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mario62 [17]
3 years ago
6

How many atoms are in 11.5g of Hg

Chemistry
1 answer:
Shtirlitz [24]3 years ago
8 0
5. is your answer thank you!
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Can someone help me with this
Flura [38]

Answer:

I don't know how to get the answer sorry

Explanation:

db

6 0
3 years ago
If 3.289 x 10^23 atoms of potassium react with excess water, how many grams of hydrogen gas would be produced?
GarryVolchara [31]

Answer:

The amount in grams of hydrogen gas produced is 0.551 grams

Explanation:

The parameters given are;

Number of atoms of potassium, aₙ = 3.289 × 10²³ atoms

Chemical equation for the reaction is given as follows;

2K + 2H₂O \rightarrow KOH + H₂

Avogadro's number, N_A, regarding the number of molecules or atom per mole is given s follows;

N_A = 6.02 × 10²³ atoms/mole

Therefore;

The number of moles of potassium present = 3.289 × 10²³/(6.02 × 10²³) = 0.546 moles

2 moles of potassium produces one mole of hydrogen gas, therefore;

1 moles of potassium produces 1/2 mole of hydrogen gas, and 0.546 moles of potassium will produce 0.546/2 moles of hydrogen which is 0.273 moles of hydrogen gas

The molar mass of hydrogen gas = 2.016 grams

Therefore, 0.273 moles will have a mass of 0.273×2.016 = 0.551 grams.

The amount in grams of hydrogen gas produced = 0.551 grams.

8 0
3 years ago
A chlorine atom has a diameter of 175 picometers (pm). Please write the diameter, in metered, in scientific notation.
Flauer [41]
Picometre is equal to 1·10⁻¹²m
175·10⁻¹² m = 1,75·10⁻¹⁰ m
7 0
3 years ago
A sample of an ionic compound NaA, where A- is the anion of a
vitfil [10]

Answer:

a) Kb = 10^-9

b) pH = 3.02

Explanation:

a) pH 5.0 titration with a 100 mL sample containing 500 mL of 0.10 M HCl, or 0.05 moles of HCl. Therefore we have the following:

[NaA] and [A-] = 0.05/0.6 = 0.083 M

Kb = Kw/Ka = 10^-14/[H+] = 10^-14/10^-5 = 10^-9

b) For the stoichiometric point in the titration, 0.100 moles of NaA have to be found in a 1.1L solution, and this is equal to:

[A-] = [H+] = (0.1 L)*(1 M)/1.1 L = 0.091 M

pKb = 10^-9

Ka = 10^-5

HA = H+ + A-

Ka = 10^-5 = ([H+]*[A-])/[HA] = [H+]^2/(0.091 - [H+])

[H+]^2 + 10^5 * [H+] - 10^-5 * 0.091 = 0

Clearing [H+]:

[H+] = 0.00095 M

pH = -log([H+]) = -log(0.00095) = 3.02

6 0
3 years ago
Rewrite 76.00 to have 1 sig fig.
Luba_88 [7]

Answer:

76.00

sig fig:4

decmials:2

scientific notation:7.600 x 10 1

words: seventy-six

p.s that one is on the right top side of ten

Explanation:

4 0
4 years ago
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