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Tema [17]
3 years ago
15

in the laboratory, a group of students was assigned to determine the density of an unknown liquid. They used a buret to find a v

olume of 2.04 mL. The mass was determined on an analytical balance to be 2.260 g. How should they report the density of the liquid?
Chemistry
1 answer:
Daniel [21]3 years ago
4 0

They should report the density as 1.11 g/L .

Density = mass/volume = 2.260g/2.04 mL

My calculator says the density <em>1.107 843 137 g/mL</em>

However, the answer can have <em>no more</em> significant figures than are in the number with the <em>fewest </em>significant figures.

The volume measurement has only three significant figures, so we must round off the density to three significant figures.

We drop all the digits after the zero.

The digit to be dropped is 7, so we <em>round up </em>the last significant figure of the answer.

1.10<u>7 843 137</u> → 1.11

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There are 100.0 grams of each reactant available determine the limiting reactant in this equation
Romashka [77]
Since you have not included the chemical reaction I will explain you in detail.

1) To determine the limiting agent you need two things:

- the balanced chemical equation

- the amount of every reactant involved as per the chemical equation

2) The work is:

- state the mole ratios of all the reactants: these are the ratios of the coefficientes of the reactans in the balanced chemical equation.


- determine the number of moles of each reactant with this formula:

number of moles = (mass in grams) / (molar mass)

- set the proportion with the two ratios (theoretical moles and actual moles)


- compare which reactant is below than the stated by the theoretical ratio.

3) Example: determine the limiting agent in this reaction if there are 100 grams of each reactant:

i) Chemical equation: H₂ + O₂ → H₂O

ii) Balanced chemical equation: 2H₂ + O₂ → 2H₂O

iii) Theoretical mole ration of the reactants: 2 moles H₂ : 1 mol O₂

iv) Covert 100 g of H₂ into number of moles

n = 100g / 2g/mol = 50 mol of H₂

v) Convert 100 g of O₂ to moles: 

n = 100 g / 32 g/mol = 3.125 mol

vi) Actual ratio: 50 mol H₂ / 3.125 mol O₂

vii) Compare the two ratios:

2 mol H₂ / 1 mol O ₂ < 50 mol H₂ / 3.125 mol O₂

Conclusion: the actual ratio of H₂ to O₂ is greater than the theoretical ratio, meaning that the H₂ is in excess respect to the O₂. And that means that O₂ will be consumed completely while some H₂ will remain without react.

Therefore, the O₂ is the limiting reactant in this example.

7 0
3 years ago
4. A student started with a 0.032 g sample of copper which he took through the series of reactions described in this experiment.
Elodia [21]

Answer:

Y=48.6\%

Explanation:

Hello,

In this case, we can consider the following chemical reaction for the oxidation of copper which only occurs at high temperatures:

2Cu+O_2\rightarrow 2CuO

In such a way, for 0.032 grams of copper, the following grams of copper (II) oxide (black product) are yielded:

m_{CuO}=0.032gCu*\frac{1molCu}{63.546gCu} *\frac{2molCuO}{2molCu}*\frac{79.546gCuO}{1molCuO}  =0.078gCuO

Therefore, the percent yield is:

Y=\frac{0.038g}{0.078g}*100\%\\ \\Y=48.6\%

Best regards.

6 0
3 years ago
Which two kinds of information describe weather and not climate?
Fynjy0 [20]

Answer: b & c

Explanation: A-P-E-X

5 0
3 years ago
Read 2 more answers
A sample of sodium reacts completely with 0.497 kg of chlorine, forming 819 g of sodium chloride. what mass of sodium reacted?
Slav-nsk [51]

Sodium reacts to chlorine and gives NaCl. The balanced reaction is given below:

2Na + Cl₂→ 2NaCl. Two moles Na reacts with one mole Cl₂ and produces two moles of NaCl. Atomic mass of Na= 23, Molar mass of Cl₂= 71, molar mass of NaCl=58.5.

So, 46 g Na reacts with 71 g of Cl₂ and produces (2 X 58.5)g = 117 g of NaCl. As per question Na reacts completely which means Na is the limiting reagent. So, number of moles of Na reacts = number moles of NaCl produced.

NaCl produced= 819 g= (819/58.5) moles= 15.69 moles. Therefore, 15.69 moles = 15.69 X 23 g=360.87 g of Na reacted.

8 0
3 years ago
Which of the following best explains how the results of Rutherford’s experiment affected Thomson’s widely-accepted atomic model?
hichkok12 [17]

Explanation:

Thomson's suggested the plum pudding model of the atom in which the atomic space is made up of electrons surround by positive charges.

Rutherford in his gold foil experiment revised the plum pudding model of the atom;

  • He discovered that most of the alpha particles passed through the foil while a few of them were deflected back.
  • To explain this observation, he suggested the atomic model of the atom.
  • In this model, an atom is made up of a small positively charged center where nearly all the mass is concentrated.
  • Surrounding the nucleus is the extranuclear part made up of electrons.
5 0
2 years ago
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