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4 years ago

Why can the positive ions be considered to be fixed during the electrons’ oscillations?

Chemistry

1 answer:

spayn [35]4 years ago

5 0

Answer:

An ion is thousands of times more massive than an electron.

Explanation:

positive ions can be considered to be fixed during the electrons’ oscillations as an ion is thousands of times more massive than an electron and will not respond quickly due to large inertia.

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Use the periodic table to identify the element indicated by each electron configuration by typing in the chemical symbol for the

Schach [20]

Answer:

1s22s22p6: Ne 1s22s22p63s23p3: P 1s22s22p63s23p64s1: K 1s22s22p63s23p64s23d8: Ni 1s22s22p63s23p64s23d104p65s24d3: Nb

6 0

2 years ago

Calculate the mass of water produced when 1.36 g of butane reacts with excess oxygen.

ANTONII [103]

Answer:

Mass of water produced= 1.8 g

Explanation:

Given data:

Mass of water produced = ?

Mass of butane = 1.36 g

Mass of oxygen = excess

Solution:

Chemical equation:

2C₄H₁₀ +13 O₂ → 8CO₂ + 10H₂O

Number of moles of butane:

Number of moles = mass/molar mass

Number of moles = 1.36 g/ 58.12 g/mol

Number of moles = 0.02 mol

Now we will compare the moles of butane with water.

C₄H₁₀ : H₂O

2 : 10

0.02 : 10/2×0.02 = 0.1 mol

Mass of water produced:

Mass = molar mass × molar mass

Mass = 0.1 mol × 18 g/mol

Mass = 1.8 g

6 0

3 years ago

Which is the best description of a chain of custody?

leonid [27]

D) A log of who has handled the evidence and when so the integrity of the evidence is upheld.

7 0

3 years ago

Cyclopropane, a substance used with oxygen as a general anesthetic, contains only two elements, carbon and hydrogen. When 1.00 g

ELEN [110]

Answer:

CH₂

Explanation:

From the question given above, the following data were obtained:

Mass of compound = 1 g

Mass of CO₂ = 3.14 g

Mass of H₂O = 1.29 g

Empirical formula =?

Next, we shall determine the mass of Carbon and hydrogen present in the compound. This can be obtained as follow:

For Carbon, C:

Mass of CO₂ = 3.14 g

Molar mass of CO₂ = 12 + (2×16)

= 12 + 32

= 44 g/mol

Molar mass of C = 12 g/mol

Mass of C =?

Mass of C = molar mass of C/ Molar mass of CO₂ × Mass of CO₂

Mass of C = 12/44 × 3.14

Mass of C = 0.86 g

For hydrogen, H:

Mass of C = 0.86 g

Mass of compound = 1 g

Mass of H =?

Mass of H = (Mass of compound) – (mass of C)

Mass of H = 1 – 0.86

Mass of H = 0.14 g

Finally, we shall determine the empirical formula of the cyclopropane. This can be obtained as follow:

Mass of C = 0.86 g

Mass of H = 0.14 g

Divide by their molar mass

C = 0.86 / 12 = 0.07

H = 0.14 / 1 = 0.14

Divide by the smallest

C = 0.07 / 0.07 = 1

H = 0.14 / 0.07 = 2

Thus, the empirical formula of cyclopropane is CH₂

8 0

3 years ago

Consider the balanced equation for the following reaction:

Zanzabum

Answer: The amount of carbon dioxide formed in the reaction is 5.663 grams

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of oxygen gas = 8 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{8g}{32g/mol}=0.25mol$

For the given chemical equation:

$7O_2(g)+2C_2H_6(g)\rightarrow 4CO_2(g)+6H_2O(l)$

By Stoichiometry of the reaction:

7 moles of oxygen gas produces 4 moles of carbon dioxide

So, 0.25 moles of oxygen gas will produce = $\frac{4}{7}\times 0.25=0.143mol$ of carbon dioxide

Now, calculating the mass of carbon dioxide from equation 1, we get:

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = 0.143 moles

Putting values in equation 1, we get:

$0.143mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=(0.143mol\times 44g/mol)=6.292g$

To calculate the experimental yield of carbon dioxide, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Percentage yield of carbon dioxide = 90 %

Theoretical yield of carbon dioxide = 6.292 g

Putting values in above equation, we get:

$90=\frac{\text{Experimental yield of carbon dioxide}}{6.292g}\times 100\\\\\text{Experimental yield of carbon dioxide}=\frac{90\times 6.292}{100}=5.663g$

Hence, the amount of carbon dioxide formed in the reaction is 5.663 grams

7 0

3 years ago