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3 years ago

What is a nanomaterial that is often used with other compounds to desalinate and decontaminate water?

Chemistry

1 answer:

Katarina [22]3 years ago

3 0

Answer:

Silver ions.

Explanation:

The most extensively studied nanomaterials for water purification and treatment mainly includes zero-valent metal nanoparticles, metal oxides nanoparticles, carbon nanotubules and nanocomposites.

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How is ice-cream melting a physical change?

sashaice [31]

Because it goes from a somewhat solid state to a liquid state.

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4 years ago

"A nuclear reaction of significant historical note occurred in 1932, when a beryllium target was bombarded with alpha particles.

rodikova [14]

Answer: $_{0}^{1}\textrm{e}$ is missing

Explanation:

In a nuclear reaction, the total mass and total atomic number remains the same.

For the given reaction:

$^{4}_{2}\textrm{He}+^9_4\textrm{Be}\rightarrow ^{12}_{6}\textrm{C}+^A_Z\textrm{X}$

To calculate A:

Total mass on reactant side = total mass on product side

4 + 9 = 12 + A

A = 1

To calculate Z:

Total atomic number on reactant side = total atomic number on product side

2 + 4 = 6 + Z

Z = 0

$^{4}_{2}\textrm{He}+^9_4\textrm{Be}\rightarrow ^{12}_{6}\textrm{C}+^1_0\textrm{e}$

Hence, missing is $_{0}^{1}\textrm{e}$ .

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3 years ago

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Gnoma [55]

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3 years ago

The balanced combustion reaction for C6H6 is 2C6H6(l)+15O2(g)⟶12CO2(g)+6H2O(l)+6542 kJ If 8.600 g C6H6 is burned and the heat pr

jenyasd209 [6]

Answer : The final temperature of the water is, $22.5166^oC$

Explanation : Given,

Mass of benzene = 8.600 g

Molar mass of benzene = 78 g/mole

First we have to calculate the moles of benzene.

$\text{Moles of benzene}=\frac{\text{Mass of benzene}}{\text{Molar mass of benzene}}=\frac{8.600g}{78g/mol}=0.1103mole$

Now we have to calculate the energy of combustion.

The given balanced chemical reaction is:

$2C_6H_6(l)+15O_2(g)\rightarrow 12CO_2(g)+6H_2O(l)+6542 kJ$

According to reaction,

As, 2 moles of benzene gives 6542 kJ of energy on combustion.

So, 0.1103 mole of benzene gives $\frac{6542 kJ}{2}\times 0.1103=360.7913kJ$ of energy on combustion.

Now we have to calculate the final temperature of the water.

Formula used : $q_w=m_w\times c_w\times \Delta T=m_w\times c_w\times (T_{final}-T_{initial})$

where,

$q_w$ = heat released = 360.7913 kJ = 36079.13 J

$m_w$ = mass of water = 5691 g

$c_w$ = specific heat of water= $4.18J/g^oC$

$T_{final}$ = final temperature = ?

$T_{initial}$ = initial temperature = $21^oC$

Now put all the given values in the above formula, we get:

$36079.13J=5691g\times 4.18J/g^oC\times (T_{final}-21^oC)$

$T_{final}=22.5166^oC$

Therefore, the final temperature of the water is, $22.5166^oC$

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3 years ago

Read 2 more answers