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andre [41]
3 years ago
15

How many electrons does each gold atom gain? How many electrons does each iodine atom lose? What is the total number of electron

s that are moved in the oxidation-reduction reaction? Complete the final balanced equation based on the half-reactions.
Chemistry
2 answers:
gulaghasi [49]3 years ago
7 0

Answer:

3 , 1 , 6 , 2 , 6 , 2 , 3 , 6

forsale [732]3 years ago
6 0

Answer:

Explanation:

Half equation:

I. Au3+ + 3e- --> Au

II. 2I- --> I2 + 2e-

Au3+ + 3e- --> Au. ×2

2I- --> I2 + 2e- ×3

2Au3+ + 6e- --> 2Au

6I- --> 3I2 + 6e-

Overall equation

AuI3 --> 2Au + 3I2

A.

3 electrons

B.

2 electrons

C.

6 electrons

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What type of elements form metallic bonds?​
astra-53 [7]

Answer:

Aluminum , iron , gold , silver , nickels etc are the type of elements form metallic bonds .

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3 years ago
1
Alex Ar [27]
River Banks? I'm not completely sure but hope this helps!
6 0
3 years ago
Read 2 more answers
you have been given 250.0 mL of an aqueous solution containing 18.7grams of AgNO3. What is the molarity of solution?
Reil [10]

Hey there!:

Molar mass AgNO3 = 169.87 g/mol

Number of moles:

moles of solution = mass of solute / molar mass

moles of solution = 18.7 / 169.87

moles of solution = 0.110084 moles of AgNO3

Volume in liters:

250.0 mL / 1000 => 0.25 L

Therefore:

Molarity = moles of solution / Volume of solution ( L )

Molarity = 0.110084 / 0.25

=> 0.440 M

Hope that helps!

8 0
3 years ago
A sample of 0.3283 g of an ionic compound containing the bromide ion (Br−) is dissolved in water and treated with an excess of A
Pavel [41]

Answer:

92.49 %

Explanation:

We first calculate the number of moles n of AgBr in 0.7127 g

n = m/M where M = molar mass of AgBr = 187.77 g/mol and m = mass of AgBr formed = 0.7127 g

n = m/M = 0.7127g/187.77 g/mol = 0.0038 mol

Since 1 mol of Bromide ion Br⁻ forms 1 mol AgBr, number of moles of Br⁻ formed = 0.0038 mol and

From n = m/M

m = nM . Where m = mass of Bromide ion precipitate and M = Molar mass of Bromine = 79.904 g/mol

m = 0.0038 mol × 79.904 g/mol = 0.3036 g

% Br in compound = m₁/m₂ × 100%

m₁ = mass of Br in compound = m = 0.3036 g (Since the same amount of Br in the compound is the same amount in the precipitate.)

m₂ = mass of compound = 0.3283 g

% Br in compound = m₁/m₂ × 100% = 0.3036/0.3283 × 100% = 0.9249 × 100% = 92.49 %

4 0
3 years ago
Calculate the ph of a solution in which [oh–] = 4.5 × 10–9m.
frosja888 [35]

pOH = -LOG([OH])  

pOH = -LOG(4.5*10^-9)  

pOH = 8.34  

pH + pOH = 14  

pH = 14 - 8.34 = 5.65

hope this helps!

(:

5 0
3 years ago
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