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Drupady [299]
3 years ago
8

Sodium and chlorine react to form sodium chloride. If there are 1.4 moles of sodium used, how many moles of chlorine were requir

ed?
Chemistry
1 answer:
sammy [17]3 years ago
3 0

Answer:

0.7 moles of Chlorine gas is required.

Explanation:

The reaction is balanced as follows:

2Na + Cl_{2} ------>  2NaCl

According to stoichiometry,

2 mole Sodium reacts with 1 mole Chlorine gas

∴1 mole Sodium reacts with 1/2 moles of Chlorine gas

∴1.4 mole sodium reacts with 1.4/2 moles of Chlorine gas

                                                = 0.7 moles of Chlorine gas

So, 0.7 moles of Chlorine gas is required.

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A cylinder with a moving piston expands from an initial volume of 0.250 L against an external pressure of 2.00 atm. The expansio
kondor19780726 [428]

Answer:

The final volume of the cylinder is 1.67 L

Explanation:

Step 1: Data given

Initial volume = 0.250 L

external pressure = 2.00 atm

Expansion does 288 J of work on the surroundings

Step 2: Definition of reversible work:

Wrev = -P(V2-V1) = -288 J

The gas did work, so V2>V1  (volume expands) and the work has a negative sign.(Wrev<0)

V2 = (-Wrev/P)  + V1

⇒ with Wrev = reverse work (in J)

⇒ with P = the external pressure (in atm)

⇒ with V1 = the initial volume

We can see that your pressure is in  atm  and energy in J

To convert from J to L * atm we should use a convenient conversion unit using the universal gas constants :

R = 8.314472 J/mol *K and R= 0.08206 L*atm/K*mol

V2 =- (-288 J * (0.08206 L*atm/K*mol  /8.314 J/mol *K))/2.00 atm  + 0.250L

V2 = 1.67 L

The final volume of the cylinder is 1.67 L

8 0
3 years ago
If you mix a 25.0mL sample of a 1.20m potassium sulfide solution with 15.0 mL of a 0.900m basium nitrate solution and the reacti
ddd [48]

Answer:

Limiting reagent: barium nitrate

Theoretical yield: 2.29 g BaS

Percent yield: 87%

Explanation:

The corrected balanced reaction equation is:

K₂S + Ba(NO₃)₂ ⇒ 2KNO₃ + BaS

The amount of potassium sulfide added is:

(25.0 mL)(1.20mol/L) = 30 mmol

The amount of barium nitrate added is:

(15.0mL)(0.900mol/L = 13.5 mmol

Since the reactant react in a 1:1 molar ratio, barium nitrate is the limiting reagent. If all of the limiting reagent reacts, the amount of barium sulfide produced is:

(13.5 mmol Ba(NO₃)₂)(BaS/Ba(NO₃)₂ ) = 13.5 mmol BaS

Converting this amount to grams gives the theoretical yield of BaS (molar mass 169.39 g/mol).

(13.5 mmol)(169.39 g/mol)(1g/1000mg) = 2.29 g BaS

The percent yield is calculated as follows:

(actual yield) / (theoretical yield) x 100%

(2.0 g) / (2.29 g) x 100% = 87%

6 0
3 years ago
what is the density of an object that has a mass of 550 grams and displaces 25 ml of water? select all that apply. 22 ml/g .04 m
Black_prince [1.1K]
Density = mass / volume

D = 550 / 25

D = 22 g/mL

hope this helps!
3 0
3 years ago
Read 2 more answers
What element is represented by [Rn]7s1?
lara [203]

Answer:

Radon

Explanation:

4 0
3 years ago
What mass of F2 is needed to produce 180 g of PF3 if the reaction has a 78.1% yield?
nexus9112 [7]

Answer:

91.26 g

Explanation:

Given data:

Mass of PF₃ = 180 g

Mass of F₂ required = ?

Solution:

Chemical equation:

P₄ + 6F₂   → 4PF₃

Moles of PF₃:

Number of moles = mass/ molar mass

Number of moles = 180 g/ 88 g/mol

Number of moles = 2.05 mol

Now we will compare the moles of PF₃ with F₂.

                        PF₃            :           F₂

                          4               :           6

                          2.05         :           6/4×2.05 = 3.075

Mass of  F₂:

Mass of F₂ = moles × molar mass

Mass of F₂ = 3.075 mol × 38 g/mol

Mass of F₂ =  116.85 g

If reaction yield is 78.1%:

116.85 /100 ×78.1 = 91.26 g

6 0
3 years ago
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