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kari74 [83]
3 years ago
5

When 1.98 g of NaOH (molar mass= 40 g/mole) are added to 75.0 mL of a 0.350 M HCl solution in a coffee-cup calorimeter, the temp

erature changes from 21.7°C to 25.8°C. If the resulting solution has a mass of 78.2 grams and a specific heat capacity of 3.97 J/g°C, what is the enthalpy change (ΔHrxn) for this reaction in kJ/mole.
Chemistry
1 answer:
Wittaler [7]3 years ago
3 0

Answer:

(ΔHrxn) = 1.273kJ/mol

Explanation:

Applying

m=78.2, c= 3.97, ∆t= 25.8-21.7

(ΔHrxn)= mc∆t

=78.2× 3.97× 4.1= 1.273kJ/mol

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Convert grams to moles
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A. 2.99g
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Why is it important for scientists to use a classification
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So they can tell what exact species it is. 
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CH4 + 202 → CO2 + 2H2O<br> How many grams of O2 needed to produce 36 grams of H2O?
Kipish [7]
<h3>Answer:</h3>

64 g O₂

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced]   CH₄ + 2O₂ → CO₂ + 2H₂O

[Given]   36 g H₂O

[Solve]   x g O₂

<u>Step 2: Identify Conversions</u>

[RxN] 2 mol O₂ → 2 mol H₂O

[PT] Molar Mass of O - 16.00 g/mol

[PT] Molar Mas of H - 1.01 g/mol

Molar Mass of O₂ - 2(16.00) = 32.00 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

<u>Step 3: Stoichiometry</u>

  1. Set up conversion:                     \displaystyle 36 \ g \ H_2O(\frac{1 \ mol \ H_2O}{18.02 \ g \ H_2O})(\frac{2 \ mol \ O_2}{2 \ mol \  H_2O})(\frac{32.00 \ g \ O_2}{1 \ mol \ O_2})
  2. Divide/Multiply [Cancel Units]:                                                                       \displaystyle 63.929 \ g \ O_2

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

63.929 g O₂ ≈ 64 g O₂

8 0
3 years ago
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How many moles of NH3 can be produced from 30.0 mol of H2 and excess N2?
VMariaS [17]
20 mol of NH, can be produce from 30 mol o H2
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What mass of Sodium Chloride is required to make 100.0 mL of 3.0 M solution?
Julli [10]

Answer:

17.55 g of NaCl

Explanation:

The following data were obtained from the question:

Molarity = 3 M

Volume = 100.0 mL

Mass of NaCl =..?

Next, we shall convert 100.0 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

100 mL = 100/1000

100 mL = 0.1 L

Therefore, 100 mL is equivalent to 0.1 L.

Next, we shall determine the number of mole NaCl in the solution. This can be obtained as follow:

Molarity = 3 M

Volume = 0.1 L

Mole of NaCl =?

Molarity = mole /Volume

3 = mole of NaCl /0.1

Cross multiply

Mole of NaCl = 3 × 0.1

Mole of NaCl = 0.3 mole

Finally, we determine the mass of NaCl required to prepare the solution as follow:

Mole of NaCl = 0.3 mole

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Mass of NaCl =?

Mole = mass /Molar mass

0.3 = mass of NaCl /58.5

Cross multiply

Mass of NaCl = 0.3 × 58.5

Mass of NaCl = 17.55 g

Therefore, 17.55 g of NaCl is needed to prepare the solution.

5 0
3 years ago
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