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SCORPION-xisa [38]
2 years ago
5

Why is it important to record all accidents and breakages that occur in the laboratory

Chemistry
1 answer:
Svet_ta [14]2 years ago
8 0

It is important to record all accidents and breakages that occur in the laboratory to prevent them from happening again.

<h3>What is laboratory accident?</h3>

Laboratory accident is defined as the type of accident that occurs in the laboratory leading to harm.

Example of laboratory accidents include the following:

  • chemical burns,

  • cuts from broken glass,

  • inhalation of toxic fumes,

  • absorption of chemicals through the skin, and

  • ingestion of toxic chemicals.

A record of these type of laboratory accidents would hel prevent it's reoccurrence.

Learn more about laboratory here:

brainly.com/question/26264740

#SPJ1

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Which objects refract light? (Select all that apply.)
Free_Kalibri [48]
Concave mirrors, magnifying lens... sorry not sure about the rest
3 0
3 years ago
Read 2 more answers
Consider 2.4 moles of a gas contained in a 4.0 L bulb at a constant temperature of 32°C. This bulb is connected to an evacuated
Luden [163]

Complete Question

The complete question is shown on the first uploaded image

Answer

a

As the valve is opened , the gas will flow into the empty container until the both containers have the same pressure

b

\Delta H = 0\ , \Delta E = 0 , q= 0 , w= 0

c

The driving force for this process is the increase in entropy this is because the movement of the internal energy of the gas into a larger volume, what this does is that it increases the amount of disorder(entropy).

Explanation

In order to obtain the parameter in the part B of the question we are first obtain the initial pressure, using the ideal gas equation  

                      P = \frac{nRT}{V}

                     P = \frac{(2.4mol)(0.0821\frac{1 atm}{K \cdot \ mol} )}{4.0L}

                         P =15 \ atm

The next thing is to obtain the new pressure of the gas , using boyle's law

              P_1V_1 = P_2V_2

                  P_2 = \frac{P_1 V_1}{V_2}

                  P_2 = \frac{(15 \ atm)(4.0L)}{24.0 L}

                  P_2 = 2.5 \ atm    

Since the this process is isothermal , the change in heat is equal to zero

                      i.e  q = 0 J

  The workdone to move  the gas to the other container is zero because the  the pressure at this second container is zero due to the fact that it is a vacuum

    i.e  w = -P_{external} \Delta V

              =-(0 \ atm) (24.0 - 4.0L)

              = 0L \cdot atm

  Since the change in heat is zero and the workdone is zero then the change in internal energy is equal to 0

     This is because the change in internal energy is equal to a summation of change in heat and the workdone

                i.e \Delta E = q + w

                            = 0J

Generally the change in enthalpy is mathematically represented as

                \Delta H = n C_p \Delta T

Since the temperature is zero this means that the change in temperature is zero , substituting this value for change in temperature into the equation for  change in enthalpy

           \Delta H = n C_p (0)

                   = 0J

 

7 0
4 years ago
The half-life of cobalt-60 is 5. 20 yr. how many milligrams of a 2. 000 mg sample remain after 6. 55 years?
Stels [109]

0.84 milligrams of a 2. 000 mg sample remain after 6. 55 years, according to radioactive decay.

Given data,

t\frac{1}{2} of Co-60 = 5.20years

amount of sample = 2.000mg initially = 0.002grams

According to radioactive decay,

N_{t} = N_{0}e^{-λt}

(N_{0} - 0.002 )λ = \frac{0.693}{t\frac{1}{2} } = \frac{0.693}{5.20}  = 0.133

According to radioactive decay,

N_{t} = N_{0}e^{-λt}

lnN_{t}  = lnN_{0} - λt

lnN_{t} = ln0.002 - (0.133×6.55)

       = -6.21 - 0.87 = -7.08 = 0.00084g = 0.84mg

Therefore, 0.84 milligrams of a 2. 000 mg sample remain after 6. 55 years.

Learn more about radioactive decay here:

brainly.com/question/1770619

#SPJ4

5 0
2 years ago
Classify the substances according to the strongest solute-solvent interaction that will occur between the given substances and w
MArishka [77]

Answer:

A.  Dipole-dipole forces; B. dipole-dipole forces;

C. ion-dipole forces; D. ion-dipole forces

Explanation:

A. HF

HF is a weak acid but a highly polar molecule. The strongest intermolecular force with water is an especially strong dipole-dipole force (hydrogen bonding) of the type

H-F· · ·H-OH and H₂O· · ·H-F

B. CH₃OH

CH₃OH has a highly polar O-H bond. The strongest intermolecular force with water is the especially strong dipole-dipole force (hydrogen bonding):

CH₃(H)O· · ·H-OH and CH₃O-H· · ·OH₂

C. CaCl₂

CaCl₂(s) ⟶Ca²⁺(aq) + 2Cl⁻(aq)

CaCl₂ separates into hydrated ions in solution. The strongest intermolecular force with water is ion-dipole attraction.

Ca²⁺· · ·OH₂ and Cl⁻· · ·H-OH

D. FeBr₃

FeBr₃(s) ⟶Fe³⁺(aq) + 3Br⁻(aq)

FeBr₃ separates into hydrated ions in solution. The strongest intermolecular force with water is ion-dipole attraction.

Fe³⁺· · ·OH₂ and Br⁻· · ·H-OH

6 0
3 years ago
What is the first step in replication
levacccp [35]
<span>the first major step for the dan replication to take place is the breaking of hydrogen bonds between bases of the two antiparallel strands </span>
4 0
3 years ago
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