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bixtya [17]
3 years ago
7

Is a molecule the same as a compound ?

Chemistry
1 answer:
Alenkasestr [34]3 years ago
5 0
No a molecule is 2 different atoms bond
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Simora [160]

15 mins

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3 years ago
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3 years ago
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4 0
2 years ago
In a coffee-cup calorimeter, 1 mol NaOH and 1 mol HBr initially at 22.5 oC (Celsius) are mixed in 100g of water to yield the fol
zavuch27 [327]

Answer:

ΔH = -55.92 kJ

Explanation:

<u>Step 1:</u> Data given

1 mol NaOH and 1 mol HBr initially at 22.5 °C are mixed in 100g of water

After mixing the temperature rises to 83 °C

Specific heat of the solution = 4.184 J/g °C

Molar mass of NaOH = 40 G/mol

Molar mass of HBr = 80.9 g/mol

<u>Step 2: </u>The balanced equation

NaOH + HBr → Na+(aq) + Br-(aq) + H2O(l)

<u>Step 3:</u> mass of NaOH

Mass = moles * Molar mass

Mass NaOH = 1 * 40 g/mol

Mass NaOH = 40 grams

Step 4: Mass of HBr

Mass HBr = 1 mol * 80.9 g/mol

Mass HBr = 80.9 grams

Step 5: Calculate ΔH

ΔH = m*c*ΔT

ΔH= (100 + 40 + 80.9) * 4.184 * (83-22.5)

ΔH= 220.9 * 4.184 * 60.5

ΔH= 55916.86 J = 55.92 kJ

Since this is an exothermic reaction, the change in enthalpy is negative.

ΔH = -55.92 kJ

4 0
4 years ago
How many grams of carbon dioxide will be produced if 76.4 grams of
goldfiish [28.3K]

Answer:

Assuming that all of the oxygen is used up, 1.53×4111.53×411 or 0.556 moles of C2H3Br3 are required. Because there are only 0.286 moles of C2H3Br3 available, C2H3Br3 is the limiting reagent.

Limiting Reagent What is the limiting reagent if 76.4 grams of C2H3Br3 were reacted with 49.1 grams of O2? C2H3Br3 + 11O2 → 8CO2 + 6H2O + 6Br2 SOLUTION Using Approach 1: A. 76.4g &times; (1 mol/ 266.72 g) = 0.286 moles C2H3Br3 49.1g &times; (1 mole/ 32 g) = 1.53 moles O2 B.

Explanation:

MRK ME BRAINLIEST PLZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ

https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Map%3A_Introductory_Chemistry_(Tro)/08%3A_Quantities_in_Chemical_Reactions/8.04%3A_Limiting_Reactant_and_Theoretical_Yield

5 0
3 years ago
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