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AleksAgata [21]
1 year ago
6

3. Scott and James work at a grocery store. After the grocery store closed, they were playing a game with a shopping cart and Sc

ott’s skateboard. They crashed the skateboard and the shopping cart two times. The shopping cart had more mass in Crash 2 than it did in Crash 1. Use the information in the diagram to answer.
In which crash did the skateboard experience a stronger force? Why?
Chemistry
1 answer:
weqwewe [10]1 year ago
6 0

Answer: B) Crash 2; the force on the cart was stronger in this crash, so the force on the skateboard was also stronger.

Explanation:

You might be interested in
You've just solved a problem and the answer is the mass of an electron, me=9.11×10−31kilograms. How would you enter this number
irina1246 [14]

Answer: 9.11\times 10^{-31}kg

Explanation:

Significant figures : The figures in a number which express the value or the magnitude of a quantity to a specific degree of accuracy is known as significant digits.

Rules for significant figures:

Digits from 1 to 9 are always significant and have infinite number of significant figures.

All non-zero numbers are always significant.

All zero’s between integers are always significant.

All zero’s after the decimal point are always significant.

All zero’s preceding the first integers are never significant.

Thus 9.11\times 10^{-31}kg has three significant figures

7 0
2 years ago
For the reaction A + B − ⇀ ↽ − C + D A+B↽−−⇀C+D , assume that the standard change in free energy has a positive value. Changing
AURORKA [14]

Answer:

a. Not change the free energy value

b. Increase the free energy value

c. Decrease the free energy value

d. Decrease the free energy value

Explanation:

a. Adding a catalyst:

A catalyst is a substance that will reduce the activation energy of a reaction, it means that the reaction will occur fast. The values of enthalpy, entropy, and free energy are not affected by a catalyst, so ΔG remains the same.

b. Increasing [C] and [D]:

For a reversible reaction, the value of free energy can be calculated by:

ΔG = ΔG° + RT*lnK

Where ΔG° is the standard value for free energy, R is the gas constant, T is the temperature, and K is the constant of equilibrium, which in this case:

K = ([C]*[D])/([A]*[B])

When [C] and [D] increase, the value of K increases, and lnK also increases, then, the value of ΔG increases.

c. Coupling with ATP hydrolysis:

The free energy can be calculated by:

ΔG = ΔH - TΔS

Where ΔH is the change in enthalpy, and ΔS the change in entropy. The ATP hydrolysis is an exothermic reaction, so ΔH <0. When it is coupled, it will reduce the total value of ΔH, and because of that, the value of ΔG will decrease.

d. Increasing [A] and [B]:

As explained above, the increasing at [A] and [B] will decrease the value of K, so the value of lnK will decrease, and ΔG value will also decrease.

4 0
1 year ago
Read 2 more answers
What is the hybridization of the central atom in each of the following? 1. Beryllium chloride 2. Nitrogen dioxide 3. Carbon tetr
Lina20 [59]

Answer :

(1) The hybridization of central atom beryllium in BeCl_2  is, sp

(2) The hybridization of central atom nitrogen in NO_2  is, sp^2

(3) The hybridization of central atom carbon in CCl_4  is, sp^3

(4) The hybridization of central atom xenon in XeF_4  is, sp^3d^2

Explanation :

Formula used  :

\text{Number of electron pair}=\frac{1}{2}[V+N-C+A]

where,

V = number of valence electrons present in central atom

N = number of monovalent atoms bonded to central atom

C = charge of cation

A = charge of anion

Now we have to determine the hybridization of the following molecules.

(1) The given molecule is, BeCl_2

\text{Number of electrons}=\frac{1}{2}\times [2+2]=2

The number of electron pair are 2 that means the hybridization will be sp and the electronic geometry of the molecule will be linear.

(2) The given molecule is, NO_2

\text{Number of electrons}=\frac{1}{2}\times [4]=2

If the sum of the number of sigma bonds, lone pair of electrons and odd electrons present is equal to three then the hybridization will be, sp^2.

In nitrogen dioxide, there are two sigma bonds and one lone electron pair. So, the hybridization will be, sp^2.

(3) The given molecule is, CCl_4

\text{Number of electrons}=\frac{1}{2}\times [4+4]=4

The number of electron pair are 4 that means the hybridization will be sp^3 and the electronic geometry of the molecule will be tetrahedral.

(4) The given molecule is, XeF_4

\text{Number of electrons}=\frac{1}{2}\times [8+4]=6

Bond pair electrons = 4

Lone pair electrons = 6 - 4 = 2

The number of electrons are 6 that means the hybridization will be sp^3d^2 and the electronic geometry of the molecule will be octahedral.

But as there are four atoms around the central xenon atom, the fifth and sixth position will be occupied by lone pair of electrons. The repulsion between lone and bond pair of electrons is more and hence the molecular geometry will be square planar.

3 0
1 year ago
Submit Test
scoray [572]

The correct answer is Hot water increases the collision rate of molecules, causing the reaction to occur faster.

Explanation:

Temperature is directly related to the kinetic energy or movement of molecules in a substance. In this context, a higher temperature leads to more kinetic energy or more collision between molecules. At the same time, a chemical reaction involves molecules of two or more substances colliding and creating bonds to form new substances. This implies an increase in temperature means molecules colliding faster, new substances forming in a shorter time, and therefore a faster chemical reaction. According to this, the first answer is correct.

5 0
1 year ago
For a single component system, why do the allotropes stable at high temperatures have higher enthalpies than allotropes stable a
solniwko [45]

Answer:

The difference in the magnetic orientation influences the thermal stability of the allotropes of iron.

Explanation:

It is known that the allotropes of iron exist in three phases: α - phase, β- phase, and γ-phase. However, two prominent structures are the  α - phase and γ-phase. Now, let us look at the two phrases:

α - phase

This structure is a body-centered cube. It means that the unit cell structure resembles a cube. The lattice points are in the face of the cube. This subsequently affects the magnetic structure of the iron allotrope.

γ-phase

This allotrope has a lattice structure. It simply means that the structure has lattice points on the face of the cube. The structure generally affects the magnetic properties of the transitional metal; hence the stability of the γ-phase compared to α-phase.

8 0
2 years ago
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