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tester [92]
4 years ago
5

Tim and Jim finished their lab experiment earlier than their classmates. What should they do while they wait for the others to f

inish?
A.
look for an interesting experiment in their lab manual and perform that experiment
B.
grab a snack out of their backpacks and eat it in the lab area
C.
look over their answers to the lab questions and check for errors
D.
bring their math textbooks to the lab area and finish tonight's homework
Chemistry
1 answer:
Alisiya [41]4 years ago
3 0

The answer to your question is,

C. Look over their answers to the lab questions and check for errors.

-Mabel <3

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5. Why doesn't the air at a frontal boundary mix?​
Vsevolod [243]

Answer:

At a front, the two air masses have different densities, based on temperature, and do not easily mix. One air mass is lifted above the other, creating a low pressure zone.

Explanation:

Hope this helps!

4 0
3 years ago
Question 1(Multiple Choice Worth 4 points) (04.05 LC) What is the oxidation number of manganese in MnO41−? +3 +4 +7 +8 Question
Elan Coil [88]

1) Answer is: the oxidation number of manganese in MnO₄⁻ is +7.

Permanganate anion has negative charge 1-.

Oxygen (O) in permanganate anion has oxidation number -2.

x + 4 · (-2) = -1.

x - 8 = -1.

x = +7; oxidation nzmber of manganese.

Oxidation number shows the degree of oxidation of an atom in a chemical compound.

2) Answer is: Zn + Cu2+ → Zn2+ + Cu.

In this chemical reaction, there is transfer of electrons from zinc (Zn) to copper (Cu). Zinc change oxidation number from 0 to +2 (lost electrons) and copper change oxidation number from +2 to 0 (gain electrons).

Oxidation half reaction: Zn⁰ → Zn²⁺ + 2e⁻.

Reduction half reaction: Cu²⁺ + 2e⁻ → Cu⁰.

In other chemical reactions, there is no change of oxidation number of elements.

3) Answer is: Br2 is the oxidizing agent because its oxidation number decreases.

Balanced chemical reaction: 2Al + 3Br₂ → 2AlBr₃.

In this chemical reaction, aluminium change oxidation number from 0 to +3 (lose electrons) and bromine change oxidation number from 0 to -1 (gain electrons, reduced).

Oxidizing agent is a substance that has the ability to oxidize other substances, to cause them to lose electrons.

In oxido-reduction reaction, at least one element lose and one element gain electrons.

4) Answer is: Br2 + 2Cl− → Cl2 + 2Br⁻.

Oxidation is increase of oxidation number.

In this balanced chemical reaction, chlorine change oxidation number from -1 (Cl⁻) to oxidation number 0 (Cl₂).

Oxidation half reaction: 2Cl⁻ → Cl₂⁰ + 2e⁻.

Reduction half reaction: Br₂ + 2e⁻ → 2Br⁻.

In other chemical reactions, chlorine is reduced.

5) Answer is: Observation 1 is a result of copper ions moving into the solution.

The reactivity series is a series of metals from highest to lowest reactivity. Metal higher in the reactivity series will displace another.

Copper (Cu) is higher in activity series than silver (Ag), so copper lose electron and silver gain electrons.

Copper is oxidized (increase of oxidation number) and silver is reduced.

3 0
3 years ago
1. Hydrogen and nitrogen combine to form ammonia. When nitrogen and hydrogen bond, nitrogen pulls the electrons from hydrogen to
UNO [17]

Answer:

did you do it yet?

Explanation:

4 0
3 years ago
According to the following reaction, how many grams of potassium sulfate will be formed upon the complete reaction of 23.8 grams
Alexxandr [17]

<u>Answer:</u> The mass of potassium sulfate that can be produced is 73.88 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For KOH:</u>

Given mass of KOH = 23.8 g

Molar mass of KOH = 56.1 g/mol

Putting values in equation 1, we get:

\text{Moles of KOH}=\frac{23.8g}{56.1g/mol}=0.424mol

The chemical equation for the reaction of KOH and potassium hydrogen sulfate follows:

KHSO_4+KOH\rightarrow K_2SO_4+H_2O

As, potassium hydrogen sulfate is present in excess. It is considered as an excess reagent.

KOH is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of KOH produces 1 mole of potassium sulfate

So, 0.424 moles of KOH will produce = \frac{1}{1}\times 0.424=0.424moles of potassium sulfate

Now, calculating the mass of potassium sulfate from equation 1, we get:

Molar mass of potassium sulfate = 174.26 g/mol

Moles of potassium sulfate = 0.424 moles

Putting values in equation 1, we get:

0.424mol=\frac{\text{Mass of potassium sulfate}}{174.26g/mol}\\\\\text{Mass of potassium sulfate}=(0.424mol\times 174.26g/mol)=73.88g

Hence, the mass of potassium sulfate that can be produced is 73.88 grams

8 0
3 years ago
If a single gold atom has a diameter of 2.9×x10−8 cm, how many atoms thick was rutherford's foil
ale4655 [162]

27,586

<h3>Further explanation</h3>

<u>Given:</u>

A single gold atom has a diameter of \boxed{ \ 2.9 \times 10^{-8} \ cm. \ }

From a reference, the Rutherford gold foil used in his scattering experiment had a thickness of approximately \boxed{ \ 8 \times 10^{-3} \ mm. \ }

<u>Question:</u>

How many atoms thick were Rutherford's foil?

<u>The Process:</u>

Convert thickness from mm to cm.

\boxed{ \ 8 \times 10^{-3} \ mm = 8 \times 10^{-3} \times 10{-1} \ cm \ } \rightarrow \boxed{ \ 8 \times 10^{-4} \ cm \}

The number of atoms is calculated from gold foil thickness divided by the atomic diameter.

\boxed{ \ = \frac{8 \times 10^{-4} \ cm}{2.9 \times 10^{-8} \ cm} \ }

\boxed{ \ =2.7586 \times 10^4 \ atoms \ }

Therefore, we get an atomic thickness of 27,586 atoms.

<u>Notes:</u>

  • In 1909-1910, Ernest Rutherford with two of his assistants, namely Hans Geiger and Ernest Marsden, conducted a series of experiments to find out more about the arrangement of atoms. They fired at a very thin gold plate with high-energy alpha particles.
  • One of their observations is that a small portion of alpha particles are reflected. This greatly surprised Rutherford. The reflected alpha particle must have hit something very dense in the atom. This fact is incompatible with the atomic model proposed by J.J. Thomson where the atoms are described as homogeneous in all parts with electrons and protons evenly distributed.
  • In 1911, Rutherford was able to explain the scattering of alpha rays by proposing ideas about atomic nuclei. According to him, most of the mass and positive charge of atoms are concentrated at the center of the atom, hereinafter referred to as the nucleus.
<h3>Learn more</h3>
  1. The energy density of the stored energy  brainly.com/question/9617400
  2. The theoretical density of platinum which has the FCC crystal structure. brainly.com/question/5048216
  3. Compound microscope brainly.com/question/4000241

Keywords: if a single gold atom, has a diameter of 2.9 x 10⁻⁸ cm, how many, atoms thick, Rutherford's foil, his scattering experiment

7 0
3 years ago
Read 2 more answers
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