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2 years ago

The major ways people are exposed to toxins are by ingesting them, inhaling them, injecting

1 answer:

7 0

It is true that the major ways people are exposed to toxins are by ingesting them, inhaling them, injecting them, or absorbing them through the skin.

<u>Explanation:</u>

Toxins can enter in four various ways into our body often referred as "route of exposure". It is one among many factors affecting toxicity.

Inhalation: When the toxin is a gas, it is usually absorbed into the body by inhalation. Inhalation of toxins may cause respiratory and lung damage. Eg: cigarette smoke

Absorption: If a toxin enters the human eye or skin, it can be absorbed into the bloodstream. This is called absorption. The skin can prevent some toxins from entering the body. For example: work stations (gas stations) where people work with toxins

Ingestion means that the person has ingested the toxin. Some chemicals cannot get into blood from digestive tract. For example: drinking contaminated water.

The injected toxin enters the body when the person carrying the toxin breaks the skin. Injections can end in veins, muscles or just under the skin. Eg: Botulinum (used to get rid of wrinkles)

In bio-chemistry, poison, a natural or synthetic substance that damages living tissue and has a harmful or lethal effect on the body, regardless of whether it is absorbed, inhaled or injected through the skin.

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Hospital patients are administered oxygen from an pressurized

Nataly_w [17]

Answer:

110L

Explanation:

Boyle's Law states that P1×V1=P2×V2

Volume is indirectly proportional to Pressure so P×V is constant

P1=55atm

V1=6L

P2=3atm

V2 is to be found

P1×V1=P2×V2

6×55=3×V2

330=3×V2

Answer: V2=110L

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2 years ago

The pictures showed an organ system in the human body

LenaWriter [7]

The answer is D okay

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3 years ago

Equation Given : Al^(3+) + Na3PO4 ==> 3Na^+ + AlPO4

Helga [31]

1 mols of Aluminium ion forms 1 mol aluminium phosphate

Molar mass of AlPO_4

27+31+16(4)
58+48
106u

Moles of AlPO_4

61µg/106
0.000061/106
5.75×10^{-7}
57.5µmol

Moles of Al3+=57.5µmol

3 0

2 years ago

The copper(I) ion forms a chloride salt (CuCl) that has Ksp = 1.2 x 10-6. Copper(I) also forms a complex ion with Cl-:Cu+ (aq) +

Mnenie [13.5K]

Answer: (a) The solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) The solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

Explanation:

(a) Chemical equation for the given reaction in pure water is as follows.

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$

Initial: 0 0

Change: +x +x

Equilibm: x x

$K_{sp} = 1.2 \times 10^{-6}$

And, equilibrium expression is as follows.

$K_{sp} = [Cu^{+}][Cl^{-}]$

$1.2 \times 10^{-6} = x \times x$

x = $1.1 \times 10^{-3} M$

Hence, the solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) When NaCl is 0.1 M,

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$ , $K_{sp} = 1.2 \times 10^{-6}$

$Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq)$ , $K = 8.7 \times 10^{4}$

Net equation: $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

$K' = K_{sp} \times K$

= 0.1044

So for, $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

Initial: 0.1 0

Change: -x +x

Equilibm: 0.1 - x x

Now, the equilibrium expression is as follows.

K' = $\frac{CuCl_{2}}{Cl^{-}}$

0.1044 = $\frac{x}{0.1 - x}$

x = $9.5 \times 10^{-3} M$

Therefore, the solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

7 0

3 years ago

Calculate the standard reaction enthalpy for the reaction NO2(g) → NO(g) + O(g) given +142.7 kJ/mol for the standard enthalpy of

bulgar [2K]

Answer:

The standard reaction enthalpy for the given reaction is 235.15 kJ/mol.

Explanation:

$O_2(g) \rightarrow \frac{2}{3}O_3(g),\Delta H^o_{1}=142.7 kJ/mol$ ..[1]

$O_2(g) \rightarrow 2 O(g),\Delta H^o_{2}=498.4 kJ/mol$ ..[2]

$NO(g) + O_3(g)\rightarrow NO_2(g) + O_2(g) ,\Delta H^o_{3} = -200 kJ/mol$ ..[3]

$NO_2(g)\rightarrow NO(g) + O(g),\Delta H^o_{4}=?$ ..[4]

Using Hess's law:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

2 × [4] = [2]- (3 ) × [1] - (2) × [3]

$2\times \Delta H^o_{4}=\Delta H^o_{2} -3\times \Delta H^o_{1}-2\times \Delta H^o_{3}$

$2\times \Delta H^o_{4}=498.4 kJ/mol-3\times 142.7 kJ/mol-2\times -200 kJ/mol$

$2\times \Delta H^o_{4}=470.3 kJ/mol$

$\Delta H^o_{4}=\frac{470.3 kJ/mol}{2}=235.15 kJ/mol$

The standard reaction enthalpy for the given reaction is 235.15 kJ/mol.

7 0

2 years ago