answer.
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
Tcecarenko
2 months ago
11

How many atoms of zirconium are in 0.3521 mol of zirconium?

Chemistry
1 answer:
KiRa [2.9K]2 months ago
3 0

Jawaban:

2.12×10²³ atom.

Penjelasan:

Dari hipotesis Avogadro, kita belajar bahwa 1 mol dari suatu zat mengandung 6.02×10²³ atom. Ini berarti bahwa 1 mol zirconium juga 6.02×10²³ atom.

Dengan demikian, kita dapat menghitung jumlah atom dalam 0.3521 mol zirconium sebagai berikut:

1 mol zirconium juga 6.02×10²³ atom.

Jadi, 0.3521 mol zirconium akan mengandung = 0.3521 × 6.02×10²³ = 2.12×10²³ atom.

Artinya, 0.3521 mol zirconium memiliki 2.12×10²³ atom.

You might be interested in
The standard cell potential (E°cell) for the reaction below is +0.63 V. The cell potential for this reaction is ________ V when
castortr0y [3046]

Answer: The cell potential for the given reaction stands at 0.50 V

Explanation:

The provided cell reaction is:

Pb^{2+}(aq)+Zn(s)\rightarrow Zn^{2+}(aq)+Pb(s)

The half-reactions are:

Anode oxidation half reaction:  Zn\rightarrow Zn^{2+}+2e^-

Cathode reduction half reaction:  Pb^{2+}+2e^-\rightarrow Pb

Initially, we need to find the cell potential for this reaction.

Utilizing the Nernst equation:

E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Zn^{2+}]}{[Pb^{2+}]}

where,

F = Faraday's constant = 96500 C

R = gas constant = 8.314 J/mol·K

T = room temperature = 25^oC=273+25=298K

n = electrons exchanged in oxidation-reduction = 2

E^o_{cell} = standard electrode potential for the cell = +0.63 V

E_{cell} = cell potential for the reaction =?

[Zn^{2+}] = concentration of Zn²⁺ = 3.5 M

[Pb^{2+}] = 2.0\times 10^{-4}M

Now substituting all known values into the equation, we arrive at:

E_{cell}=(+0.63)-\frac{2.303\times (8.314)\times (298)}{2\times 96500}\log \frac{3.5}{2.0\times 10^{-4}}

E_{cell}=0.50V

So, the resulting cell potential for this reaction is 0.50 V

5 0
28 days ago
The researcher performed a follow-up experiment to measure the rate of oxygen consumption by muscle and brain cells. Predict the
Alekssandra [3086]

Answer:

Mitochondria are plentiful in mammalian cells, with their proportions varying across different tissues, from less than 1% in white blood cells to as high as 35% in heart muscle cells. It is essential to understand that mitochondria are not static structures but instead form a dynamic network that frequently undergoes processes of fission and fusion. In skeletal muscle, they exist as part of a reticular membrane network. The two subpopulations, subsarcolemmal (SS) and intermyofibrillar (IMF) mitochondria, occupy different subcellular regions and exhibit slight differences in their biochemical and functional characteristics tied to their anatomical context. The SS mitochondria are positioned just beneath the sarcolemma, while IMF mitochondria are found closely associated with myofibrils. Their distinct properties likely play a role in their adaptability. SS mitochondria make up about 10-15% of the total mitochondrial volume and are believed to be more adaptable than their IMF counterparts, despite the latter displaying higher levels of protein synthesis, enzyme activity, and respiration (1).

Explanation:

0 0
2 months ago
Read 2 more answers
How many milliliters of 0.200 M NH4OH are needed to react with 12.0 mL of 0.550 M FeCl3?
eduard [2782]

Response:

9.9 ml of 0.200M NH₄OH(aq)

Reasoning:

3NH₄OH(Iaq) + FeCl₃(aq) => NH₄Cl(aq) + Fe(OH)₃(s)

What volume in ml of 0.200M NH₄OH(aq) will fully react with 12ml of 0.550M FeCl₃(aq)?

