All categories

2 years ago

Chemicals are classified as health hazards when they pose which of the following hazardous effects?

Chemistry

1 answer:

AlexFokin [52]2 years ago

4 0

Answer:

poisoning, breathing problems, skin rashes, allergic reactions, allergic sensitisation, cancer, and other health problems from exposure.

Explanation:

many hazardous chemicals are also classified as dangerous goods.

You might be interested in

What is the calculated value of the cell potential at 298K for an

Tpy6a [65]

The question is incomplete, here is the complete question:

What is the calculated value of the cell potential at 298 K for an electrochemical cell with the following reaction, when the H₂ pressure is 6.56 x 10⁻² atm, the H⁺ concentration is 1.39 M, and the Sn²⁺ concentration is 9.35 x 10⁻⁴ M?

$2H^+(aq)+Sn(s)\rightarrow H_2(g)+Sn^{2+}(aq)$

Answer: The cell potential of the given electrochemical cell is 0.273 V

Explanation:

For the given chemical equation:

$2H^+(aq)+Sn(s)\rightarrow H_2(g)+Sn^{2+}(aq)$

The half cell reactions for the given equation follows:

Oxidation half reaction: $Sn(s)\rightarrow Sn^{2+}(aq)+2e^-;E^o_{Sn^{2+}/Sn}=-0.14V$

Reduction half reaction: $H_2+2e^-\rightarrow H_2(g);E^o_{2H^{+}/H_2}=0.0V$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=0.0-(-0.14)=0.14V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Sn^{2+}]\times p_{H_2}}{[H^+]^2}$

where,

$E_{cell}$ = electrode potential of the cell = ?

$E^o_{cell}$ = standard electrode potential of the cell = +0.14 V

n = number of electrons exchanged = 2

$[H^{+}]=1.39M$

$[Sn^{2+}]=9.35\times 10^{-4}M$

$p_{H_2}=6.56\times 10^{-2}atm$

Putting values in above equation, we get:

$E_{cell}=0.14-\frac{0.059}{2}\times \log(\frac{(9.35\times 10^{-4})\times (6.56\times 10^{-2})}{(1.39)^2})\\\\E_{cell}=0.273V$

Hence, the cell potential of the given electrochemical cell is 0.273 V

4 0

2 years ago

If a solution containing 51.429 g of mercury(ii) perchlorate is allowed to react completely with a solution containing 16.642 g

OLga [1]

Hg(No3)2 +NaSO4 --->2NaNO3 + HgSO4(s)
calculate the moles of each reactant
moles=mass/molar mass

moles of Hg(NO3)2= 51.429g/ 324.6 g/mol(molar mass of Hg(NO3)2)=0.158 moles

moles Na2SO4 16.642g/142g/mol= 0.117 moles of Na2SO4

Na2SO4 is the limiting reagent in the equation and by use mole ratio Na2So4 to HgSO4 is 1:1 therefore the moles of HgSO4 =0.117 moles

mass of HgSO4=moles x molar mass of HgSo4= 0.117 g x 303.6g/mol= 35.5212 grams

7 0

3 years ago

Insertion mutation definition ?

blondinia [14]

Answer:

Insertion is a type of mutation involving the addition of genetic material. An insertion mutation can be small, involving a single extra DNA base pair, or large, involving a piece of a chromosome.

Explanation:

4 0

2 years ago

Read 2 more answers

What element is an atom that has 7 protons 8 neutrons and 6 electrons?

andreev551 [17]

Answer:

I think it should be Carbon.

6 0

3 years ago

Read 2 more answers

Predict which of the following metals reacts with hydrochloric acid.