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insens350 [35]
3 years ago
8

What is the mass of 6.02 x 1023 particles of rubidium carbonate

Chemistry
1 answer:
Firdavs [7]3 years ago
4 0

Answer:

Explanation:

84.97

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Given the three equations below, what is the heat of reaction for the production of glucose, C6H12O6, as described by this equat
UkoKoshka [18]

Answer:

- 1273.02 kJ.

Explanation:

This problem can be solved using Hess's Law.

Hess's Law states that <em>regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes. This law is a manifestation that enthalpy is a state function.</em>

  • We should modify the given 3 equations to obtain the proposed reaction:

<em>6C(s) + 6H₂(g) + 3O₂(g) → C₆H₁₂O₆(s),</em>

<em></em>

  • We should multiply the first equation by (6) and also multiply its ΔH by (6):

6C(s) + 6O₂(g) → 6CO₂(g), ∆H₁ = (6)(–393.51 kJ) = - 2361.06 kJ,

  • Also, we should multiply the second equation and its ΔH by (6):

6H₂(g) + 3O₂(g) → 6H₂O(l), ∆H₂ = (6)(–285.83 kJ) = - 1714.98 kJ.

  • Finally, we should reverse the first equation and multiply its ΔH by (- 1):

6CO₂(g) + H₂O(l) → C₆H₁₂O₆(s) + 6O₂(g), ∆H₃ = (-1)(–2803.02 kJ) = 2803.02 kJ.

  • By summing the three equations, we cam get the proposed reaction:

<em>6C(s) + 6H₂(g) + 3O₂(g) → C₆H₁₂O₆(s),</em>

<em></em>

  • And to get the heat of reaction for the production of glucose, we can sum the values of the three ∆H:

<em>∆Hrxn = ∆H₁ + ∆H₂ + ∆H₃ =</em> (- 2361.06 kJ) + (- 1714.98 kJ) + (2803.02 kJ) = <em>- 1273.02 kJ.</em>

6 0
2 years ago
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Which of the following pieces of laboratory equipment is not directly used to make measurements?
pychu [463]
Test are not directly used to make measurements, I mean you could use them to put 4 mL of a substance in, but not to measure out 4 mL, like with graduated cylinders. Rulers can be used to measure the length of something, like the number of inches. And a thermometer will help you measure the temperature of something.
8 0
3 years ago
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Predict the mass of iron (III) sulfide produced when 3.0 g of iron filings react completely with 2.5 g of yellow sulfur solid, S
777dan777 [17]
1) Chemical equation

16Fe(s) + 3S8(s) ---> 8Fe2S3

2) Molar ratios:

16 mol Fe : 3 mole S8 : 8 mol Fe2S3

3) Convert masses in grams to number of moles

number of moles = mass in grams / molar mass

a) iron, Fe

mass = 3.0 g
atomic mass = 55.845 g/mol

=> number of moles of Fe = 3.0g / 55.845 g/mol = 0.0537 mol

b) Sulfur, S8

mass = 2.5 g
molar mass = 8*32.065 g/mol = 256.52 g/mol

=> number of moles of S8 = 2.5g / 256.52 g/mol = 0.009746 mol

4) Limiting reactant

Theoretical ratio                           actual ratio

16 mol Fe / 3 mol S8                 0.0537 mol Fe / 0.009746 mol S8

5.33                                               5.50

So, there is a little bit more Fe than the theoretical needed to react all the S8, which means the S8 is the limiting reactant.

5) Calculate the number of moles of iron (III) produced with 2.5 g (0.009746 moles) of S8

3moles S8 / 8 moles Fe2S3 = 0.009746 moles S8 / x

=> x = 0.009746 * 8 / 3 moles Fe2S3 = 0.026 moles Fe2S3

6) Convert 0.026 moles Fe2S3 into grams

mass in grams = number of moles * molar mass

molar mass of Fe2S3 = 207.9 g/mol

mass = 0.026 mol * 207.9 g/mol = 5.40 g

7) Answer: option D)




3 0
3 years ago
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WHAT IS THE CIENTIFIC RESULT WHEN YOU MAKE MARSHMELLOWS!?!?!??!?!??!!? pls its for 2morrow! :(
soldi70 [24.7K]
If you made the sugar and corn syrup solutions you heated, the sugar made the solutions' boiling points higher than that of pure liquid water.
7 0
3 years ago
What is the percentage composition of each element in hydrogen peroxide, H2O2?
Degger [83]

Answer:

The percentage composition of each element in H2O2 is 5.88% H and 94.12% O (Option D).

Explanation:

Step 1: Data given

Molar mass of H = 1.0 g/mol

Molar mass of O = 16 g/mol

Molar mass of H2O2 = 2*1.0 + 2*16 = 34.0 g/mol

Step 2: Calculate % hydrogen

% Hydrogen = ((2*1.0) / 34.0) * 100 %

% hydrogen = 5.88 %

Step 3: Calculate % oxygen

% Oxygen = ((2*16)/34)

% oxygen = 94.12 %

We can control this by the following equation

100 % - 5.88 % = 94.12 %

The percentage composition of each element in H2O2 is 5.88% H and 94.12% O (Option D).

8 0
2 years ago
Read 2 more answers
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