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Arlecino [84]
2 years ago
6

All isotopes of a given element must have the same

Chemistry
2 answers:
Irina-Kira [14]2 years ago
7 0
Isotopes of any given element all contain the same number of protons, so they have the same atomic number (for example, the atomic number of helium is always 2). Isotopes of a given element contain different numbers of neutrons, therefore, different isotopes have different mass numbers.
mafiozo [28]2 years ago
5 0

Answer:

number of protons

Explanation:

Isotopes of any given element all contain the <u>same number of protons</u>, so they have the same atomic number

<em>e.g:</em> the atomic number of helium is always 2.

Isotopes of a given element contain <u>different numbers of neutrons</u>, therefore, <em><u>different isotopes have different mass numbers.</u></em>

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Usually, the more something ionizes, the better it conducts electricity. NH3 is molecular. CH3OH is an alcohol and doesn't ionize well. I'd say MgCl2 was more ionizable than H2O2, just because it's less likely to cause metathesis.
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Balance equations<br> FeS + O^2 - Fe^2O^3 + SO^2
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Explanation:

2FeS + 7/2O2 ---> Fe2O3 + 2SO2

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Magnesium reacts with hydrochloric acid, HCl, to form magnesium chloride and hydrogen gas. The reaction is:
Hunter-Best [27]
5.2/ 24.3 = 0.214
0.214 x 2 = 0.428
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3 years ago
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Hydrogen sulfide (H2S) is a gas that smells like rotten eggs. It can be produced by bacteria in your mouth and contributes to ba
Doss [256]

Answer:

See explanation.

Explanation:

Hello!

In this case, we consider the questions:

a. Ideal gas at:

 i. 273.15 K and 22.414 L.

 ii. 500 K and 100 cm³.

b. Van der Waals gas at:

 i. 273.15 K and 22.414 L.

 ii. 500 K and 100 cm³.

Thus, we define the ideal gas equation and the van der Waals one as shown below:

P^{id}=\frac{nRT}{V}\\\\ P^{vdW}=\frac{RT}{v-b}-\frac{a}{v^2}

Whereas b and a for hydrogen sulfide are 0.0434 L/mol and 4.484 L²*atm / mol² respectively, therefore, we proceed as follows:

a.

 i. 273.15 K and 22.414 L.

P^{id}=\frac{1.00mol*0.08206\frac{atm*L}{mol*K}*273.15K}{22.414L}=1 .00 atm

 ii. 500 K and 100 cm³ (0.1 L).

P^{id}=\frac{1.00mol*0.08206\frac{atm*L}{mol*K}*500K}{0.100L}=410.3 atm

b.

 i. 273.15 K and 22.414 L: in this case, v = 22.414 L / 1.00 mol = 22.414 L/mol

P^{vdW}=\frac{0.08206\frac{atm*L}{mol*K}*273.15K}{22.414L/mol-0.0434L/mol}-\frac{4.484 atm*L^2/mol^2}{(22.414L/mol)^2}=0.993atm

 ii. 500 K and 100 cm³: in this case, v = 0.1 L / 1.00 mol = 0.100 L/mol

P^{vdW}=\frac{0.08206\frac{atm*L}{mol*K}*500K}{0.100L/mol-0.0434L/mol}-\frac{4.484 atm*L^2/mol^2}{(0.100L/mol)^2}=276.5atm

Whereas we can see a significant difference when the gas is at 500 K and occupy a volume of 0.100 L.

Best regards!

5 0
3 years ago
How should the answer be reported using the correct number of significant figures?
valentina_108 [34]
What are the calculation like answer
5 0
4 years ago
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