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Andreas93 [3]
4 years ago
12

Mass to mass stoichiomatry

Chemistry
1 answer:
7nadin3 [17]4 years ago
4 0
Carbon-12 has a molar mass of 12 g/mol. So to calculate the stoichiometry by mass, the number of molecules required for each reactant is expressed in moles and multiplied by the molar mass of each to give the mass of each reactant per mole of reaction.
-Hope that helps!! :)
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Two moles of neon gas enclosed in a constant volume system receive 4250 J of heat. If the gas was initially at 293 K, what is th
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Answer:

<u><em>=355.5K</em></u>

Explanation:

Specific heat, Q = mcΔT

where

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  • ΔT= change in temp = final temp - initial temp
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[1 mole of Ne = 20g; 2 moles of Ne = 2 × 20 = 40g]

4250 = 40 × 1.7 × (final - 293K)

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I hope this steps are simple to follow and understand.

3 0
3 years ago
Oh, no! You just spilled 85.00 mL of 1.500 M sulfuric acid on your lab bench and need to clean it up immediately! Right next to
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Explanation:

We will balance equation which describes the reaction between sulfuric acid and sodium bicarbonate: as follows.

   H_2SO_4(aq) + 2NaHCO_3(s) \rightarrow Na2SO_4(aq) + 2H_2O(l) + 2CO_2(g)

Next we will calculate how many moles of H_2SO_4 are present in 85.00 mL of 1.500 M sulfuric acid.

As,       Molarity = \frac{\text{moles of solute}}{\text{liters of solution&#10;}}

            1.500 M = \frac{n}{0.08500 L&#10;}

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Now set up and solve a stoichiometric conversion from moles of H_2SO_4  to grams of NaHCO_3. As, the molar mass of NaHCO_3 is 84.01 g/mol.

 0.1275 mol H_2SO_4 \times (\frac{2 mol NaHCO_3}{1 mol H_2SO_4}) \times (\frac{84.01 g NaHCO_3}{1 mol NaHCO_3})

                 = 21.42 g NaHCO_3

So unfortunately, 15.00 grams of sodium bicarbonate will "not" be sufficient to completely neutralize the acid. You would need an additional 6.42 grams to complete the task.

4 0
3 years ago
2.247 liter to milliliter
gladu [14]

Answer:

2247 mililiters

Explanation:

6 0
4 years ago
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Answer:

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