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Svetlanka [38]
3 years ago
5

What is the kinetic energy of a 478 kg object that is moving with a speed of 15m/s

Chemistry
1 answer:
AVprozaik [17]3 years ago
8 0

Answer:Kinetic energy = (1/2)*mass*velocity^2

KE = (1/2)mv^2

KE = (1/2)(478)(15)^2

KE = 53775J

Explanation:

Kinetic energy = (1/2)*mass*velocity^2

KE = (1/2)mv^2

KE = (1/2)(478)(15)^2

KE = 53775J

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In the decomposition of water, why is twice as much hydrogen as oxygen formed?(1 point)
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Answer:

1. Which statement correctly describes the classification of chemical reactions into different categories?

-Not all reactions fit into a category, and some reactions can fit into more than one category.

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-Two

3. In the decomposition of water, why is twice as much hydrogen as oxygen formed?

- There are two atoms of hydrogen and one atom of oxygen in a water molecule.

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5 0
2 years ago
In this section, you learned about pressure in fluids. Answer the question that follows.
Serga [27]

I think that the answer might be B.

7 0
3 years ago
Consider the following equilibrium reaction having gaseous reactants and products.4HCl + O2 ⇌ 2H2O + Cl2Which of the following w
Eddi Din [679]

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From the given choices, the one that goes according to this reason is the third one: The volume of water vapor increases.

6 0
1 year ago
Plz someone help, really struggling
frez [133]

Answer:

22.9 Liters CO(g) needed

Explanation:

2CO(g)     +   O₂(g)    =>    2CO₂(g)

? Liters          32.65g

                 = 32.65g/32g/mol

                 =   1.02 moles O₂

Rxn ratio for CO to O₂ = 2 mole CO(g) to 1 mole O₂(g)

∴moles CO(g) needed = 2 x 1.02 moles CO(g) = 2.04 moles CO(g)

Conditions of standard equation* is STP (0°C & 1atm) => 1 mole any gas occupies 22.4 Liters.

∴Volume of CO(g) = 1.02mole x 22.4Liters/mole = 22.9 Liters CO(g) needed

___________________

*Standard Equation => molecular rxn balanced to smallest whole number ratio coefficients is assumed to be at STP conditions (0°C & 1atm).

6 0
2 years ago
Why can gasses change volume?​
soldi70 [24.7K]

Answer:

b. I've seen a question like this before lol

7 0
3 years ago
Read 2 more answers
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