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alexgriva [62]
2 years ago
12

Predict the final temperature of 3.50 kg of water in a calorimeter if the water is at 27.5°C before 0.77 oz of noodles containin

g 81.0 kcal are burned
Chemistry
1 answer:
BaLLatris [955]2 years ago
6 0

The final temperature of the water is determined as 50.55 ⁰C.

<h3>Final temperature of the water</h3>

The final temperature of the water is determined from the following calculations;

Q = mcΔθ

Δθ = Q/mc

where;

  • Q is the amount of energy = 81 kcal = 338904 J
  • c is specific heat capacity of water = 4,200 J/kgC

Δθ = 338904 /(3.5 x 4200)

Δθ = 23.05 °C

Final temperature = T₁ + Δθ

Final temperature = 27.5°C + 23.05 °C = 50.55 ⁰C.

Learn more about final temperature here: brainly.com/question/16559442

#SPJ1

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A sample of pure lithium chloride contains 16% lithium by mass. What is the % lithium by mass in a sample of pure lithium carbon
Anna35 [415]

Answer:

Percentage lithium by mass in Lithium carbonate sample = 19.0%

Explanation:

Atomic mass of lithium = 7.0 g; atomic mass of Chlorine = 35.5 g; atomic mass of carbon = 12.0 g; atomic mass of oxygen = 16.0 g

Molar mass of lithium chloride, LiCl = 7 + 35.5 = 42.5 g

Percentage by mass of lithium in LiCl = (7/42.5) * 100% = 16.4 % aproximately 16%

Molar mass of lithium carbonate, Li₂CO₃ = 7 * 2 + 12 + 16 * 3 =74.0 g

Percentage by mass of lithium in Li₂CO₃ = (14/74) * 100% = 18.9 % approximately 19%

Mass of Lithium carbonate sample = 2 * 42.5 = 85.0 g

mass of lithium in 85.0 g Li₂CO₃ = 19% * 85.0 g = 16.15 g

Percentage by mass of lithium in 85.0 g Li₂CO₃ = (16.15/85.0) * 100 % = 19.0%

Percentage lithium by mass in Lithium carbonate sample = 19.0%

3 0
3 years ago
0.2640 g of sodium oxalate is dissolved in a flask and requires 30.74 mL of potassium
Free_Kalibri [48]

Moles of potassium permanganate = 0.0008

<h3>Further explanation  </h3>

Titration is a procedure for determining the concentration of a solution by reacting with another solution which is known to be concentrated (usually a standard solution). Determination of the endpoint/equivalence point of the reaction can use indicators according to the appropriate pH range  

Reaction

5Na2C2O4(aq) + 2KMnO4(aq) + 8H2SO4(aq) ---> 2MnSO4(aq) + K2SO4(aq) + 5Na2SO4(aq) +  10CO2(g) + 8H2O(1)

The end point ⇒titrant and analyte moles equal

titrant : potassium  permanganate-KMnO4

analyte : sodium oxalate - Na2C2O4

so moles of KMnO4 = moles of Na2C2O4

moles of Na2C2O4(mass = 0.2640 g, MW=134 g/mol) :

\tt mol=\dfrac{mass}{MW}\\\\mol=\dfrac{0.264}{134 g/mol}\\\\mol=0.002

From equation, mol ratio  Na2C2O4 : KMnO4 = 5 : 2, so mol KMnO4 :

\tt \dfrac{2}{5}\times 0.002=0.0008

6 0
3 years ago
Sodium carbonate (Na2CO3) is used to neutralize the sulfuric acid spill. How many kilograms of sodium carbonate must be added to
valina [46]

5.451 X 10³ kg of sodium carbonate must be added to neutralize 5.04×103 kg of sulfuric acid solution.

<u>Explanation</u>:

  • Sodium carbonate is used to neutralized sulfuric acid, H₂SO₄. Sodium carbonate is the salt of a strong base (NaOH) and weak acid (H₂CO₃). The balanced chemical reaction for neutralization is as follows:

                        Na₂CO₃ + H₂SO₄ ----> Na₂SO₄ + H₂CO₃

  • From a balanced chemical equation, it is clear that one mole of Na₂CO₃ is required to neutralize one mole of  H₂SO₄.
  • Molar mass of Na₂CO₃= 106 g/mol = 0.106 kg/mol and                                Molar mass of H₂SO₄= 98 g/mol = 0.098 kg/mol.
  • To neutralize 0.098 kg of H₂SO₄ amount of Na₂CO₃ required is 0.106 kg, so, To neutralize 5.04×10³ kg of H₂SO₄, Na₂CO₃ required is =  5.451 X 10³ kg.

 

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