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MatroZZZ [7]
3 years ago
6

What makes an ionic difference than a covalent bond

Chemistry
1 answer:
Eva8 [605]3 years ago
3 0
I’m not completely sure what you mean here but if your talking about charges in covalently bonded molecules then here is something:

Charges in covalent bonds are partial because the ions are shared when bonding.

The charges are no longer full and strong they are half charges, they are not as strong.

You might want to just google a little bit. :)

Hope that helps
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Combustion of hydrocarbons such as butane () produces carbon dioxide, a "greenhouse gas." Greenhouse gases in the Earth's atmosp
Dahasolnce [82]

Answer:

1. C₄H₁₀ + ¹³/₂O₂ → 4CO₂ + 5H₂O

2. V = 596L

Explanation:

Butane (C₄H₁₀) reacts with oxygen (O₂) to produce carbon dioxide (CO₂) and water (H₂O) thus:

C₄H₁₀ + O₂ → CO₂ + H₂O

1. The balanced chemical equation is:

C₄H₁₀ + ¹³/₂O₂ → 4CO₂ + 5H₂O

2. 0,360kg of butane are:

360g×\frac{1mol}{58,12g}=<em>6,19moles of butane</em>

These moles of butane are:

6,19moles of butane×\frac{4CO_2}{1molButane}= <em>24,8 moles CO₂</em>

Using V=nRT/P

Where:

n are moles (24,8 moles CO₂); R is gas constant (0,082atmL/molK); T is temperature, 20°C (293,15K); and P is pressure (1atm).

Volume (V) is:

<em>V = 596L</em>

I hope it helps!

8 0
3 years ago
What best describes the degree to which a material can transmit Heat A:melting point B:boiling point C: thermal conductivity D:
Oduvanchick [21]
Your answer is going to be C
7 0
3 years ago
Read 2 more answers
Someone answer the limited regent...
OleMash [197]

Answer:

20 g Ag

General Formulas and Concepts:

<u>Chemistry - Stoichiometry</u>

  • Using Dimensional Analysis

<u>Chemistry - Atomic Structure</u>

  • Reading a Periodic Table

Explanation:

<u>Step 1: Define</u>

[RxN]   Cu (s) + AgNO₃ (aq) → CuNO₃ (aq) + Ag (s)

[Given]   10 g Cu

<u>Step 2: Identify Conversions</u>

[RxN]   1 mol Cu = 1 mol Ag

Molar Mass of Cu - 63.55 g/mol

Molar Mass of Ag - 197.87 g/mol

<u>Step 3: Stoichiometry</u>

<u />10 \ g \ Cu(\frac{1 \ mol \ Cu}{63.55 \ g \ Cu})(\frac{1 \ mol \ Ag}{1 \ mol \ Cu} )(\frac{197.87 \ g \ Ag}{1 \ mol \ Ag} ) = 16.974 g Ag

<u>Step 4: Check</u>

<em>We are given 1 sig fig. Follow sig fig rules and round.</em>

16.974 g Ag ≈ 20 g Ag

6 0
3 years ago
The elements silicon (atomic number 14) and chlorine (atomic number 17) are both in Period 3. Which is the more reactive element
Svetlanka [38]

Answer;

-Chlorine is more reactive than silicon

Explanation;

-As you move across a period, the nuclear charge will increase; the number of energy levels will stay the same, so there is a stronger and stronger attraction for the electrons.

-The electrons are being held more tightly as you move across a period. It becomes more and more difficult to lose electrons and consequently the reactivity of non metals increases as you go from left to right across the periodic table; Therefore; chlorine is more reactive than silicon.

5 0
3 years ago
Read 2 more answers
Calculate the unit cell edge length for an 78 wt% Ag- 22 wt% Pd alloy. All of the palladium is in solid solution, and the crysta
BaLLatris [955]

Answer:

The unit cell edge length for the alloy is 0.405 nm

Explanation:

Given;

concentration of Ag, C_{Ag} = 78 wt%

concentration of Pd, C_P_d = 22 wt%

density of Ag = 10.49 g/cm³

density of Pd = 12.02 g/cm³

atomic weight of Ag, A_A_g = 107.87 g/mol

atomic weight of iron, A_P_d = 106.4 g/mol

Step 1: determine the average density of the alloy

\rho _{Avg.} = \frac{100}{\frac{C_A_g}{\rho _A_g} + \frac{C__{Pd}}{\rho _{Pd}} }

\rho _{Avg.} = \frac{100}{\frac{78}{10.49} + \frac{22}{12.02} } = 10.79 \ g/cm^3

Step 2: determine the average atomic weight of the alloy

A _{Avg.} = \frac{100}{\frac{C_v}{A _A_g} + \frac{C__{Pd}}{A _{Pd}} }

A _{Avg.} = \frac{100}{\frac{78}{107.87} + \frac{22}{106.4} } = 107.54 \ g/mole

Step 3: determine unit cell volume

V_c=\frac{nA_{avg.}}{\rho _{avg.} N_a}

for a FCC crystal structure, there are 4 atoms per unit cell; n = 4

V_c=\frac{4*107.54}{ 10.79*6.022*10^{23}} = 6.620*10^{-23} \ cm^3/cell

Step 4: determine the unit cell edge length

Vc = a³

a = V_c{^\frac{1}{3} }\\\\a = (6.620*10^{-23}}){^\frac{1}{3}}\\\\a = 4.05 *10^{-8} \ cm= 0.405 nm

Therefore, the unit cell edge length for the alloy is 0.405 nm

7 0
3 years ago
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