1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Nutka1998 [239]
3 years ago
10

The two isotopes of chlorine are 3517Cl and 3717Cl. Which isotope is the most abundant?

Chemistry
1 answer:
nikklg [1K]3 years ago
6 0

Answer:

  • The most abundant isotope is ³⁵Cl₁₇.
  • Since the average atomic mass of Cl is 35.45 g/mole, so the most abundant one is that one near this number, ³⁵Cl₁₇.
  • ³⁵Cl₁₇ is found by 75 % and ³⁷Cl₁₇ is found by 25 % by a ratio 3:1.

You might be interested in
I need to know the element in group 15 in the periodic table .
gogolik [260]

Answer: 5

Explanation:

3 0
3 years ago
A small cube of iron and a large flat sheet of iron contain the same volume. Which one will completely rust first? Explain why.
allochka39001 [22]
Flat as more oxygen and water can react over it think of it like this would a cube rust faster than a sheet
6 0
3 years ago
This is something easy can someone please finish this one. i'll give brainliest.
Ainat [17]

Answer:

B B C A C B A A C C

Explanation:

jk it was a big guess

4 0
3 years ago
Read 2 more answers
100.00 mL of 0.15 M nitrous acid (HNO2) are titrated with a 0.15 M NaOH solution. (a) Calculate the pH for the initial solution.
wolverine [178]

Answer:

a. pH = 2.04

b. pH = 3.85

c. pH = 8.06

d. pH = 11.56

Explanation:

The nitrous acid is a weak acid (Ka = 5.6x10⁻⁴) that reacts with NaOH as follows:

HNO₂ + NaOH → NaNO₂(aq) + H₂O(l)

a. At the beginning there is just a solution of 0.12M HNO₂. As Ka is:

Ka = [H⁺] [NO₂⁻] / [HNO₂]

Where [H⁺] and [NO₂⁻] ions comes from the same equilibrium ([H⁺] = [NO₂⁻] = X):

5.6x10⁻⁴ = X² / 0.15M

8.4x10⁻⁵ = X²

X = [H⁺] = 9.165x10⁻³M

As pH = -log [H⁺]

<h3>pH = 2.04</h3><h3 />

b. At this point we have HNO₂ and NaNO₂ (The weak acid and the conjugate base), a buffer. The pH of a buffer is obtained using H-H equation:

pH = pKa + log [NaNO₂] / [HNO₂]

<em>Where pH is the pH of the buffer,</em>

<em>pKa is -log Ka = 3.25</em>

<em>And [NaNO₂] [HNO₂] could be taken as the moles of each compound.</em>

<em />

The initial moles of HNO₂ are:

0.100L * (0.15mol / L) = 0.015moles

The moles of base added are:

0.0800L * (0.15mol / L) = 0.012moles

The moles of base added = Moles of NaNO₂ produced = 0.012moles.

And the moles of HNO₂ that remains are:

0.015moles - 0.012moles = 0.003moles

Replacing in H-H equation:

pH = 3.25 + log [0.012moles] / [0.003moles]

<h3>pH = 3.85</h3><h3 />

c. At equivalence point all HNO2 reacts producing NaNO₂. The volume added of NaOH must be 100mL. That means the concentration of the NaNO₂ is:

0.15M / 2 = 0.075M

The NaNO₂ is in equilibrium with water as follows:

NaNO₂(aq) + H₂O(l) ⇄ HNO₂(aq) + OH⁻(aq) + Na⁺

The equilibrium constant, kb, is:

Kb = Kw/Ka = 1x10⁻¹⁴ / 5.6x10⁻⁴ = 1.79x10⁻¹¹ = [OH⁻] [HNO₂] / [NaNO₂]

<em>Where [OH⁻] = [HNO₂] = x</em>

<em>[NaNO₂] = 0.075M</em>

<em />

1.79x10⁻¹¹ = [X] [X] / [0.075M]

1.34x10⁻¹² = X²

X = 1.16x10⁻⁶M = [OH⁻]

pOH = -log [OH-] = 5.94

pH = 14-pOH

<h3>pH = 8.06</h3><h3 />

d. At this point, 5mL of NaOH are added in excess, the moles are:

5mL = 5x10⁻³L * (0.15mol / L) =7.5x10⁻⁴moles NaOH

In 100mL + 105mL = 205mL = 0.205L. [NaOH] = 7.5x10⁻⁴moles NaOH / 0.205L =

3.66x10⁻³M = [OH⁻]

pOH = 2.44

pH = 14 - pOH

<h3>pH = 11.56</h3>
5 0
2 years ago
Iodine-131 is administered orally in the form of NaI(aq) as a treatment for thyroid cancer. The half-life of iodine-131 is 8.04
evablogger [386]

Answer:

16.6 mg

Explanation:

Step 1: Calculate the rate constant (k) for Iodine-131 decay

We know the half-life is t1/2 = 8.04 day. We can calculate the rate constant using the following expression.

k = ln2 / t1/2 = ln2 / 8.04 day = 0.0862 day⁻¹

Step 2: Calculate the mass of iodine after 8.52 days

Iodine-131 decays following first-order kinetics. Given the initial mass (I₀ = 34.7 mg) and the time elapsed (t = 8.52 day), we can calculate the mass of iodine-131 using the following expression.

ln I = ln I₀ - k × t

ln I = ln 34.7 - 0.0862 day⁻¹ × 8.52 day

I = 16.6 mg

8 0
3 years ago
Other questions:
  • How many times faster will oxygen gas diffuse than carbon dioxide
    10·1 answer
  • Avoltic cell prepared using aluminium and nickle has the flowing cell notation ​
    6·1 answer
  • For a 1.00 m ( 100 cm) crumble zone, how much deformation do you think is needed in order to keep the passenger the safest? Expl
    12·1 answer
  • What is the effect of day lenght in plant growth? dependent and independent
    8·1 answer
  • Based on VSEPR theory and your observations from the Molecular Geometry lab consider the following questions What is the predict
    11·1 answer
  • Which type of wind is responsible for moving state to state
    10·1 answer
  • BRAINLIST!!
    6·2 answers
  • Purpose Question:
    14·2 answers
  • At the end we stopped the distillation with some material left in the vial. Why is it considered unsafe to distill until dryness
    8·1 answer
  • There are three different dichloroethylenes (molecular formula C₂H₂Cl₂), which we can designate X, Y, and Z. Compound X has no d
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!