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3 years ago

Bonds in which atoms share electrons are defined as _____ bonds.

Chemistry

1 answer:

Kaylis [27]3 years ago

5 0

Ionic, think about the way metals share electrons with other metals, resulting in malleability.

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Can you help me with this ?

kherson [118]

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the support web thing, i'm confused with. But I hope this helped.

8 0

2 years ago

What mass of CaCO3 is required to react completely with 0.56 L of HCl?

Lina20 [59]

The balanced equation for the above reaction is as follows;
CaCO₃ + 2HCl ----> CaCl₂ + H₂O + CO₂
stoichiometry of CaCO₃ to HCl is 1:2
molar volume states that 1 mol of any gas occupies a volume of 22.4 L at STP.
volume of 22.4 L occupied by 1 mol
therefore 0.56 L occupied by - 0.56 L / 22.4 L/mol = 0.025 mol
number of HCl moles reacted - 0.025 mol
2 mol of HCl reacts with 1 mol of CaCO₃
therefore 0.025 mol reacts with - 0.025/2 = 0.0125 mol
mass of CaCO₃ required - 0.0125 mol x 100 g/mol = 1.25 g
1.25 g of CaCO₃ is required

8 0

2 years ago

Ag2S + Al(s) = Al2S3 + Ag(s) (unbalanced)

Dovator [93]

Answer:

1. 0.97 V

2. $Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/Ag_(_s_)$

Explanation:

In this case, we can start with the half-reactions:

$Ag^+~_(_a_q_)->~Ag_(_s_)$

$Al_(_s_)~->~Al^+^3~_(_a_q_)$

With this in mind we can add the electrons:

$Ag^+~_(_a_q_)+~e^-~->~Ag_(_s_)$ Reduction

$Al_(_s_)~->~Al^+^3~_(_a_q_)+~3e^-~$ Oxidation

The reduction potential values for each half-reaction are:

$Ag_2S~+~e^-~->~Ag_(_s_)~+~S^-^2~_(_a_q_)$ - 0.69 V

$Al^+^3~_(_a_q_)+~2e^-~->~Al_(_s_)$ -1.66 V

In the aluminum half-reaction, we have an oxidation reaction, therefore we have to flip the reduction potential value:

$Al_(_s_)~->~Al^+^3~+~2e^-~$ +1.66 V

Finally, to calculate the overall potential we have to add the two values:

1.66 V - 0.69 V = 0.97 V

For the second question, we have to keep in mind that in the cell notation we put the anode (the oxidation half-reaction) in the left and the cathode (the reduction half-reaction) in the right. Additionally, we have to use "//" for the salt bridge, therefore:

$Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/~Ag_(_s_)$

I hope it helps!

3 0

3 years ago

How many moles of gold, Au, are in 3.60 x 10^-5 g of gold?

zlopas [31]

<h3>Answer:</h3>

1.83 × 10⁻⁷ mol Au

<h3>General Formulas and Concepts:</h3>

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Chemistry

Atomic Structure

Reading a Periodic Table
Using Dimensional Analysis

<h3>Explanation:</h3>

Step 1: Define

3.60 × 10⁻⁵ g Au (Gold)

Step 2: Identify Conversions

Molar Mass of Au - 196.97 g/mol

Step 3: Convert

Set up: $\displaystyle 3.60 \cdot 10^{-5} \ g \ Au(\frac{1 \ mol \ Au}{196.97 \ g \ Au})$
Multiply: $\displaystyle 1.82769 \cdot 10^{-7} \ mol \ Au$

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

1.82769 × 10⁻⁷ mol Au ≈ 1.83 × 10⁻⁷ mol Au

4 0

2 years ago

Which of these is true for natural selection but not artificial selection?

ElenaW [278]

Can you show a picture so I can help??? :)

8 0

2 years ago