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vlabodo [156]
1 year ago
10

The radius of an indium atom is 0.163 nm. If indium crystallizes in a face-centered unit cell, what is the length of an edge of

the unit cell
Chemistry
1 answer:
adoni [48]1 year ago
8 0

<u>Answer:</u> The edge length of the unit cell is 0.461 nm

<u>Explanation:</u>

We are given:

Atomic radius of iridium = 0.163 nm

To calculate the edge length, we use the relation between the radius and edge length for FCC lattice:

a=2\sqrt{2}R

Putting values in above equation, we get:

a=2\sqrt{2}\times 0.163=0.461nm

Hence, the edge length of the unit cell is 0.461 nm

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If you add 25.0 mL of water to 125 mL of a 0.150 M LiOH solution, what will be the molarity of the resulting diluted solution?
Alborosie

Concentration is the number of moles of solute in a fixed volume of solution

Concentration(c) = number of moles of solute(n) / volume of solution (v)

25.0 mL of water is added to 125 mL of a 0.150 M LiOH solution and solution becomes more diluted.

original solution molarity - 0.150 M

number of moles of LiOH in 1 L - 0.150 mol

number of LiOH moles in 0.125 L  - 0.150 mol/ L x 0.125 L = 0.01875 mol

when 25.0 mL is added the number of moles of LiOH will remain constant but volume of the solution increases

new volume -  125 mL + 25 mL = 150 mL

therefore new molarity is

c = 0.01875 mol / 0.150 L  = 0.125 M

answer is 0.125 M

7 0
2 years ago
The equilibrium reaction below has the Kc = 4.00 at 25°C. If the temperature of the system at equilibrium is increased to 100°C,
professor190 [17]

Answer:

Following are the answer to this question:

Explanation:

In the given question information is missing, that is equation which can be defined as follows:

CH_4(g)+H_2O(g)\rightarrow CO(g)+3H_2(g)   \ \ \bigtraingleup H^0=+206.2KJ

  • Growing temperatures may change its connection to just the way which consumes thermal energy in accordance with Le chatelier concepts Potential connection is endothermic. Answer: shifts to the right  
  • Kc are described as a related to the concentration by the intensity of both the reaction for each phrase which reaches a power equal towards its stoichiometric equation coefficient  Kc = \frac{product}{reactant}   It increases [product] but reduces [reactant] Therefore, Kc increases
6 0
2 years ago
A proton transfer reaction can occur when an aldehyde is placed in strong base, such as an alkoxide ion, producing an alcohol an
Pani-rosa [81]

Hi, you have not provided structure of the aldehyde and alkoxide ion.

Therefore i'll show a mechanism corresponding to the proton transfer by considering a simple example.

Explanation: For an example, let's consider that proton transfer is taking place between a simple aldehyde e.g. acetaldehyde and a simple alkoxide base e.g. methoxide.

The hydrogen atom attached to the carbon atom adjacent to aldehyde group are most acidic. Hence they are removed by alkoxide preferably.

After removal of proton from aldehyde, a carbanion is generated. As it is a conjugated carbanion therefore the negative charge on carbon atom can conjugate through the carbonyl group to form an enolate which is another canonical form of the carbanion.

All the structures are shown below.

7 0
2 years ago
the half life of the radioactive element strontium-90 is 29.1 years. If 16 grams of strontium-90 are initially present, how many
8_murik_8 [283]

Answer:

4 g after 58.2 years

0.0156 After 291 years

Explanation:

Given data:

Half-life of strontium-90 = 29.1 years

Initially present: 16g

mass present after 58.2 years =?

Mass present after 291 years =?

Solution:

Formula:

how much mass remains =1/ 2n (original mass) ……… (1)

Where “n” is the number of half lives

to find n

For 58.2 years

n = 58.2 years /29.1 years

n= 2

or  291 years

n = 291 years /29.1 years

n= 10

Put values in equation (1)

Mass after 58.2 years

mass remains =1/ 22 (16g)

mass remains =1/ 4 (16g)

 mass remains = 4g

Mass after 58.2 years

mass remains =1/ 210 (16g)

mass remains =1/ 1024 (16g)

mass remains = 0.0156g

8 0
2 years ago
Read 2 more answers
The volume of distilled water that should be added to 10.0 mL of 6.00 M HCl(aq) in order to prepare a 0.500 M HCl(aq) solution i
bija089 [108]

Answer:

110ml

Explanation:

<em>Using the dilution equation, C1V1 = C2V2</em>

<em>Where C1 is the initial concentration of solution</em>

<em>C2 is final concentration of solution</em>

<em>V1 is intital volume of solution</em>

<em>V2 is final volume of solution.</em>

From the question , C1=6M, C2=0.5M, V1=10ml, V2=?

V2 =\frac{C1V1}{C2}

V2 =\frac{10*6}{0.5}

V2 =120ml

volume of water added = final volume -initial volume

                                    = 120-10

                                   =110ml

3 0
2 years ago
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