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Usimov [2.4K]
3 years ago
12

PLEASE HELP

Chemistry
2 answers:
statuscvo [17]3 years ago
7 0

Answer:

The Answer is Protons C.

Explanation:

Its quite simple, u see, according to my calculations its c.

Sergio [31]3 years ago
5 0

Answer:

C. The Protons

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8). How many moles of oxygen gas must be placed in a 35.0 L container in order
klemol [59]

Answer:

4.1 moles

Explanation:

Applying

PV = nRT................ equation 1

Where P = pressure, V = volume, n = number of moles, R = molar gas constant, T = Temperature.

make n the subject  of the equation

n = PV/RT.............. Equation 2

From the question,

Given: V = 35 L , P = 2.8 atm, T = 15 °C  = (15+273) = 288 K, R = 0.083 L.atm/K.mol

Substitute these values into equation 2

n = (35×2.8)/(0.083×288)

n = 4.1 moles

5 0
3 years ago
Hafnium has six naturally occurring isotopes: 0.16% of 174Hf, with an atomic weight of 173.940 amu; 5.26% of 176Hf, with an atom
ser-zykov [4K]

Answer:

177.277amu

Explanation:

the total occuring isotopes for Hafnium is =6.

First isotope had an atomic weight of 173.940amu

Second isotope =175.941amu

Third isotope =176.943amu

Fourth isotope=177.944amu

Fifth isotope. =178.946amu

sixth isotope .179.947amu

<em>Avera</em><em>ge</em><em> </em><em>ato</em><em>mic</em><em> </em><em>wei</em><em>ght</em><em> </em><em>of</em><em> </em><em>Haf</em><em>nium</em><em>=</em><em> </em><em>sum</em><em> </em><em>of</em><em> </em><em>all</em><em> </em><em>the </em><em>atomi</em><em>c</em><em> </em><em>weights</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>iso</em><em>topes</em><em>/</em><em> </em><em>Tota</em><em>l</em><em> </em><em>occu</em><em>ring</em><em> </em><em>isotopes</em>

Thus, 173.940amu+175.941amu+176.943amu+177.944amu+178.946amu+179.947amu.= 1063.661amu

Average atomic weight= 1063.661amu /6 = 177.2768333amu

= 177.277amu to 3 decimal places.

6 0
3 years ago
How do chemists solve problems?
Westkost [7]

Answer:Chemistry problems can be solved using a variety of techniques.

Explanation:  Many chemistry teachers and most introductory chemistry texts illustrate problem solutions using the factor-label method. ... The use of analogies and schematic diagrams results in higher achievement on problems involving moles, stoichiometry, and molarity. Hope this helped!

4 0
3 years ago
200.0 mL of 3.85 M HCl is added to 100.0 mL of 4.6 M barium hydroxide. The reaction goes to completion. What is the concentratio
Ede4ka [16]

Answer:

2.387 mol/L

Explanation:

The reaction that takes place is:

  • 2HCl + Ba(OH)₂ → BaCl₂ + 2H₂O

First we <u>calculate how many moles of each reagent were added</u>:

  • HCl ⇒ 200.0 mL * 3.85 M = 203.85 mmol HCl
  • Ba(OH)₂ ⇒ 100.0 mL * 4.6 M = 460 mmol Ba(OH)₂

460 mmol of Ba(OH)₂ would react completely with (2*460) 920 mmol of HCl. There are not as many mmoles of HCl so Ba(OH)₂ will remain in excess.

Now we <u>calculate how many moles of Ba(OH)₂ reacted</u>, by c<em>onverting the total number of HCl moles to Ba(OH)₂ moles</em>:

  • 203.85 mmol HCl * \frac{1mmolBa(OH)_{2}}{2mmolHCl}= 101.925 mmol Ba(OH)₂

This means the remaining Ba(OH)₂ is:

  • 460 mmol - 101.925 mmol = 358.075 mmoles Ba(OH)₂

There are two OH⁻ moles per Ba(OH)₂ mol:

  • OH⁻ moles = 2 * 358.075 = 716.15 mmol OH⁻

Finally we <u>divide the number of OH⁻ moles by the </u><u><em>total</em></u><u> volume</u> (100 mL + 200 mL):

  • 716.15 mmol OH⁻ / 300.0 mL = 2.387 M

So the answer is 2.387 mol/L

7 0
3 years ago
Help!!! chemistry electrolysis
Tpy6a [65]
Electrolysis can be used to separate a substance into its original components/elements and it was through this process that a number of elements have been discovered and are still produced in today's industry.
5 0
3 years ago
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