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3 years ago

Which source of water pollution would be the hardest to trace and control?

Chemistry

1 answer:

Alex Ar [27]3 years ago

5 0

Answer:

Nonpoint pollution is harder to control

Explanation:

Nonpoint pollution is harder to control because they include runoff from farms, golf courses, construction sites, roads. Whereas the single origin of the pollution is harder to trace.

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A renowned laboratory reports quadruple-point coordinates of 10.2 Mbar and 24.1°C for four-phase equilibrium of allotropic solid

Inessa [10]

Answer:

The claim is untrue.

Explanation:

Given the information from the question. We need to evaluate the claim.

According to the phase rule we have F= C-P+2

In this particular situation, the forms are completely allotropic .In order words, they are conjured through the same chemical composition. Thus constituents C= 1, P=4 for four phases and the number variables is 2. As a result, F= C-P+2 =1-4+2= -1. Therefore, the claim is untrue.

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3 years ago

A substance joined by this type of bond has a low melting point

cestrela7 [59]

Covalent compounds have bonds where electrons are shared between atoms. Due to the sharing of electrons, they exhibit characteristic physical properties that include lower melting points and electrical conductivity compared to ionic compounds.

5 0

3 years ago

Read 2 more answers

Calculate the masses of oxygen and nitrogen that are dissolved in of aqueous solution in equilibrium with air at 25 °C and 760 T

shtirl [24]

Explanation:

Let us assume that the volume of given aqueous solution is 7.5 L.

Therefore, according to Henry's law, the relation between concentration and pressure is as follows.

C = $\frac{P}{K_{h}}$

where, pressure (P) = 760 torr = 1 atm

According to Henry's law, constants for gases in water at $25^{o}C$ are as follows.

$p(O_{2})$ = 0.21 atm = 0.21 bar

$p(N_{2})$ = 0.78 atm = 0.78 bar

$K_{h}$ for $O_{2}$ = $7.9 \times 10^{2} bar/mol$

$K_{h}$ for $N_{2}$ = $1.6 \times 10^{3} bar /mol$

Since, 21% oxygen is present in air so, its mass will be 0.21 g. Similarly, 78% nitrogen means the mass of nitrogen is 0.78 g.

Therefore, concebtrations will be calculated as follows.

$C(O_{2}) = \frac{0.21}{7.9 \times 10^{2}} = 2.66 \times 10^-4 mol/L$

$C(N_2) = \frac{0.78}{1.6 \times 10^3} = 4.875 \times 10^-4 mol/L$

Now, we will calculate the number of moles as follows.

$n(O_{2}) = 7.5 \times 2.66 \times 10^-4 = 1.995 \times 10^-3$ mol

$n(N_2) = 7.5 \times 4.875 \times 10^-4 = 3.66 \times 10^-3$ mol

As the molar mass of $O_2$ = 32 g/mol

Hence, mass of oxygen will be as follows.

Mass of $O_2 = 32 \times 1.995 \times 10^-3 \times 1000$

= 63.84 mg

As the molar mass of $N_{2}$ = 28

Mass of $N_{2} = 28 \times 3.66 \times 10^-3$ = 102.5 mg

Thus, we can conclude that mass of oxygen is 63.84 mg and nitrogen is 102.5 mg.

5 0

4 years ago

Read 2 more answers

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NISA [10]

Answer:

ladidididadaladaddididi

Explanation:

5 0

3 years ago

21 Which process is used to determine the concentration of an acid?

Inessa05 [86]

Answer:

4) Titration

Explanation:

Titration is a common process used to determine the concentration of acids. It does this by adding a solution of base with a known concentration to the acid until it reaches neutralization.

8 0

4 years ago