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NARA [144]
3 years ago
9

A cube has a length of 1.02 cm and a mass of .001012 kg what is its density ​

Chemistry
1 answer:
slega [8]3 years ago
8 0

Answer:

d = 9.5× 10⁻⁴ Kg/cm³

Explanation:

Given data:

Length of cube = 1.02 cm

Mass of cube = 0.001012 Kg

Density of cube = ?

Solution:

The length , height and width of cube is always same. Thus the volume of cube is,

Volume = length × height× width

Volume = 1.02 cm × 1.02 cm × 1.02 cm

Volume = 1.06 cm³

Density of cube:

d = 0.001012 Kg /1.06 cm³

d = 9.5× 10⁻⁴ Kg/cm³

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A 17.6-g sample of ammonium carbonate contains ________ mol of ammonium ions.
postnew [5]
These problems are a bit interesting. :)

First let's write the molecular formula for ammonium carbonate. 

NH4CO3 (Note! The 4 and 3 are subscripts, and not coefficients)

17.6 gNH4CO3

Now to convert to mol of one of our substances we take the percent composition of that particular part of the molecule and multiply it by our starting mass. This is what it looks like using dimensional analyse. 

17.6 gNH4CO3 * (Molar Mass of NH4 / Molar Mass of NH4CO3)

Grab a periodic table (or look one up) and find the molar masses for these molecules! Well. In this case I'll do it for you. (Note: I round the molar masses off to two decimal places)

NH4 = 14.01 + 4*1.01 = 18.05 g/mol
NH4CO3 = 14.01 + 4*1.01 + 12.01 + 3*16.00 = 78.06 g/mol


17.6 gNH4CO3 * (18.05 molNH4 / 78.06 molNH4CO3)
= 4.07 gNH4

Now just take the molar mass we found to convert that amount into moles!

4.07 gNH4 * (1 molNH4 / 18.05 gNH4) = 0.225 molNH4

4 0
2 years ago
How do I round 10.25 to only 3 significant digits
Lady_Fox [76]
10.3 is good the correct rounding of three sig figures

3 0
3 years ago
Which of these elements exist as a liquid at normal Earth temperatures? Select all that apply. bromine sodium oxygen mercury
Anni [7]
Mercury exist as liquid at normal earth tempertures
4 0
3 years ago
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Neptunium-237 undergoes a series of α-particle and β-particle productions to end up as thallium-205. How many α particles and β
Fantom [35]
<h3>Answer:</h3>

8 alpha particles

4 beta particles

<h3>Explanation:</h3>

<u>We are given;</u>

  • Neptunium-237
  • Thallium-205
  • Neptunium-237 undergoes beta and alpha decay to form Thallium-205.

We are required to determine the number of beta and alpha particles produced to complete the decay series.

  • We need to know that when a radioisotope emits an alpha particle the mass number reduces by 4 while the atomic number decreases by 2.
  • When a beta particle is emitted the mass number of the radioisotope increases by 1 while the atomic number remains the same.

In this case;

Neptunium-237 has an atomic number 93, while,

Thallium-205 has an atomic number 81.

Therefore;

²³⁷₉₃Np → x⁴₂He + y⁰₋₁e + ²⁰⁵₈₁Ti

We can get x and y

237 = 4x + y(0) + 205

237-205 = 4x

4x = 32

 x = 8

On the other hand;

93 = 2x + (-y) + 81

but x = 8

93 = 16 -y + 81

y = 4

Therefore, the complete decay equation is;

²³⁷₉₃Np → 8⁴₂He + 4⁰₋₁e + ²⁰⁵₈₁Ti

Thus, Neptunium-237 emits 8 alpha particles and 4 beta particles to become Thallium-205.

5 0
2 years ago
What is a property in chemistry
Crazy boy [7]
DescriptionA chemical property is any of a material's properties that becomes evident during, or after, a chemical reaction; that is, any quality that can be established only by changing a substance's chemical identity.
3 0
3 years ago
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