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Art [367]
4 years ago
13

Which statement describes a chemical property that many elements might share?

Chemistry
1 answer:
nata0808 [166]4 years ago
6 0

ANSWER:

I think its B.

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How many moles of chlorine gas is contained in 4.81x10^24 molecules
ziro4ka [17]

Answer:

7.99 mol

Explanation:

Step 1: Given data

Molecules of chlorine gas: 4.81 × 10²⁴ molecules

Step 2: Calculate how many moles of chlorine gas corresponds to 4.81 × 10²⁴ molecules of chlorine gas

In order to convert molecules to moles, we will use Avogadro's number: there are 6.02 × 10²³ molecules of chlorine gas in 1 mole of molecules of chlorine gas.

4.81 × 10²⁴ molecules × (1 mol/6.02 × 10²³ molecules) = 7.99 mol

6 0
3 years ago
The valency of calcium is2 ? what does it mean​
nalin [4]

Answer:

hope you like it

Explanation:

because ouytermost shell contains two electrons

calcium is in group 2 so its valency is two valency is the number of outer shelll electrons it needs to have an stable outer shell . it needs 2 more .

5 0
3 years ago
Student moves to a new school and refuses to leave their room on the first day to go to school.
zvonat [6]

i would say Psychoanalysis so you need to call Frasier Crane lol


        hope it helps

5 0
3 years ago
Technetium (Tc; Z = 43) is a synthetic element used as a radioactive tracer in medical studies. A Tc atom emits a beta particle
Sedaia [141]

Question in incomplete, complete question is:

Technetium (Tc; Z = 43) is a synthetic element used as a radioactive tracer in medical studies. A Tc atom emits a beta particle (electron) with a kinetic energy (Ek) of 4.71\times 10^{-15}J . What is the de Broglie wavelength of this electron (Ek = ½mv²)?

Answer:

6.762\times 10^{-12} m is the de Broglie wavelength of this electron.

Explanation:

To calculate the wavelength of a particle, we use the equation given by De-Broglie's wavelength, which is:

\lambda=\frac{h}{\sqrt{2mE_k}}

where,

= De-Broglie's wavelength = ?

h = Planck's constant = 6.624\times 10^{-34}Js

m = mass of beta particle = 9.1094\times 10^{-31} kg

E_k = kinetic energy of the particle = 4.71\times 10^{-15}J

Putting values in above equation, we get:

\lambda =\frac{6.624\times 10^{-34}Js}{\sqrt{2\times 9.1094\times 10^{-31} kg\times 4.71\times 10^{-15}J}}

\lambda = 6.762\times 10^{-12} m

6.762\times 10^{-12} m is the de Broglie wavelength of this electron.

3 0
3 years ago
How many grams of carbonic acid were produced by the 3.00g sample of NaHCO3
Alexxx [7]
I dont know what subject is this
8 0
4 years ago
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