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lidiya [134]
2 years ago
5

The MSDS for chloroform indicates that it is a clear liquid that has a pleasant smell and substantial vapor pressure. People sho

uld avoid inhaling its vapors, and it is sensitive to light. Malik needs 10 mls of chloroform for an experiment. According to this information, how should he safely pour the chloroform?
A.He should locate the chloroform stored in a transparent container in chemical storage and pour directly into his beaker from that location.
B.He should locate the chloroform stored in a transparent container in chemical storage and should take it to the fume hood to pour.
C.He should locate the chloroform stored in a dark container in chemical storage and should take it to the fume hood to pour.
D.He should locate the chloroform stored in a dark container in chemical storage and pour directly into his beaker from that location.
Chemistry
1 answer:
aleksandr82 [10.1K]2 years ago
8 0

Answer:C.He should locate the chloroform stored in a dark container in chemical storage and should take it to the fume hood to pour.

Explanation:

The statement of the question clearly states that chloroform is sensitive to light and it's vapour is toxic.

If a substance is sensitive to light, then it must be stored in a dark bottle. This is because. If a substance that is sensitive to light is stored in a transparent container, it may be decomposed by light.

Being a substance whose fumes are toxic, Malik should pour the liquid in a fumes hood so that he does not inhale the fumes.

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When a complex ion forms, ______________ arrange themselves around _________, creating a new ion with a charge equal to the sum
alisha [4.7K]

Answer:

lignands, the central atom/metal ion

Explanation:

7 0
2 years ago
What is the concentration of Iodine I2 molecules in a solution containing 2.54 g of iodine 250 cm3 of solution? A 0.01mol/dm3 B
gulaghasi [49]

Answer:

C. 0.04 moles per cubic decimeter.

Explanation:

The molar mass of the Iodine is 253.809 grams per mole and a cubic decimeter equals 1000 cubic centimeters. The concentration of Iodine (c), measured in moles per cubic decimeter, can be determined by the following formula:

c = \frac{m}{M\cdot V} (1)

Where:

m - Mass of iodine, measured in grams.

M - Molar mass of iodine, measured in grams per mol.

V - Volume of solution, measured in cubic decimeters.

If we know that m = 2.54\,g, M = 253.809\,\frac{g}{mol} and V = 0.25\,dm^{3}, then the concentration of iodine in a solution is:

c = \frac{2.54\,g}{\left(253.809\,\frac{g}{mol} \right)\cdot (0.25\,dm^{3})}

c = 0.04\,\frac{mol}{dm^{3}}

Hence, the correct answer is C.

3 0
2 years ago
A solution is prepared by dissolving 10.0 g of NaBr and 10.0 g of Na2SO4 in water to make a 100.0 mL solution. This solution is
Colt1911 [192]

Answer:

M_{Na^+}=1.36M

M_{Br^-}=1.58M

Explanation:

Hello,

At first, it turns out convenient to compute the total moles of sodium that will be dissolved into the solution by considering the added amounts of sodium bromide and sodium sulfate:

n_{Na^+}=n_{Na^+,NaBr}+n_{Na^+,Na_2SO_4}\\n_{Na^+,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molNa^+}{1molNaBr}=0.0971molNa^+\\n_{Na^+,Na_2SO_4}=10.0gNa_2SO_4*\frac{1molNa_2SO_4}{142gNa_2SO_4}*\frac{2molNa^+}{1molNa_2SO_4} =0.141molNa^+\\n_{Na^+}=0.0971molNa^++0.141molNa^+\\n_{Na^+}=0.238molNa^+

Once we've got the moles we compute the final volume via:

V=100.0mL+75.0mL=175.0mL*\frac{1L}{1000mL}=0.1750L

Thus, the molarity of the sodium atoms turn out into:

M_{Na^+}=\frac{0.238mol}{0.1750L} =1.36M

Now, we perform the same procedure but now for the bromide ions:

n_{Br^-}=n_{Br^-,NaBr}+n_{Br^-,AlBr_3}\\n_{Br^-,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molBr^-}{1molNaBr}=0.0971molBr^-\\n_{Br^-,AlBr_3}=0.0750L*0.800\frac{molAlBr_3}{L} *\frac{3molBr^-}{1molAlBr_3}=0.180molBr^- \\n_{Br^-}=0.0971molBr^-+0.180molBr^-\\n_{Br^-}=0.277molBr^-

Finally, its molarity results:

M_{Br^-}=\frac{0.277molBr^-}{0.1750L}=1.58M

Best regards.

7 0
2 years ago
What is the hybridization of the central atom in each of the following? 1. Beryllium chloride 2. Nitrogen dioxide 3. Carbon tetr
Lina20 [59]

Answer :

(1) The hybridization of central atom beryllium in BeCl_2  is, sp

(2) The hybridization of central atom nitrogen in NO_2  is, sp^2

(3) The hybridization of central atom carbon in CCl_4  is, sp^3

(4) The hybridization of central atom xenon in XeF_4  is, sp^3d^2

Explanation :

Formula used  :

\text{Number of electron pair}=\frac{1}{2}[V+N-C+A]

where,

V = number of valence electrons present in central atom

N = number of monovalent atoms bonded to central atom

C = charge of cation

A = charge of anion

Now we have to determine the hybridization of the following molecules.

(1) The given molecule is, BeCl_2

\text{Number of electrons}=\frac{1}{2}\times [2+2]=2

The number of electron pair are 2 that means the hybridization will be sp and the electronic geometry of the molecule will be linear.

(2) The given molecule is, NO_2

\text{Number of electrons}=\frac{1}{2}\times [4]=2

If the sum of the number of sigma bonds, lone pair of electrons and odd electrons present is equal to three then the hybridization will be, sp^2.

In nitrogen dioxide, there are two sigma bonds and one lone electron pair. So, the hybridization will be, sp^2.

(3) The given molecule is, CCl_4

\text{Number of electrons}=\frac{1}{2}\times [4+4]=4

The number of electron pair are 4 that means the hybridization will be sp^3 and the electronic geometry of the molecule will be tetrahedral.

(4) The given molecule is, XeF_4

\text{Number of electrons}=\frac{1}{2}\times [8+4]=6

Bond pair electrons = 4

Lone pair electrons = 6 - 4 = 2

The number of electrons are 6 that means the hybridization will be sp^3d^2 and the electronic geometry of the molecule will be octahedral.

But as there are four atoms around the central xenon atom, the fifth and sixth position will be occupied by lone pair of electrons. The repulsion between lone and bond pair of electrons is more and hence the molecular geometry will be square planar.

3 0
2 years ago
Moving from boron to carbon, the intensity of the bulb __
AleksandrR [38]

Explanation:

The thickness of the frosting Moving from boron to carbon, the intensity of the bulb blank because Z increases from blank to blank. The thickness of the frosting blank because the core electron configuration is the same for both atoms. because the core electron configuration is the same for both atoms.

7 0
2 years ago
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