1 x Molarity of NH₄OH x Volume of NH₄OH Solution(L) = 2 x Molarity of FeCl₃ x Volume of FeCl₃ Solution

1(0.200M)(Volume of NH₄OH Soln) = 3(0.550M)(0.012L)

=> Volume of NH₄OH Soln = 3(0.550M)(0.012L)/1(0.200M) = 0.0099 Liters = 9.9 milliliters

5 0
2 months ago
How many sulfur atoms are present in 25.6 g of Al2(S2O3)3
castortr0y [3046]

Given the mass of Al_{2}(S_{2}O_{3})_{3}=25.6 g

The molar mass of Al_{2}(S_{2}O_{3})_{3}=390.35g/mol

Conversion of mass of Al_{2}(S_{2}O_{3})_{3} to moles:

25.6 g Al_{2}(S_{2}O_{3})_{3}*\frac{1molAl_{2}(S_{2}O_{3}}{390.35 gAl_{2}(S_{2}O_{3}} =0.0656molAl_{2}(S_{2}O_{3}

Conversion of moles Al_{2}(S_{2}O_{3})_{3} to moles of sulfur:

0.0656mol Al_{2}(S_{2}O_{3})_{3}*\frac{6molS}{1mol Al_{2}(S_{2}O_{3})_{3}}=0.3936 molS

Conversion of moles of sulfur to individual sulfur atoms using Avogadro's number:

1 mol = 6.022*10^{23}atoms

0.3936mol S *\frac{6.022*10^{23}atoms S}{1 mol S}=2.37*10^{23} S atoms

5 0
1 month ago
Los automóviles actuales tienen “parachoques de 5 mi/h (8 km/h)” diseñados para comprimirse y rebotar elásticamente sin ningún d
KiRa [2933]

Response: k = 23045 N/m

Clarification:

To determine the spring constant, one must consider the maximum elastic potential energy that the spring can withstand. The kinetic energy of the vehicle should equal at minimum the elastic potential energy of the spring when it is fully compressed. Hence, we express it as:

K=U\\\\\frac{1}{2}Mv^2=\frac{1}{2}kx^2    (1)

M: mass of the vehicle = 1050 kg

k: spring constant =?

v: car speed = 8 km/h

x: maximum spring compression = 1.5 cm = 0.015m

You need to resolve equation (1) for k. Beforehand, convert the speed v to meters per second:

v=8\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s}=2.222\frac{m}{s}

k=\frac{Mv^2}{x^2}=\frac{(1050kg)(2.222m/s)^2}{(0.015m)^2}=23045\frac{N}{m}

The spring constant calculates to 23045 N/m

3 0
1 month ago
Other questions:
  • Suppose you need of Grade 70 tow chain, which has a diameter of and weighs , to tow a car. How would you calculate the mass of t
    11·1 answer
  • Assuming complete dissociation of the solute, how many grams of KNO3 must be added to 275 mL of water to produce a solution that
    12·1 answer
  • Predict what will observe in below mention experiment.
    10·1 answer
  • How many molecules of ethane (C2H6) are present in 0.334 g of C2H6?
    10·1 answer
  • The discharge of chromate ions (CrO42-) to sewers or natural waters is of concern because of both its ecological impacts and its
    5·1 answer
  • Which example best demonstrates stewardship of the atmosphere
    8·1 answer
  • The properties of two elements are listed below. Element Atomic radius (pm) Ionic radius (pm) First ionization energy (kJ/mol) E
    7·2 answers
  • The atomic weight of iodine is less than the atomic weight of tellurium. However, Mendeleev listed iodine after tellurium in his
    10·1 answer
  • Element X reacts with element Y to give a product containing X3+ ions and Y2− ions.
    10·1 answer
  • How many moles are 25.0 mL of benzene (density 0.876 g/mL)?
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!