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nasty-shy [4]
3 years ago
8

Physical and psychological dependence, high-risk behaviour, and chronic high blood

Chemistry
1 answer:
antiseptic1488 [7]3 years ago
6 0

Answer:

Physical and psychological dependence is high, and withdrawal symptoms include watery eyes, runny nose, loss of appetite, irritability, tremors, panic, abdominal cramps and diarrhea, nausea, chills, and sweating. Use of contaminated syringes/needles to inject drugs may result in serious blood borne infections such as HIV-AIDS and hepatitis.

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Eyeglasses provide as much protection as safety goggles. * true or false
Artyom0805 [142]
False. Eyeglasses do not cover all around the eye however safety goggles do
7 0
3 years ago
Read 2 more answers
What is the value of the equilibrium constant at 25 oC for the reaction between the pair: I2(s) and Br-(aq) Give your answer usi
insens350 [35]

Explanation:

Formula to calculate standard electrode potential is as follows.

          E^{o}_{cell} = E^{0}_{cathode} - E^{0}_{anode}

                             = 0.535 - 1.065

                             = - 0.53 V

Also, it is known that relation between E^{o}_{cell} and K is as follows.

            E^{o}_{cell} = \frac{RT}{nF} \times ln K

                 ln K = \frac{nFE^{0}_{cell}}{RT}      

Substituting the given values into the above formula as follows.

                 ln K = \frac{nFE^{0}_{cell}}{RT}    

                        =  \frac{2 \times 96485 C mol^{-1} \times -0.53 V}{8.314 l atm/mol K \times 298 K} \times \frac{1 J}{1 V C}  

                ln K = -41.28

                    K = e^{-41.28}    

                        = 1 \times 10^{-18}

Thus, we can conclude that the value of the equilibrium constant for the given reaction is 1 \times 10^{-18}.      

5 0
3 years ago
Estimate the air pressure at an altitude of 5km
r-ruslan [8.4K]
The answer is 100 kg
6 0
3 years ago
A mixture of gases with a pressure of 800.0 mm hg contains 60% nitrogen and 40% oxygen by volume. What is the partial pressure o
larisa86 [58]

Answer:

  • <em>The partial pressure of oxygen in the mixture is</em><u> 320.0 mm Hg</u>

Explanation:

<u>1) Take a base of 100 liters of mixture</u>:

  • N: 60% × 100 liter  = 60 liter

  • O: 40 % × 100 liter = 40 liter.

<u>2) Volume fraction:</u>

At constant pressure and temperature, the volume of a gas is proportional to the number of molecules.

Then, the mole ratio is equal to the volume ratio. Callin n₁ and n₂, the number of moles of nitrogen and oxygen, respectively, and V₁, V₂ the volume of the respective gases you can set the proportion:

  • V₁ / V₂ = n₁ / n₂

That means that the mole ratio is equal to the volume ratio, and the mole fraction is equal to the volume fraction.

Then, since the law of partial pressures of gases states that the partial pressure of each gas is equal to the mole fraction of the gas multiplied by the total pressure, you can draw the conclusion that the partial pressure of each gas is equal to the volume fraction of the gas in the mixture multiplied by the total pressure.

Then calculate the volume fractions:

  • Volume fraction of a gas = volume of the gas / volume of the mixture

  • N: 60 liter / 100 liter = 0.6 liter

  • V: 40 liter / 100 liter = 0.4 liter

<u>3) Partial pressures:</u>

These are the final calculations and results:

  • Partial pressure = volume fraction × total pressure

  • Partial pressure of N = 0.6 × 800.0 mm Hg = 480.0 mm Hg

  • Partial pressure of O = 0.4 × 800.0 mm Hg = 320.0 mm Hg
8 0
3 years ago
PLEASE ANSWER CORRECTLY FOLLOWING GUIDLINES SO ANSWER WONT GET DELETED I REALLY NEED HELP
blagie [28]

Chemical reaction

equation

reactants

products

yields

mass

balanced

atoms

coefficients

numbers

element

correct

substance

two

Explanation:

did my best lol I'm like 98.69% confident

8 0
2 years ago
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