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Romashka-Z-Leto [24]
3 years ago
6

How many valence electrons does an atom of Iodine have? a) 5 b) 6 c) 7 d) 8

Chemistry
2 answers:
baherus [9]3 years ago
8 0
The answer is 7 !! :)
BabaBlast [244]3 years ago
7 0

Answer: c) 7

Explanation:

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I’m driving down the highway at 75.0 miles/hour, how fast is that in m/seconds?
Aliun [14]

Answer:

1.25 miles/ 1 minute

60x1.25=7.50

Explanation:

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4 years ago
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Advocard [28]

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Explanation:

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You could add solid KCl to the solution to precipitate out AgCl(s). What mass of KCl is needed to precipitate the silver ions fr
padilas [110]

Answer:

0.143 g of KCl.

Explanation:

Equation of the reaction:

AgNO3(aq) + KCl(aq) --> AgCl(s) + KNO3(aq)

Molar concentration = mass/volume

= 0.16 * 0.012

= 0.00192 mol AgNO3.

By stoichiometry, 1 mole of AgNO3 reacts with 1 mole of KCl to form a precipitate.

Number of moles of KCl = 0.00192 mol.

Molar mass of KCl = 39 + 35.5

= 74.5 g/mol

Mass = molar mass * number of moles

= 74.5 * 0.00192

= 0.143 g of KCl.

4 0
3 years ago
Question 5 of 10
pychu [463]
The answer is D. S2O6
4 0
3 years ago
2NO (g) + O2 (g) →2NO2 (g) At equilibrium [NO] = 2.4 × 10 -3 M, [O2] = 1.4 × 10 -4 M, and [NO2] = 0.95 M.
azamat

Answer:

K=1.12x10^9

Explanation:

Hello there!

Unfortunately, the question is not given in the question; however, it is possible for us to compute the equilibrium constant as the problem is providing the concentrations at equilibrium. Thus, we first set up the equilibrium expression as products/reactants:

K=\frac{[NO_2]^2}{[NO]^2[O_2]}

Then, we plug in the concentrations at equilibrium to obtain the equilibrium constant as follows:

K=\frac{(0.95)^2}{(0.0024)^2(0.00014)}\\\\K=1.12x10^9

In addition, we can infer this is a reaction that predominantly tends to the product (NO2) as K>>>>1.

Best regards!

4 0
3 years ago
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