O2(g)->10H2O(g)+8CO2(g)

Calculate the mass of water produced when 1.77 grams of butane reacts with excessive oxygen?
1.77 g C4H10 (1 mol C4H10/58.14 g C4H10) (10 mol H2O / 2 mol C4H10) ( 1.01 g H2O / 1 mol H2O ) = 0.15 g H2O

Calculate the mass of butane needed to produce 71.6 of carbon dioxide.
71.6 g CO2 (1 mol CO2/ 44.01 g CO2) ( 2 mol C4H10 / 8 mol CO2 ) (58.14 g C4H10 / 1 mol C4H10 ) = 23.65 g C4H10

4 0

3 years ago

Read 2 more answers

Calculate the molality of a solution that contains 51.2 g of naphthalene, C10H8, in 500 mL of carbon tetrachloride. The density

PtichkaEL [24]

Answer: The molality of naphthalene solution is 0.499 m

Explanation:

Density is defined as the ratio of mass and volume of a substance.

$\text{Density}=\frac{\text{Mass}}{\text{Volume}}$ ......(1)

Given values:

Volume of carbon tetrachloride = 500 mL

Density of carbon tetrachloride = 1.60 g/mL

Putting values in equation 1, we get:

$\text{Mass of carbon tetrachloride}=(1.60g/mL\times 500mL)=800g$

Molality is defined as the amount of solute expressed in the number of moles present per kilogram of solvent. The units of molarity are mol/kg. The formula used to calculate molarity:

$\text{Molality of solution}=\frac{\text{Given mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent (in g)}}$ .....(2)

Given values:

Given mass of naphthalene = 51.2 g

Molar mass of naphthalene = 128.17 g/mol

Mass of solvent = 800 g

Putting values in equation 2, we get:

$\text{Molality of naphthalene}=\frac{51.2\times 1000}{128.17\times 800}\\\\\text{Molality of naphthalene}=0.499m$

Hence, the molality of naphthalene solution is 0.499 m

4 0

3 years ago

PLEASE HELP ME ASAPPPP

sattari [20]

Boyle’s law gives the relationship between pressure and volume of gases. It states that at constant temperature the pressure of gas is inversely proportional to volume of gas.
PV = k
Where P is pressure V is volume and k is constant
P1V1 = P2V2
Parameters at STP are on the left side and parameters for the second instance are on the right side of the equation
P1 - standard pressure - 1.0 atm
Substituting the values in the equation
1.0 atm x 5.00 L = P x 15.0 L
P = 0.33 atm
New pressure is 0.33 atm

5 0

3 years ago