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Rainbow [258]
3 years ago
11

20 points! So I got this lab report to do and I need some more info it's about lab safety got any info I can use! :) If so, Than

k you!
Chemistry
1 answer:
FromTheMoon [43]3 years ago
8 0
Don’t eat or drink in labs

Dress for the lab; don’t wear open toed shoes

Dispose of lab waste properly

No horse play

Don’t taste or sniff things in the lab

Tire your hair back
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2) Show the calculation of Kc for the following reaction if an initial reaction mixture of 0.800 mole of CO and 2.40 mole of H2
nadezda [96]

Answer:

Kc = 3.90

Explanation:

CO reacts with H_2 to form CH_4 and H_2O. balanced reaction is:

CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g)  +  H_2O(g)

No. of moles of CO = 0.800 mol

No. of moles of H_2 = 2.40 mol

Volume = 8.00 L

Concentration = \frac{Moles}{Volume\ in\ L}

Concentration of CO = \frac{0.800}{8.00} = 0.100\ mol/L

Concentration of H_2 = \frac{2.40}{8.00} = 0.300\ mol/L

                 CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g)  +  H_2O(g)

Initial            0.100      0.300             0   0

equi.            0.100 -x    0.300 - 3x     x    x

It is given that,

at equilibrium H_2O (x) = 0.309/8.00 = 0.0386 M

So, at equilibrium CO = 0.100 - 0.0386 = 0.0614 M

At equilibrium H_2 = 0.300 - 0.0386 × 3 = 0.184 M

At equilibrium CH_4 = 0.0386 M

Kc=\frac{[H_2O][CH_4]}{[CO][H_2]^3}

Kc=\frac{0.0386 \times 0.0386}{(0.184)^3 \times 0.0614} =3.90

8 0
3 years ago
At what temperature (in C) will a sample of gas occupy 91.3 L if it occupies 45.0 L at 70.0°C? Assume constant pressure.)
sdas [7]

Solution is here,

for initial case,

temperature(T1)=70°C=70+ 273=343K

vloume( V1) =45 L

for final case,

temperature( T2)=?

volume(V2)= 91.3 L

at constant pressure,

V1/V2 = T1/T2

or, 45/91.3 = 343/ T2

or, T2= (343×91.3)/45

or, T2=695.9 K = (695.9-273)°C=422.9°C

5 0
3 years ago
Which of the following elements is a transition metal?
GenaCL600 [577]

Answer:

Scandium

Titanium

Vanadium

Chromium

Manganese

Iron

Cobalt

Nickel

Copper

Zinc

Yttrium

Zirconium

Niobium

Molybdenum

Technetium

Ruthenium

Rhodium

Palladium

Silver

Cadmium

Lanthanum

Hafnium

Tantalum

Tungsten

Rhenium

Osmium

Iridium

Platinum

Gold

Mercury

Actinium

Rutherfordium

Dubnium

Seaborgium

Bohrium

Hassium

Meitnerium

Darmstadtium

Roentgenium

Copernicium

Explanation:

all of those are transition metals lol

5 0
3 years ago
Read 2 more answers
Why wouldn't the concorde travel lower
Alika [10]

Supersonic flight over land is typically forbidden to minimize disruption to people on the ground. This meant that until it was over open water, Concorde could not operate at its maximum Mach 2 capability.

<h3>What is Concorde?</h3>
  • The Concorde is a tailless aircraft with a drooping nose, an oval delta wing, and a narrow fuselage that allows for seating for 92 to 128 passengers.
  • It's improbable that Concorde will ever fly for an airline again in a commercial capacity. The economics of reactivating Concorde and incorporating it into the fleet of a contemporary airline would be challenging to justify, even if the technical and regulatory concerns could be handled.
  • Supersonic flight over land is usually not allowed to minimize the impact on people on the ground. This meant that Concorde could not operate at its maximum Mach 2 speed until it was over an open body of water.

To learn more about Concorde refer to:

brainly.com/question/14354579

#SPJ1

5 0
1 year ago
A chamber with a fixed volume is shown above. The temperature of the gas inside the chamber before heating is 25.2 C and it’s pr
rusak2 [61]

Answer:

Explanation:

Given parameters:

Initial temperature T₁  = 25.2°C  = 25.2 + 273  = 298.2K

Initial pressure  = P₁  = 0.6atm

Final temperature = 72.4°C   = 72.4 + 273  = 345.4K

Unknown:

Final pressure = ?

Solution:

To solve this problem, we use an adaption of the combined gas law where the volume gas is fixed. This simplification results into:

                  \frac{P_{1} }{T_{1} }   = \frac{P_{2} }{T_{2} }

where P and T are temperatures, 1 and 2 are initial and final temperatures.

 Input the parameters and solve;

          \frac{0.6}{298.2}   = \frac{P_{2} }{345.4}  

          P₂   = 0.7atm

         

3 0
3 years ago
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