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nasty-shy [4]
2 years ago
12

When a meteor falls on earth. a streak of light is seen. will this streak of light be seen if a meteor falls on the moon? Explai

n??
​
Chemistry
1 answer:
Lera25 [3.4K]2 years ago
7 0

Answer:

No streak of light will appear.

Reason I given below.

Explanation:

No streak of light will be seen.

This is because as it falls to the moon, the gravitational pull plus the dense atmosphere which was necessary to produce friction when it fell to the earth are not present in this case and as such the whole meteor is not burnt like it did when it fell to the earth.

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What may be expected when K > 1.0?
lora16 [44]

The two correct statements are:

1. The concentration of one or more of the products is small

2. The reaction will proceed to the right and favor the formation of products.

LeChatelier's Principle states that when a stress is imposed on a system at equilibrium, the equilibrium will shift to counteract the change. A reduction in the number of products cause the system to respond by making more product, which leads to K>1 and products being more favorable.

4 0
2 years ago
Read 2 more answers
The concentration of ozone in ground-level air can be determined by allowing the gas to react with an aqueous solution of potass
Natali5045456 [20]

Answer:

a) 2KI + O₃ + H₂O ----> I₂ + O₂ + 2KOH

b) Ozone concentration = 0.246 ppb

Explanation:

a) The balanced equation for the reaction is

2KI + O₃ + H₂O ----> I₂ + O₂ + 2KOH

b) We first convert 17 μg of KI to number of moles

Number of moles = (mass)/(molar mass)

Molar mass of KI = 166 g/mol

Mass of KI that reacted = 17 μg = (17 × 10⁻⁶) g

Number of moles = (17 × 10⁻⁶)/166

Number of moles of KI that reacted = (1.0241 × 10⁻⁷) moles

From the stoichiometric balance of the reaction,

2 moles of KI reacts with 1 mole of O₃

Then, (1.0241 × 10⁻⁷) moles of KI will react with (1.0241 × 10⁻⁷ × 1/2) moles of O₃

Number of moles of O₃ that reacted = (5.12 × 10⁻⁸) moles.

To express the amount of O₃ in 10.0 L of air in ppb, we need to convert the amount of O₃ that reacted.

Mass = (number of moles) × (molar mass)

Molar mass of O₃ = 48 g/mol

Mass of O₃ that reacted = (5.12 × 10⁻⁸) × 48 = 0.0000024578 g = (2.46 × 10⁻⁶) g

Concentration in ppb = (Mass of solute in μg)/(volume of solution in L)

Mass of solute = Mass of O₃ = (2.46 × 10⁻⁶) g = 2.46 μg

Volume of solution = 10.0 L

Concentration of O₃ in air in ppb = 2.46/10 = 0.246 ppb

4 0
2 years ago
20 POINTS!!
Andrews [41]

Answer:

B. Interferon

Explanation:

interferon targets any lingering melanoma cells and prevents them from spreading and growing.

3 0
3 years ago
A 0.055 mol sample of formaldehyde vapor, CH2O, was placed in a heated 500 mL vessel and some of it decomposed. The reaction is
Elis [28]

Answer:

Kc for this reaction is 0.06825

Explanation:

Step 1: Data given

Number of moles formaldehyde CH2O = 0.055 moles

Volume = 500 mL = 0.500 L

At equilibrium, the CH2O(g) concentration = 0.051 mol

Step 2: The balanced equation

CH2O  <=>  H2 + CO

Step 3: Calculate the initial concentrations

Concentration = moles / volume

[CH2O] = 0.055 moles . 0.500 L

[CH2O] = 0.11 M

[H2] = 0M

[CO] = 0M

Step 4: The concentration at the equilibrium

[CH2O] = 0.11 - X M = 0.051 M

[H2] = XM

[CO] = XM

[CH2O] = 0.11 - X M = 0.051 M

X = 0.11 - 0.051 = 0.059

[H2] = XM = 0.059 M

[CO] = XM = 0.059 M

Step 5: Calculate Kc

Kc = [H2][CO]/[CHO]

Kc = (0.059 * 0.059) / 0.051

Kc = 0.06825

Kc for this reaction is 0.06825

7 0
3 years ago
PLEASE HELP ME SOLVE THIS.Thank you so much!
Tju [1.3M]

Answer: The coefficients for the given reaction species are 1, 6, 2, 3.

Explanation:

The given reaction equation is as follows.

Cr_{2}O^{2-}_{7} + Cl^{-} \rightarrow Cr^{3+} + Cl_{2}

Now, the two half-reactions can be written as follows.

Reduction half-reaction: Cr_{2}O^{2-}_{7} + 3e^{-} \rightarrow Cr^{3+}

This will be balanced as follows.

Cr_{2}O^{2-}_{7} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + 7H_{2}O ... (1)

Oxidation half-reaction: Cl^{-} \rightarrow Cl_{2} + 1e^{-}

This will be balanced as follows.

6Cl^{-} \rightarrow 3Cl_{2} + 6e^{-} ... (2)

Adding both equation (1) and (2) we will get the resulting equation as follows.

Cr_{2}O^{2-}_{7} + 14H^{+} + 6Cl^{-} \rightarrow 2Cr^{3+} + 3Cl_{2} + 7H_{2}O

Thus, we can conclude that coefficients for the given reaction species are 1, 6, 2, 3.

6 0
2 years ago
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