All categories

3 years ago

A. 0.5 m/sb. 0.75 m/sc. 2 m/sd. 3 m/s

Chemistry

1 answer:

GenaCL600 [577]3 years ago

4 0

Answer:

Option a. 0.5 m/s

Explanation:

This graph shows a straight line, where "Y" axis would be "Position" and "X" graph would be "Time". The ecuation that would describe this straight line is Y= aX + 1 , where "a" is the slope or inclination for this graph, and would give us the speed of the object

How do we find the slope (and hence, the speed)?: if you notice this graph, you will check that:

-When X (Time) is zero, Y (Position) is 1

-When X (Time) is 2, Y (Position) is 2

With these 4 points, you can calculate the slope (which will call "m") for this graph with:

m = (Y2-Y1)/(X2-X1) so: Y2=2, Y1=1, X2=2, X1=0

Which gives us: m=1/2 (0.5), the slope or speed of the object: 0.5 m/s

You might be interested in

In the reaction, A → Products, the rate constant is 3.6 × 10−4 s−1. If the initial concentration of A is 0.548 M, what will be t

Arada [10]

Answer:

$\large\boxed{\large\boxed{0.529M}}$

Explanation:

Since the rate constant has units of s⁻¹, you can tell that the order of the reaction is 1.

Hence, the rate law is:

$r=d[A]/dt=-k[A]$

Solving that differential equation yields to the well known equation for the rates of a first order chemical reaction:

$[A]=[A]_0e^{-kt}$

You know [A]₀, k, and t, thus you can calculate [A].

$[A]=0.548M\times e^{-3.6\cdot 10^{-4}/s\times99.2s}$

$[A]=0.529M$

7 0

3 years ago

The diameter of the He He atom is approximately 0.10 nm nm . Calculate the density of the He atom in g/cm 3 g/cm3 (assuming that

Sladkaya [172]

Answer:

Density of the He atom = 12.69 g/cm³

Explanation:

From the information given:

Since 1 mole of an atom = 6.022x 10²³ atoms)

1 atom of He = $1 \ atom \times (\dfrac{1 \ mole}{ 6.022 \times 10^{23} \ atoms}) \times ( \dfrac{4.003 \ grams}{ 1 \ mole})$

$=6.647 \times 10^{-24} \ grams$

The volume can be determined as folows:

since the diameter of the He atom is approximately 0.10 nm

the radius of the He = $\dfrac{0.10}{2}$ = 0.05 nm

Converting it into cm, we have:

$0.05 nm \times \dfrac{10^{-9} \ meters}{ 1 nm} \times \dfrac{ 1 cm }{10^{-2} \ meters}$

$=5 \times 10^{-9} \ cm$

Assuming that it is a sphere, the volume of a sphere is

= $\dfrac{4}{3}\pi r^3$

= $\dfrac{4}{3}\pi \times (5\times 10^{-9})^3$

= $5.236 \times 10^{-25} \ cm^3$

Finally, the density can be calcuated by using the formula :

$Density = \dfrac{mass}{volume}$

$D = \dfrac{6.647 \times 10^{-24} \ grams }{ 5.236 \times 10^{-25} \ cm^3}$

D = 12.69 g/cm³

Density of the He atom = 12.69 g/cm³

4 0

3 years ago

Read 2 more answers

Calculate the ratio of the velocity of helium atoms to the velocity of neon atoms at the same temperature.

o-na [289]

Answer:

vHe / vNe = 2.24

Explanation:

To obtain the velocity of an ideal gas you must use the formula:

v = √3RT / √M

Where R is gas constant (8.314 kgm²/s²molK); T is temperature and M is molar mass of the gas (4x10⁻³kg/mol for helium and 20,18x10⁻³ kg/mol for neon). Thus:

vHe = √3×8.314 kgm²/s²molK×T / √4x10⁻³kg/mol

vNe = √3×8.314 kgm²/s²molK×T / √20.18x10⁻³kg/mol

The ratio is:

vHe / vNe = √3×8.314 kgm²/s²molK×T / √4x10⁻³kg/mol / √3×8.314 kgm²/s²molK×T / √20.18x10⁻³kg/mol

vHe / vNe = √20.18x10⁻³kg/mol / √4x10⁻³kg/mol

vHe / vNe = 2.24

I hope it helps!

8 0

3 years ago

Increasing the temperature, as noted in the graph, increases the rate of this chemical reaction. What is the relationship betwee

Soloha48 [4]

Answer: (C) Although the average kinetic energy of the colliding substances increases, this has no influence on activation energy.

Explanation:

After increasing the temperature of the reaction , the rate of the chemical reaction increases due to increase in the average kinetic energy of the particles. At increased temperature high proportions of particles can react making the reaction faster.

8 0

3 years ago

Which scenario involves a reaction that is at equilibrium?

musickatia [10]

Answer:

the amount of products and reactants is constant

4 0

3 years ago

A. 0.5 m/sb. 0.75 m/sc. 2 m/sd. 3 m/s ​

A. 0.5 m/sb. 0.75 m/sc. 2 m/sd. 3 m/s