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Maru [420]
3 years ago
5

A buret is used to dispense standardized NaOH solution. The initial buret reading is 2.73 mL, and after dispensing a known volum

e of NaOH the buret reading is 15.67 mL. If the concentration of the NaOH is 0.125 M, how many moles of NaOH were dispensed
Chemistry
1 answer:
Novosadov [1.4K]3 years ago
3 0

Answer:

1.62x10⁻³ moles of NaOH were dispensed

Explanation:

Molarity is an unit in chemistry defined as the ratio between moles of solute (In the problem, NaOH), per liter of solution.

The concentration of the solution is 0.125moles per liter. That means 1L of solution has 0.125 moles of NaOH.

The volume you dispensed in the buret was:

15.67mL - 2.73mL =

12.94mL of the 0.125M NaOH are:

12.94mL = 0.01294L * (0.125moles / L) =

<h3>1.62x10⁻³ moles of NaOH were dispensed</h3>
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The combustion reaction described in part (b) occurred in a closed room containing 5.56 10g of air
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Answer:

Explanation:

Combustion reaction is given below,

C₂H₅OH(l) + 3O₂(g) ⇒ 2CO₂(g) + 3H₂O(g)

Provided that such a combustion has a normal enthalpy,

ΔH°rxn = -1270 kJ/mol

That would be 1 mol reacting to release of ethanol,

⇒ -1270 kJ of heat

Now,

0.383 Ethanol mol responds to release or unlock,

(c) Determine the final temperature of the air in the room after the combustion.

Given that :

specific heat c = 1.005 J/(g. °C)

m = 5.56 ×10⁴ g

Using the relation:

q = mcΔT

- 486.34 =  5.56 ×10⁴  × 1.005 × ΔT

ΔT= (486.34 × 1000 )/5.56×10⁴  × 1.005

ΔT= 836.88 °C

ΔT= T₂ - T₁

T₂ =  ΔT +  T₁

T₂ = 836.88 °C + 21.7°C

T₂ = 858.58 °C

Therefore, the final temperature of the air in the room after combustion is 858.58 °C

4 0
3 years ago
A tank of gas is found to exert 8.6 atm at 38°C. What would be the required
Vesna [10]

Answer:

36.2 K

Explanation:

Step 1: Given data

  • Initial pressure of the gas (P₁): 8.6 atm
  • Initial temperature of the gas (T₁): 38°C
  • Final pressure of the gas (P₂): 1.0 atm (standard pressure)
  • Final temperature of the gas (T₂): ?

Step 2: Convert T₁ to Kelvin

We will use the following expression.

K = °C +273.15

K = 38 °C +273.15 = 311 K

Step 3: Calculate T₂

We will use Gay Lussac's law.

P₁/T₁ = P₂/T₂

T₂ = P₂ × T₁/P₁

T₂ = 1.0 atm × 311 K/8.6 atm = 36.2 K

6 0
2 years ago
Some one please help me ​
Norma-Jean [14]

Answer:

C.4

Explanation:

6 0
3 years ago
How many grams of glucose are in 4.5x1023 molecules of glucose?:
Leokris [45]

Answer:

.5865 \\  \\  \\  \\  \\  \5865

\\  \\  5865 \inf  \inft \cos( \\  \)

7 0
3 years ago
Data was collected by students in an acid base titration lab. They used 1.63 M Ca(OH)2(AQ)
jenyasd209 [6]

Answer:

3.8 M

Explanation:

Volume of acid used VA= 57.0 - 37.5 = 19.5 ml

Volume of base used VB= 67.8 - 45.0 = 22.8 ml

Equation of the reaction

2HNO3(aq) + Ca(OH)2(aq) --------> Ca(NO3)2(aq) + 2H2O(l)

Number of moles of acid NA= 2

Number of moles of base NB= 1

Concentration of acid CA= ???

Concentration of base CB= 1.63 M

CAVA/CBVB = NA/NB

CAVANB = CBVBNA

CA= CBVBNA/VANB

CA= 1.63 × 22.8 × 2/ 19.5 × 1

CA= 3.8 M

HENCE THE MOLARITY OF THE ACID IS 3.8 M.

6 0
3 years ago
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