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Vlada [557]
3 years ago
9

Atomic fission requires temperatures of one million degrees or hotter. True False

Chemistry
1 answer:
pickupchik [31]3 years ago
8 0

<u>Answer:</u> The correct answer is false.

<u>Explanation:</u>

Atomic fission reactions are defined as reactions in which nucleus of an atom splits into two or more smaller nuclei. These reactions are carried out in nuclear reactors. The temperature range for the reactors are around 300°C.

Atomic fusion reactions are the reactions in which nucleus of two or more atoms fuse together to form one large nuclei. These reactions require a huge amount of energy to occur. The temperature required by these reactions are about one million kelvins.

Hence, the given statement is False.

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Select the lewis structure for xeo2f2 which correctly minimizes formal charges.
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Answer:

explanation and image attached

Explanation:

Our aim is to draw a structure of XeO2F2 whith the least formal charges. We must remember that the compound has 34 valence electrons.

To obtain the least formal charges then Xe must have a total of twelve electrons on its valence shell instead of eight.

The other atoms around the central Xe atom are arranged as shown in the image attached.

Image Credit: UCLA

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2 years ago
State the color of methyl orange in a sample of NaOH
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Is it possible to make new water
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No, Matter cannot be created nor deastroyed.

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3 years ago
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Submit your answer for the remaining reagent in Tutorial Assignment #1 Question 9 here, including units. Note: Use e for scienti
djverab [1.8K]
<h3>Answer:</h3>

The mass of excessive water (H₂O) is 40.815 kg

<h3>Explanation:</h3>

The Equation for the reaction is;

Li₂O(s) + H₂O(l) → 2LiOH(s)

From the question;

Mass of water removed is 80.0 kg

Mass of available Li₂O is 65.0 kg

We are required to calculate the mass of excessive reagent.

<h3>Step 1: Calculating the number of moles of water to be removed</h3>

Moles = Mass ÷ Molar mass

Molar mass of water = 18.02 g/mol

Mass of water = 80 kg (but 1000 g = 1kg)

                        = 80,000 g

Therefore;

Moles of water = \frac{80,000g}{18.02 g/mol}

                = 4.44 × 10³ moles

<h3>Step 2: Moles of Li₂O available </h3>

Moles = mass ÷ molar mass

Mass of Li₂O  available = 65.0 kg or 65,000 g

Molar mass Li₂O  = 29.88 g/mol

Moles of Li₂O  = 65,000 g ÷ 29.88 g/mol

          = 2.175 × 10³ moles Li₂O

<h3>Step 3: Mass of excess reagent </h3>

From the equation; Li₂O(s) + H₂O(l) → 2LiOH(s)

1 mole of Li₂O reacts with 1 mole of water to form two moles of LiOH

The ratio of Li₂O to H₂O is 1:1

  • Thus, 2.175 × 10³ moles of Li₂O will react with 2.175 × 10³ moles of water.
  • However, the number of moles of water to be removed is 4.44 × 10³ moles  but only 2.175 × 10³ moles will react with the available Li₂O.
  • This means, Li₂O  is the limiting reactant while water is the excessive reagent.

Therefore:

Moles of excessive water =  4.44 × 10³ moles  - 2.175 × 10³ moles

                                           = 2.265 × 10³ moles

Mass of excessive water = 2.265 × 10³ moles × 18.02 g/mol

                                          = 4.0815 × 10⁴ g or

                                          = 40.815 kg

Thus, the mass of excessive water is 40.815 kg

7 0
3 years ago
Atomic mass is typically expressed in
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Answer:

amu

Explanation:

Please mark as brainleist

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