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Zinaida [17]
3 years ago
15

Most chemicals can be cleaned up with a general spill kit, but a few chemicals require specialized spill procedures. For each su

bstance listed, determine whether a general spill kit is sufficient or if a specialized spill kit is needed.
a. hydrofluoric acid
b. special mercury
c. special phosphoric acid
d. acetone
e. general methanol
Chemistry
1 answer:
const2013 [10]3 years ago
8 0

Answer:

a. hydrofluoric acid  - Specialized spill kit is needed

b. special mercury  - Specialized spill kit is needed

c. special phosphoric acid  - General spill kit is Sufficient

d. acetone  - General spill kit is Sufficient

e. general methanol - General spill kit is Sufficient

Explanation:

a. hydrofluoric acid  -

It is a weak acid , and very corrosive therefore , it need Specialized spill kit

b. special mercury  -

At the room temperature , liquid mercury evaporates. But Small amount of mercury can cause harm so , Specialized spill kit is needed

c. special phosphoric acid  -

It is relatively compared to others is a week acid , therefore General spill kit is Sufficient

d. acetone  -

At the room temperature , acetone evaporates, therefore General spill kit is Sufficient

e. general methanol -

At the room temperature , methanol evaporates, therefore General spill kit is Sufficient

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Determine the partial pressure and number of moles of each gas in a 15.75-L vessel at 30.0 C containing a mixture of xenon and n
lawyer [7]

Answer:

The Partial pressure of Xe and Ne will be 4.95 atm and 1.55 atm. The number of moles of Xe and Ne will be 3.13 and 0.981

Explanation:

Let the total pressure of the vessel= 6.5 atm and mole fraction of Xenon= 0.761

As we know,

\chi_{Ne} + \chi_{Xe} = 1\\\chi_{Ne}= 1- 0.761\\\chi_{Ne}= 0.239

According to Dalton's Law of partial pressure-

P_i=\chi_i\times P_{total}

Where,

P_i=The pressure of the gas component in the mixture

\chi_i= Mole fraction of that gas component

P_t= The total pressure of the mixture

P_{Xe}=(0.761)\times(6.5)\\P_{Xe}= 4.95 atm\\\\\\P_{Ne}=(0.239)\imes (6.5)\\P_{Ne}= 1.55 atm

<u>Calculation: </u>

To calculate the number of moles,

PV=nRT

n=\frac{PV}{RT}

n_{Xe}= \frac{4.95\times 15.75}{0.0821\times303 }\\ n_{Xe}= \frac{77.96}{24.87} \\n_{Xe}= 3.13\,mole \\\\\\n_{Ne}= \frac{1.55\times 15.75}{0.0821\times303 }\\\\n_{Ne}=\frac{24.41}{24.87}\\ n_{Ne}=0.981 \,mole

Learn more about Dalton's Law of partial pressure here;

brainly.com/question/14119417

#SPJ4

4 0
2 years ago
How to do Lewis structure for Al?
Gnoma [55]

Notes:-

  • Al has Z=15

electron configuration:-

  • [Ne]3s²3p³

Valency is 3

So 3 dots are included

4 0
2 years ago
What is the molarity of a solution in which 58g of nacl are dissolved in 1.0 l of solution
Lady_Fox [76]

Hey there :

Molar mass of NaCl = 58.44 g/mol

Number of moles :

n = mass of solute / molar mass

n = 58 / 58.44

n = 0.9924 moles of NaCl

Volume = 1.0 L

Therefore:

Molarity = number of moles / volume ( L )

Molarity = 0.9924 / 1.0

Molarity = 0.9924 M

Hope that helps!

7 0
3 years ago
what is the specific heat of a substance if 300 j are required to raise the temperature of a 267-g sample by 12 degrees c
slava [35]

Answer : The specific heat of the substance is 0.0936 J/g °C

Explanation :

The amount of heat Q can be calculated using following formula.

Q = m \times C \times \bigtriangleup T

Where Q is the amount of heat required = 300 J

m is the mass of the substance = 267 g

ΔT is the change in temperature = 12°C

C is the specific heat of the substance.

We want to solve for C, so the equation for Q is modified as follows.

C = \frac{Q}{m \times \bigtriangleup T}

Let us plug in the values in above equation.

C = \frac{300J}{267g \times 12 C}

C = \frac{300J}{3204 g C}

C = 0.0936 J/g °C

The specific heat of the substance is 0.0936 J/g°C

3 0
3 years ago
if 15 mol C is mixed with 10 mol O2 which reactant is the limiting reactant? Which reactant would be the excess reactant?
ioda

Answer:

C is the excess reactant.

Explanation:

Reaction is C + O2 --> CO2

1mol of C required to react with 1mol O2

Therefore 15 - 10 = 5moles of C will be in excess

6 0
3 years ago
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