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1 month ago

Why can metallic elements bend, while solid but non-metallic break or crumble?

2 answers:

6 0

Metallic elements possess flexibility because their atomic structure allows atoms to move over one another without breaking the metallic bonds (this property leads to higher melting and boiling points); in contrast, nonmetals form atomic bonds that do not allow this kind of movement.

Anarel [2.9K]1 month ago

5 0

Metallic elements demonstrate ductility, enabling them to bend. This property means that a solid material can elongate when tensile stress is applied. Ductile materials can potentially be drawn into wires. Additionally, these materials often exhibit malleability.

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You want to prepare a solution with a concentration of 200.0μM from a stock solution with a concentration of 500.0mM. At your di

lorasvet [2795]

Answer:

1) This dilution plan will yield a 200μM solution.

2) This dilution plan will not yield a 200μM solution.

3) This dilution plan will not yield a 200μM solution.

4) This dilution plan will yield a 200μM solution.

5) This dilution plan will yield a 200μM solution.

Explanation:

Convert the initial molarity into molar form as shown.

$500mM = 500mM \times (\frac{1M}{1000M})= 0.5M$

Let's examine the following serial dilution processes.

Dilute 5.00 mL of the stock solution to 500 mL. Then take 10.00 mL of this new solution and dilute it further to 250 mL.

Concentration of 500 mL solution:

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5.00mL)}{500 mL}= 5 \times 10^{-3}M$

10 mL of this solution is further diluted to 250 mL

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(5 \times 10^{-3}M)(10.0mL)}{250 mL}= 2 \times 10^{-4}M$

Convert μM:

$2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M$

Thus, this dilution scheme will yield a 200μM solution.

Dilute 5.00 mL of the stock solution to 100 mL. Then take 10.00 mL of this new solution and dilute to 1000 mL.

Concentration of 100 mL solution:

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5.00mL)}{100 mL}= 2.5 \times 10^{-2}M$

10 mL of this solution is further diluted to 1000 mL

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(2.5 \times 10^{-2}M)(10.0mL)}{1000 mL}= 2.5 \times 10^{-4}M$

Convert μM:

$2.5 \times 10^{-4}M = (2.5 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 250 \mu M$

Thus, this dilution scheme will not yield a 200μM solution.

Dilute 10.00 mL of the stock solution to 100 mL, followed by diluting 5 mL of that new solution to 100 mL.

Concentration of 100 mL solution:

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(10mL)}{100 mL}= 0.05M$

5 mL of this solution is diluted to 1000 mL

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.05M)(5mL)}{1000 mL}= 0.25 \times 10^{-4}M$

Convert μM:

$0.25 \times 10^{-4}M = (0.25 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 25 \mu M$

Thus, this dilution scheme will not yield a 200μM solution.

Dilute 5 mL of the stock solution to 250 mL. Then take 10 mL of this new solution and further dilute it to 500 mL.

Concentration of 250 mL solution:

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5mL)}{250 mL}= 0.01M$

10 mL of this solution is further diluted to 500 mL

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.01M)(10mL)}{500 mL}= 2 \times 10^{-4}M$

Convert μM:

$2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M$

Thus, this dilution scheme will yield a 200μM solution.

Dilute 10 mL of the stock solution to 250 mL. Then take another 10 mL of this new solution and dilute it to 1000 mL.

Concentration of 250 mL solution:

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(10mL)}{250 mL}= 0.02M$

10 mL of this solution is further diluted to 1000 mL

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.02M)(10mL)}{1000 mL}= 2 \times 10^{-4}M$

Convert μM:

$2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M$

Thus, this dilution scheme will yield a 200μM solution.

7 0

1 month ago

The concentration of Si in an Fe-Si alloy is 0.25 wt%. What is the concentration in kilograms of Si per cubic meter of alloy?

KiRa [2933]

Answer: The mass of Si in kilograms is, $19.55kg/m^3$

Explanation:

Given that the Si concentration in an Fe-Si alloy is 0.25 weight percent, this translates to:

Mass of Si = 0.25 g = 0.00025 kg

Mass of Fe = 100 - 0.25 = 99.75 g = 0.09975 kg

Density of Si = $2.32g/cm^3=2.32\times 10^6g/m^3$

Density of Fe = $7.87g/cm^3=7.87\times 10^6g/m^3$

Next, we need to find the quantity of Si in kilograms per cubic meter of alloy.

Si concentration in kilograms = $\frac{\text{Weight of Si in 100 g of alloy}}{\text{Volume of 100 g of alloy}}$

Si concentration in kilograms = $\frac{\text{Weight of Si in 100 g of alloy}}{\frac{\text{Wight of Fe}}{\text{Density of Fe}}+\frac{\text{Wight of Si}}{\text{Density of Si}}}$

By substituting all the provided values into this formula, we arrive at:

Si concentration in kilograms = $\frac{0.00025kg}{\frac{99.75g}{7.87\times 10^6g/m^3}+\frac{0.25g}{2.23\times 10^6g/m^3}}$

Si concentration in kilograms = $19.55kg/m^3$

Hence, the mass of Si in kilograms is, $19.55kg/m^3$

5 0

1 month ago

In NMR spectroscopy, the strong magnetic field establishes an energy gap between the alpha and beta spin states, which enables t

Anarel [2989]

Answer:

In the context of NMR spectroscopy, a significant magnetic field creates an energy difference between the alpha and beta spin states, which allows nuclei to absorb RF radiation, ultimately leading to the excitation of a nucleus from a +1/2 spin state to a -1/2 spin state.

Explanation:

3 0

1 month ago

Bleach contains the active ingredient NaClO. Analysis of bleach involves two sequential redox reactions: First, bleach is reacte

VMariaS [2998]

Answer:

0.31%

Explanation:

For the chemical reaction:

I₂ + 2 S₂O₃²⁻ → 2 I⁻ + S₄O₆²⁻

0.043 L multiplied by 0.117 M of sodium thiosulfate gives 5.031x10⁻³ moles of S₂O₃²⁻

5.031x10⁻³ moles of S₂O₃²⁻ produces:

ClO⁻⁻ + 2 H⁺ + 2 I⁻ → I₂ + Cl⁻ + H2O

2.5156x10⁻³ moles of I₂ equates to moles of NaClO

2.5156x10⁻³ moles of NaClO times $\frac{74,44 g}{1mol}$ yields 0.187 g of NaClO

Thus, the mass percentage composition is:

$\frac{0,187 g of NaClO}{60 g Of Bleach} x 100$ = 0.31%

I hope this helps!

5 0

1 month ago

How much of the sulfur burnt is normally oxidized to hexavalent state?

alisha [2963]

I believe the answer is D, though I'm not entirely certain unfortunately.

7 0

15 days ago