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svetlana [45]
2 years ago
7

When metals combine with nonmetals, the metallic atoms tend to 1. lose electrons and become positive ions 2. lose electrons and

become negative ions 3. gain electrons and become positive ions 4. gain electrons and become negative ions?
Chemistry
2 answers:
Vika [28.1K]2 years ago
5 0
The correct answer is 1. Lose electrons and become positive ions.


I hope my answer was beneficial to you! c:
Andrews [41]2 years ago
4 0

Answer:

1) lose electrons and become positive ions

Explanation:

To be able to solve this question, it is crucial to understand generally the chemistry of metals and non metals.

Generally metals are found in groups 1,2,3 in the periodic table. What these  tells us is that they have 1,2,3 electrons in their outermost shells. whenever metals want to go into reactions to complete their octet, it is much more feasible energy wise to lose those few electrons than accept about 5,6,7 more electrons.

When a compound loses its electron(s), it becomes a positive ion due to the net increase of protons compared to electrons.

For instance, calcium has 20 electrons and protons. However when it loses 2 electrons to complete its octet and become Ca2+, it still retains its 20 protons but the total number of electrons left is 20 -2 = 1 8. The difference between the protons and the electrons is usually what is represented by the charge on the ion.

However, the reverse is the case when it comes to nonmetals. Non metals would rather accept the few electrons as it is much feasible energy wise to complete its octet. Also, when  non metals gain electrons they would become negative ions due to the excess in number of electrons compared to protons.

As such, the only logical answer would be 1) lose electrons and become positive ions.

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liq [111]

Answer:

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Explanation:

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3 0
2 years ago
ANSWER Soon!!
vichka [17]

Answer:

  • 1. 0.1683 mol
  • 2. 1.191 g
  • 3. 0.02695 mol
  • 4. Na₂Cl₃
  • 5. The empirical formula obtained is not correct. This is likely due to experimental errors, since much precision was required (the masses are determined in thousandths of grams).

Explanation:

<em><u></u></em>

<em><u>1. How many moles of elemental sodium were used in the reaction?</u></em>

Since all of the solid sodium is used up by the reaction, you can cancluate the number of moles of elemental sodium used dividing the mass by the molar mass:

  • number of moles = mass in grams / atomic mass

  • mass in grams = 0.3870 g (given)

  • atomic mass = 22.990 g/mol

  • number of moles = 0.3870 g / 22.990 g/mol = 0.1683 mol

<u><em>2. What is the mass of chlorine gas used in the reaction?</em></u>

a) Mass of chlorine gas introduced in the flask = mass of the stoppered flask after filling it with chlorine gas - mass of the empty flask with the sopper

  • Mass of chlorine gas introduced = 158.1743g - 156.1870g = 1.9873 g

b) Mass of chlorine gas unreacted = 0.7962 g (given)

c) Mass of chlorine gas used = mass of chorine gas introduced in the flask - mass of chlorine gass un reacted

  • Mass of chlorine gas used = 1.9873g - 0.7962g = 1.1911g

<u><em>3. How many moles of chlorine were used in the reaction?</em></u>

  • molar mass of chlorine gas, Cl₂ = 2 × 35.453 g/mol = 70.906 g/mol

  • number of moles = mass in grams / molar mass = 1.911g / 70.906g/mol = 0.02695 mol

<u><em>4. What is the empirical formula of sodium chloride based on the experimental data?</em></u>

Divide the number of moles of each element by the smalles number of moles:

  • Na: 0.01683 / 0.01683 = 1
  • Cl = 0.02695 / 0.01683 = 1.6

Multiply by 2 to obtain whole numbers:

  • Na = 2
  • Cl = 3.2 = 3

  • Empirical formula Na₂Cl₃

<u><em></em></u>

<u><em>5. Was the empirical formula you obtained correct using the chemists data correct? Why? </em></u>

<u><em></em></u>

No, the empirical formula you obtained using the chemists data is not correct, because the correct empirical formula of sodium chloride is NaCl.

That is, there is 1 atom of sodium per every atom of chlorine in one chemical formula of NaCl, but that is not reflected by the empirical formula Na₂Cl₃.

That is a demostration of big experimental errors. You can speculate that the errors are likely due to problems of procedure collecting the gas or errors in measuring the masses.

As you see, the masses are measured to thousandths of grams, which requires much precision; thus smalls absolute errors could produce huge relative errors.

7 0
2 years ago
How do acids and bases affect molecules such Proteins? ​
Taya2010 [7]

Answer:

Strong acids and bases both denature proteins by severing disulphide bonds and at higher temperatures, can break proteins into peptides, or even individual amino acids.

5 0
2 years ago
the energy to break 1 mol of C-C bonds is 348 kj/mol. what would be the minimum frequency of a single photon that would break a
mr_godi [17]

Answer:

f = 5.25 x 10³⁸ s⁻¹

Explanation:

Energy = 348KJ = 348,000 J

Frequency = ?

Energy and frequency are related by the equation below;

E = hf

where h = Planck's constant =  (6.626 x 10-34 J · s),

Upon making f subject of formular;

f = E / h

substituting the values, we have;

f = 348,000 / 6.626 x 10-34

f = 52,520 x 10 ³⁴ s⁻¹

f = 5.25 x 10³⁸ s⁻¹

4 0
2 years ago
A chemist has 2.0 mol of methanol (CH3OH). The molar mass of methanol is 32.0 g/mol. What is the mass, in grams, of the sample?
rodikova [14]

Answer:

\boxed {\boxed {\sf D. \ 64 \ grams }}

Explanation:

Given the moles, we are asked to find the mass of a sample.

We know that the molar mass of methanol is 32.0 grams per mole. We can use this number as a fraction or ratio.

\frac{32 \ g \ CH_3OH}{1 \ mol \ CH_3OH}

Multiply by the given number of moles, which is 2.0

2.0 \ mol \ CH_3OH *\frac{32 \ g \ CH_3OH}{1 \ mol \ CH_3OH}

The moles of methanol will cancel each other out.

2.0 \ *\frac{32 \ g \ CH_3OH}{1 }

The denominator of 1 can be ignored.

2.0 * 32 \ g\ CH_3OH

Multiply.

64 \ g \ CH_3OH

There are 64 grams of methanol in the sample.

3 0
2 years ago
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