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2 years ago

Ways to manage non-biodegradable waste wisely

Chemistry

1 answer:

miv72 [106K]2 years ago

6 0

Answer:

Non-biodegradable wastes can be managed by practicing the concept of 3Rs—Reduce, Reuse, and Recycle.

...

Management Of Non-Biodegradable Wastes

Use fountain pen in place of a ballpoint pen,

Use old newspapers for packaging, and.

Use cloth napkins in place of disposable ones.

Explanation:

there!

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how many liters of a 60% antifreeze solution must be added to 8L of a 10% antifreeze solution to produce a 20% antifreeze soluti

Tasya [4]

Answer: 2Liters

Explanation:

The expression used will be :

$M_1V_1+M_2V_2=M_3V_3$

where,

$M_1$ = concentration of first antifreeze= 60%

$M_2$ = concentration of second antifreeze= 10%

$V_1$ = volume of first antifreeze = x L

$V_2$ = volume of second antifreeze = 8 L

$M_3$ = concentration of final antifreeze solution= 20%

$V_3$ = volume of final antifreeze = (x+8) L

Now put all the given values in the above law, we get the volume of antifreeze added

$60\times x+10\times 8=20\times (x+8)$

$x=2L$

Therefore, the volume of 60% antifreeze solution that must be added is 2L

5 0

3 years ago

Cual es el volumen de la solución de hcl la cual se preparo con 3 gramos de soluto (k) al 14%

Elena-2011 [213]

Answer:

from which language this belongs to

3 0

3 years ago

A solution of H2SO4(aq) with a molal concentration of 4.80 m has a density of 1.249 g/mL. What is the molar concentration of thi

Naddika [18.5K]

Imagine we have mass of solvent 1kg (1000g)
According to that: $molality =n(solute) / m(solvent) =\ \textgreater \ 4.8m = 4.8mole / 1kg$
= 4.8 mole * 98 g/mole = 470g
$m(H2SO4) = n(H2SO4)*Mr(H2SO4) =\ \textgreater \$ $=\ \textgreater \ Molarity = 4.8 mole / 1.2 L = 4 M$ m(H2SO4) which is =470g

m(solution) = m(H2SO4) + m(solvent) = 470 + 1000 = 1470 g
d(solution) = m(solution) / V(solution) =>
=> 1.249 g/mL = 1470 g / V(solution) =>
$Molarity = n(solute) / V(solution) =\ \textgreater \$

5 0

3 years ago

Which of the following are weak electrolytes in aqueous solutions? Select ALL that apply.

Mandarinka [93]

Answer:

Weak electrolytes are

HgCl2 and NH₃

Explanation:

Electrolytes are compounds that conduct electricity while in molten or aqueous form.
Electrolytes may therefore be decomposed by passing electric current through them.

They may be either strong or weak electrolytes.

Strong electrolytes are those that ionize completely to form mobile ions while weak electrolytes ionize partially to generate ions.

Examples of weak electrolytes are HgCl2, NH₃
Examples of strong electrolytes are LiOH, HClO₄, and HI.

3 0

3 years ago

A buffer is composed of nh3 and nh4cl. How would this buffer solution control the ph of a solution when a small amount of a stro

Shkiper50 [21]

Answer:

According to Le-chatelier principle, equilibrium will shift towards left to minimize concentration of $OH^{-}$ and keep same equilibrium constant

Explanation:

In this buffer following equilibrium exists -

$NH_{3}(aq.)+H_{2}O(l)\rightleftharpoons NH_{4}^{+}(aq.)+OH^{-}(aq.)$

So, $OH^{-}$ is involved in the above equilibrium.

When a strong base is added to this buffer, then concentration of $OH^{-}$ increases. Hence, according to Le-chatelier principle, above equilibrium will shift towards left to minimize concentration of $OH^{-}$ and keep same equilibrium constant.

Therefore excess amount of $OH^{-}$ combines with $NH_{4}^{+}$ to produce ammonia and water. So, effect of addition of strong base on pH of buffer gets minimized.

5 0

3 years ago

Ways to manage non-biodegradable waste wisely​

Ways to manage non-biodegradable waste wisely