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2 years ago

A/an _______ is the smallest unit of matter that has the characteristics of a chemical element.

Chemistry

2 answers:

konstantin123 [22]2 years ago

8 0

The Answer is C) atom

julsineya [31]2 years ago

7 0

Answer:

The answer is atom.

Explanation:

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Find the de Broglie wavelength lambda for an electron moving at a speed of 1.00 \times 10^6 \; {\rm m/s}. (Note that this speed

masya89 [10]

(A) $7.28\cdot 10^{-10} m$

The De Broglie wavelength of an electron is given by

$\lambda=\frac{h}{p}$ (1)

where

h is the Planck constant

p is the momentum of the electron

The electron in this problem has a speed of

$v=1.00\cdot 10^6 m/s$

and its mass is

$m=9.11\cdot 10^{-31} kg$

So, its momentum is

$p=mv=(9.11\cdot 10^{-31} kg)(1.00\cdot 10^6 m/s)=9.11\cdot 10^{-25}kg m/s$

And substituting into (1), we find its De Broglie wavelength

$\lambda=\frac{6.63\cdot 10^{-34}Js}{9.11\cdot 10^{-25} kg m/s}=7.28\cdot 10^{-10} m$

(B) $1.16\cdot 10^{-34}m$

In this case we have:

m = 0.143 kg is the mass of the ball

v = 40.0 m/s is the speed of the ball

So, the momentum of the ball is

$p=mv=(0.143 kg)(40.0 m/s)=5.72 kg m/s$

And so, the De Broglie wavelength of the ball is given by

$\lambda=\frac{h}{p}=\frac{6.63\cdot 10^{-34} Js}{5.72 kg m/s}=1.16\cdot 10^{-34}m$

(C) $9.02\cdot 10^{-9}m$

The location of the first intensity minima is given by

$y=\frac{L\lambda}{a}$

where in this case we have

$y=0.492 cm = 4.92\cdot 10^{-3} m$

L = 1.091 is the distance between the detector and the slit

$a=2.00\mu m=2.00\cdot 10^{-6}m$ is the width of the slit

Solving the formula for $\lambda$ , we find the wavelength of the electrons in the beam:

$\lambda=\frac{ya}{L}=\frac{(4.92\cdot 10^{-3}m)(2.00\cdot 10^{-6} m)}{1.091 m}=9.02\cdot 10^{-9}m$

(D) $7.35\cdot 10^{-26}kg m/s$

The momentum of one of these electrons can be found by re-arranging the formula of the De Broglie wavelength:

$p=\frac{h}{\lambda}$

where here we have

$\lambda=9.02\cdot 10^{-9}m$ is the wavelength

Substituting into the formula, we find

$p=\frac{6.63\cdot 10^{-34}Js}{9.02\cdot 10^{-9}m}=7.35\cdot 10^{-26}kg m/s$

7 0

3 years ago

During science lab, Carl notices that when he adds water to his solid sample of anhydrous copper

Bond [772]

Answer:

The dissociation of copper sulfate into ions is an exothermic chemical reaction that releases heat into the surroundings.

Explanation:

Some of the potential energy stored in the solid sample of anhydrous copper sulfate is released as heat as the sample dissolves and dissociates into ions in the water. This is due to the large lattice energy of the crystalline copper sulfate.

hope this helps

4 0

2 years ago

What is the valency of aluminum in aluminum chloride? A. 1 B.2 C.3 D.4

Alinara [238K]

Answer:

C.3

13AL

2.8.3

The number of electrons from the outermost shell is the valence of the element.

5 0

2 years ago

Read 2 more answers

Which of the following pairs of elements would most likely combine to form a salt?

Gnom [1K]

I think it’s Ca and Cs

8 0

2 years ago

What is the new concentration of a solution of CaSO3 if 10.0 mL of a 2.0 M CaSO3 solution is diluted to 100 ml?

kifflom [539]

Answer: The new concentration of a solution of $CaSO_{3}$ is 0.2 M 10.0 mL of a 2.0 M $CaSO_{3}$ solution is diluted to 100 mL.

Explanation:

Given: $V_{1}$ = 10.0 mL, $M_{1}$ = 2.0 M

$V_{2}$ = 100 mL, $M_{2}$ = ?

Formula used to calculate the new concentration is as follows.

$M_{1}V_{1} = M_{2}V_{2}$

Substitute the values into above formula as follows.

$M_{1}V_{1} = M_{2}V_{2}\\10.0 mL \times 2.0 M = M_{2} \times 100 mL\\M_{2} = 0.2 M$

Thus, we can conclude that the new concentration of a solution of $CaSO_{3}$ is 0.2 M 10.0 mL of a 2.0 M $CaSO_{3}$ solution is diluted to 100 mL.

5 0

2 years ago