All categories

3 years ago

How many atoms (mol) is in 10g of sodium, sodium chloride and Iron chloride?

Chemistry

1 answer:

SashulF [63]3 years ago

8 0

Answer:

See explanation

Explanation:

Molar mass of NaCl = 58.5 g

Number of moles contained in 10 g of NaCl = 10 g/58.5 g = 0.17 moles

If 1 mole of NaCl contains 6.02 * 10^23 atoms

0.17 moles of NaCl contains 0.17 * 6.02 * 10^23 atoms = 1.02 * 10^23 atoms

Molar mass of Fe II chloride = 126.751 g/mol

Number of moles = 10 g/126.751 g/mol = 0.0789 moles

Number of atoms = 0.0789 moles * 6.02 * 10^23 atoms = 4.7 * 10^22 atoms

Molar mass of Na = 23 g/mol

Number of moles = 10g/23 g/mol = 0.43 moles

Number of atoms = 0.43 moles * 6.02 * 10^23 atoms = 2.59 * 10^ 23 atoms

You might be interested in

Any element can always be identified by its:

Alborosie

Explanation:

atomic mass

Really...i think it can also be atomic number..

3 0

4 years ago

Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 with aqueous NaI . Include phases. chemical equation: Wha

Citrus2011 [14]

Answer: The mass of lead iodide produced is 9.22 grams

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of NaI = 0.200 M

Volume of solution = 0.200 L

Putting values in above equation, we get:

$0.200M=\frac{\text{Moles of NaI}}{0.200}\\\\\text{Moles of NaI}=(0.200mol/L\times 0.200L)=0.04moles$

The chemical equation for the reaction of NaI and lead chlorate follows:

$Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)$

By Stoichiometry of the reaction:

2 moles of NaI reacts produces 1 mole of lead iodide

So, 0.04 moles of NaI will react with = $\frac{1}{2}\times 0.04=0.02mol$ of lead iodide

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of lead iodide = 461 g/mol

Moles of lead iodide= 0.02 moles

Putting values in above equation, we get:

$0.02mol=\frac{\text{Mass of lead iodide}}{461g/mol}\\\\\text{Mass of lead iodide}=(0.02mol\times 461g/mol)=9.22g$

Hence, the mass of lead iodide produced is 9.22 grams

6 0

3 years ago

Will the ph of a solution of nh4cn be >7, <7, or =7?

never [62]

PH of a solution will be higher than 7

Ammonium cyanide is a salt formed by hydrogen cyanide and ammonia. Ammonia is a weak base and hydrogen cyanide is a weak acid.
NH₄CN + H₂O ⇒ NH₃ + HCN

NH₄⁺ + H₂O -----> H₃O⁺ + NH₃

CN⁻ + H₂O -----> HCN + OH⁻

Although both compounds are weak electrolytes, NH₃ is somewhat stronger base than HCN is a strong acid, so the solution reacts alkaline. We can prove this using Ka and Kb values:

Ka(HCN) = 4.9 x × 10⁻¹⁰

Kb(NH₃) = 1.8 × 10⁻⁵
Kw= 1.0 × 10⁻¹⁴

Let's first calculate Ka for NH₄⁺:
Ka(NH₄⁺) x Kb(NH₃) = pKw

Ka(NH₄⁺) = Kw/Kb(NH₃) = 5.6 x 10⁻¹⁰

Then, Kb for CN⁻:

Kb(CN⁻) x Ka(HCN) = pKw

Kb(CN⁻) = Kw/Ka(HCN) = 2 x 10⁻⁵

From this, we can see that the acid constant NH4⁺ is much lower than the base constant of CN⁻, which will say that the solution of NH₄CN will react slightly alkaline because of the higher presence of hydroxyl ions in solution.

7 0

4 years ago

If 8.50 g of phosphorus reacts with hydrogen gas at 2.00 atm in a 10.0-L container at 298 K, calculate the moles of PH3 produced

ahrayia [7]

Answer:

The moles of PH₃ produced are 0.2742 and the total number of moles of gas present at the end of the reaction is 0.6809.

Explanation:

Phosphorus reacts with H₂ according to the balanced equation:

P₄ (s) + 6 H₂ (g) ⇒ 4 PH₃ (g)

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

P₄: 1 mole
H₂: 6 moles
PH₃:4 moles

Being the molar mass of the compounds:

P₄: 124 g/mole
H₂: 2 g/mole
PH₃: 34 g/mole

The following mass amounts of each compound participate in the reaction:

P₄: 1 mole* 124 g/mole= 124 g
H₂: 6 mole* 2 g/mole= 12 g
PH₃: 4 moles* 34 g/mole= 136 g

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

In this case you know:

P= 2 atm
V= 10 L
n= ?
R= 0.082 $\frac{atm*L}{mol*K}$
T= 298 K

Replacing:

2 atm*10 L= n*0.082 $\frac{atm*L}{mol*K}$ *298 K

and solving you get:

$n=\frac{2 atm*10 L}{0.082\frac{atm*L}{mol*K}*298 K }$

n=0.818 moles

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, you can use a simple rule of three as follows: if 6 moles of H₂ react with 124 g of P₄, 0.818 moles of H₂ with how much mass of P₄ will it react?

$mass of P_{4}=\frac{0.818 moles of H_{2}*124 grams of P_{4}}{6 moles of H_{2}}$

mass of P₄= 16.90 grams

But 16.90 grams of P₄ are not available, 8.50 grams are available. Since you have less mass than you need to react with 0.818 moles of H₂, phosphorus P₄ will be the limiting reagent.

Then you can apply the following rules of three:

If 124 grams of P₄ produce 4 moles of PH₃, 8.50 grams of P₄, how many moles do they produce?

$moles of PH_{3} =\frac{8.5 grams of P_{4}*4 moles of PH_{3} }{124grams of P_{4}}$

moles of PH₃=0.2742

If 124 grams of P₄ react with 6 moles of H₂, 8.50 grams of P₄ with how many moles of H₂ do they react?

$moles of H_{2} =\frac{8.5 grams of P_{4}*6 moles of H_{2} }{124grams of P_{4}}$

moles of H₂= 0.4113

If you have 0.818 moles of H₂, the number of moles of gas H₂ present at the end of the reaction is calculated as:

0.818 - 0.4113= 0.4067

Then the total number of moles of gas present at the end of the reaction will be the sum of the moles of PH₃ gas and H₂ gas that did not react:

0.2742 + 0.4067= 0.6809

Finally, the moles of PH₃ produced are 0.2742 and the total number of moles of gas present at the end of the reaction is 0.6809.

5 0

3 years ago

In ionic compounds, ________ lose their valence electrons to form positively charged ________. nonmetals, anions metals, polyato

Keith_Richards [23]

Answer:
In ionic compounds, Metals lose their valence electrons to form positively charged Cations.

Explanation:
Metals have the ability to loose elctrons readily. For example metals of Group IA and Group IIA readily looses electrons in order to obtain Noble Gas Configuration. On the other hand Non-metals tends to gain electrons and acquire negative charge. While Ions are made when an an element gain or loose electrons. After loosing electrons element get positive charge which is called as Cation while on gaining electron it gets negative charge called as Anion.

6 0

3 years ago

Read 2 more answers