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Cloud [144]
3 years ago
6

Which of the following properties of matter is independent of its location?

Chemistry
2 answers:
Tanzania [10]3 years ago
8 0

a is the correct answer


drek231 [11]3 years ago
7 0
The answer is mass. 
Mass is the amount of matter in an object. So no matter where the object is, it mass will stay the same unless particles are increased. 
You might be interested in
CH4 and CO₂ are the chemical formulas for:
nadya68 [22]

Answer:

CH4 = Methane

CO2 = Carbon dioxide

<h3>Hope this Helps!!</h3><h3>Hope this Helps!!:)</h3>
6 0
3 years ago
Give the n and l values and the number of orbitals for sublevel 5g.
Pepsi [2]

The n and l values and the number of orbitals for sublevel 5g is :

5g shell , n= 5 subshell g , l = 4, Number of orbitals for sublevel = 9.

There are total four quantum numbers:

1) Principal quantum number , n

2) Angular quantum number , l

3) Magnetic quantum number , ml

4) spin quantum number , ms

For 5g shell, n = 5

subshell g , l = 4     ....0 - s , 1 - p , 2 - d, 3 - f, 4 -g

number of orbitals in subshell = (2l + 1)  ( 2×4 + 1) = 9

Thus,  The n and l values and the number of orbitals for sublevel 5g is :

5g shell , n= 5 subshell g , l = 4, Number of orbitals for sublevel = 9.

To learn more about quantum numbers here

brainly.com/question/14650894

#SPJ1

6 0
1 year ago
The solubility of silver(I)phosphate at a given temperature is 1.02 g/L. Calculate the Ksp at this temperature. After you get yo
Snezhnost [94]

<u>Answer:</u> The solubility product of silver (I) phosphate is 9.57\times 10^{-10}

<u>Explanation:</u>

We are given:

Solubility of silver (I) phosphate = 1.02 g/L

To convert it into molar solubility, we divide the given solubility by the molar mass of silver (I) phosphate:

Molar mass of silver (I) phosphate = 418.6 g/mol

\text{Molar solubility of silver (I) phosphate}=\frac{1.02g/L}{418.6g/mol}=2.44\times 10^{-3}mol/L

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

The chemical equation for the ionization of silver (I) phosphate follows:

Ag_3PO_4(aq.)\rightleftharpoons 3Ag^{+}(aq.)+PO_4^{3-}(aq.)  

                            3s                  s

The expression of K_{sp} for above equation follows:

K_{sp}=(3s)^3\times s

We are given:  

s=2.44\times 10^{-3}M

Putting values in above expression, we get:

K_{sp}=(3\times 2.44\times 10^{-3})^3\times (2.44\times 10^{-3})\\\\K_{sp}=9.57\times 10^{-10}

Hence, the solubility product of silver (I) phosphate is 9.57\times 10^{-10}

4 0
3 years ago
The density of thorium, which has the FCC structure, is 11.72 g/cm3. The atomic weight of thorium is 232 g/mol. Calculate (a) th
seropon [69]

Answer:

(a) a = 5.08x10⁻⁸ cm

(b) r = 179.6 pm  

Explanation:

(a) The lattice parameter "a" can be calculated using the following equation:

\rho = \frac{(N atoms/cell)*m}{V_{c}*N_{A}}      

<em>where ρ: is the density of Th = 11.72 g/cm³, N° atoms/cell = 4, m: is the atomic weight of Th = 232 g/mol, Vc: is the unit cell volume = a³, and </em>N_{A}<em>: is the Avogadro constant = 6.023x10²³ atoms/mol. </em>

Hence the lattice parameter is:  

a^{3} = \frac{(N atoms/cell)*m}{\rho *N_{A}} = \frac{4 atoms*232 g/mol}{11.72 g/cm^{3} *6.023 \cdot 10^{23} atoms/mol} = 1.32 \cdot 10^{-22} cm^{3}

a = \sqrt[3]{1.32 \cdot 10^{-22} cm^{3}} = 5.08 \cdot 10^{-8} cm

(b) We know that the lattice parameter of a FCC structure is:

a = \frac{4r}{\sqrt{2}}

<em>where r: is the atomic radius of Th</em>

Hence, the atomic radius of Th is:

r = \frac{a*\sqrt{2}}{4} = \frac{5.08 \cdot 10^{-8} cm*\sqrt{2}}{4} = 1.796 \cdot 10^{-8} cm = 179.6 pm    

I hope it helps you!    

4 0
3 years ago
Read 2 more answers
What volume would 2.25 moles of Ne gas occupy at STP?
Dafna1 [17]
As we know that one mole of any Ideal gas at standard temperature and pressure occupies exactly 22.4 dm³ volume.

Solution for problem:

When 1 mole Neon (Ne) occupies 22.4 dm³ at STP then the volume occupied by 2.25 moles of Neon is calculated as,

                                             = ( 22.4 dm³ × 2.25 moles ) ÷ 1 mole
                
                                             = 50.4 dm³                   1dm³ = 1 L

Result:

So
, 50.4 dm³ (Liter) volume will be occupied by 2.25 moles of Neon gas if it acts ideally at STP.
8 0
3 years ago
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