26 PONITS FOR BEST ANSWER28
Which of the following is not related to the First Law of Motion?
An object in motion stays in motion unless acted on by an outside force.
An object at rest will stay at rest.
An object must overcome inertia to begin moving.
Having a heavier mass may increase the force of a moving object.
29
Which of the following is not an example of friction?
Pouring water off a bridge and watching it turn to small drops by the time it hits the ground.
A wrecking ball swinging towards its target.
A piece of ice dropping into a glass of water that sinks more slowly than it fell through the air.
Smoothing an ice rink with a Zamboni.
30
What three quantities are involved with the Second Law of Motion?
acceleration, gravity, inertia
acceleration, inertia, force
mass, gravity, acceleration
acceleration, force, mass
31
Why do phytoplanktons have projections and form long chains?
To catch more food
To respire oxygen more efficiently
To keep from sinking and capture more light
To evade predators
32
If an object's acceleration is zero, which of these could be true?
it is slowing down
it is stopped
it is not changing its speed
Both B and C could be true
Answers
C
D
C
trust me plz i am good at chemistry
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Answers
Answer:
The correct answer is option A) Oxidation
Final answer:
The release of energy through the removal of hydrogen from a molecule is referred to as oxidation. In this process, energy is often released which can be utilized by cells for different functions.
Explanation:
The release of energy from a molecule by the removal of hydrogen is known as A) Oxidation. In biochemical reactions, oxidation refers to the process in which electrons are removed from a molecule, usually by removing a hydrogen atom (which consists of one proton and one electron). This process often releases energy that can be used by cells for various functions. Thus, through oxidation, molecules such as carbohydrates or fats can be broken down to release energy used for cellular activities.
Learn more about Oxidation here:
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(2) 2 (4) 4
Answers
Answer : The correct option is, (3) 3
Explanation :
The given molecule is, Ethyne
As we know that ethyne is an unsaturated hydrocarbon which contains triple bonds between the carbon-carbon atoms.
In unsaturated compounds, there are less number of hydrogen bonded with the carbon atoms.
The formula of ethyne molecule is,. In this molecule, there are triple bond is present between the two carbon atoms that means there are 3 pair of electrons present between the two carbon atoms.
Hence, there are 3 number of electron pairs that are shared between the two carbon atoms in a molecule of ethyne.
The structure of ethyne are shown below.
Answers
Answer:
The formula for molality is m = moles of solute / kilograms of solvent. In problem solving involving molality, we sometimes need to use additional formulas to get to the final answer. One formula we need to be aware of is the formula for density, which is d = m / v, where d is density, m is mass and v is volume
Explanation:
it is a measure of the concentration of a solute in a solution in terms of amount of substance in a specified amount of mass of the solvent. This contrasts with the definition of molarity which is based on a specified volume of solution.
and a temperature of
(1) 0°C (3) 273°C
(2) 100°C (4) 373°C
Answers
Explanation:
Water undergoes phase changes at 1 atm pressure as follows:
Solidification, change from water to ice occurs at 0°C.
Evaporation, change from liquid to gas occurs at 100°C.
Answers
Answer:
It is 33.8.
Explanation:
Answers
Answer:
1.74845
Explanation:
We have the following reaction:
I2 + H2 => 2 HI
Now, the constant Kc, has the following formula:
Kc = [C] ^ c * [D] ^ d / [A] ^ a * [B] ^ b
In this case I2 is A, H2 is B and C is HI
We know that the values are:
H2 = 1 × 10 ^ -3 at 448 ° C
I2 = 2 × 10 ^ -3 at 448 ° C
HI = 1.87 × 10 ^ -3 at 448 ° C
Replacing:
Kc = [1.87 × 10 ^ -3] ^ 2 / {[2 × 10 ^ -3] ^ 1 * [1 × 10 ^ -3] ^ 1}
Kc = 1.87 ^ 2/2 * 1
Kc = 1.74845
Which means that at 448 ° C, Kc is equal to 1.74845
Answer:
Explanation:
[H2] = 10^-3
[I2] = 2*10^-3
[HI] = 0
in equilbiirum
[H2] = 10^-3 - x
[I2] = 2*10^-3 -x
[HI] = 0 + 2x
and we know
[HI] = 0 + 2x = 1.87*10^-3
x = ( 1.87*10^-3)/2 = 0.000935
then
[H2] = 10^-3 - 0.000935 = 0.000065
[I2] = 2*10^-3 -0.000935 = 0.001065
H₂ + I ⇄ 2 HI
Initially 1 × 10⁻³ 2 × 10⁻³
Change -9.35 × 10⁻⁴ -9.35 × 10⁻⁴ +1.87 × 10⁻³
At equil 6.5 × 10⁻⁵ 1.06 5 × 10⁻³ 1.87 × 10⁻³
HI increase by 1.87 × 10⁻³M