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3 years ago

How have safety concerns about chemicals changed over time?

Chemistry

1 answer:

Alex787 [66]3 years ago

8 0

Answer:

C. As scientists have learned more about chemicals, they have become more aware of their dangers

Explanation:

Chemicals are substances with standard compositions, held together by chemical bonds. Chemicals can exist in different phases such as solid, liquid, and gaseous phases. Over time, with a growing understanding of chemicals, their composition, and their reaction methods, scientists now have a better knowledge of the dangers chemicals can pose. To that effect, they have created standard safety measures for those who have to work with these chemicals.

When these guidelines are strictly adhered to, the chances of suffering accidents, burns, and explosions with these chemicals are significantly reduced.

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the diagram shows light striking a pane of glass . some of the light is reflected , and some of the light is transmitted. the le

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When light crosses any barrier

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3 years ago

0.01040 m as an integer

SpyIntel [72]

Answer:

Here,

0.01040 m as an integer= 1.04 × 10-2

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2 years ago

Why is the answer C for this problem?

pshichka [43]

Answer:

$\boxed{\text{(C) X}$_{3}$P$_{2}}$

Explanation:

Step 1. Identify the Group that contains X

We look at the consecutive ionization energies and hunt for a big jump between them

$\begin{array}{crc}n & IE_{n} & IE_{n} - IE_{n-1}\\1 & 730 & \\2 & 1450 & 720\\3 & 7700 & 6250\\4 & 10500 & 2800\\\end{array}$

We see a big jump between n = 2 and n = 3. This indicates that X has two valence electrons.

We can easily remove two electrons, but the third electron requires much more energy. That electron must be in the stable, filled, inner core.

So, X is in Group 2 and P is in Group 15.

Step 2. Identify the Compound

X can lose two valence electrons to reach a stable octet, and P can do the same by gaining three electrons.

We must have 3 X atoms for every 2 P atoms.

The formula of the compound is $\boxed{\text{X}$_{3}$P$_{2}}$}$ .

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3 years ago

Three 1.0-l flasks, maintained at 308 k, are connected to each other with stopcocks. initially the stopcocks are closed. one of

Licemer1 [7]

Answer:

0.6103 atm.

Explanation:

We need to calculate the vapor pressure of each component after the stopcocks are opened.

Volume after the stopcocks are opened = 3.0 L.

1) For N₂:

P₁V₁ = P₂V₂

P₁ = 1.5 atm & V₁ = 1.0 L & V₂ = 3.0 L.

P₂ of N₂ = P₁V₁ / V₂ = (1.5 atm) (1.0 L) / (3.0 L) = 0.5 atm.

2) For H₂O:

Pressure of water at 308 K is 42.0 mmHg.

we need to convert from mmHg to atm: (1.0 atm = 760.0 mmHg).

P of H₂O = (1.0 atm x 42.0 mmHg) / (760.0 mmHg) = 0.0553 atm.

We must check if more 2.2 g of water is evaporated,

n = PV/RT = (0.0553 atm) (3.0 L) / (0.082 L.atm/mol.K) (308 K) = 0.00656 mole.

m = n x cmolar mass = (0.00656 mole) (18.0 g/mole) = 0.118 g.

It is lower than the mass of water in the flask (2.2 g).

3) For C₂H₅OH:

Pressure of C₂H₅OH at 308 K is 102.0 mmHg.

we need to convert from mmHg to atm: (1.0 atm = 760.0 mmHg).

P of C₂H₅OH = (1.0 atm x 102.0 mmHg) / (760.0 mmHg) = 0.13421 atm.

We must check if more 0.3 g of C₂H₅OH is evaporated,

n = PV/RT = (0.13421 atm) (3.0 L) / (0.082 L.atm/mol.K) (308 K) = 0.01594 mole.

m = n x molar mass = (0.01594 mole) (46.07 g/mole) = 0.7344 g.

It is more than the amount in the flask (0.3 g), so the pressure should be less than 0.13421 atm.

We have n = mass / molar mass = (0.30 g) / (46.07 g/mole) = 0.00651 mole.

So, P of C₂H₅OH = nRT / V = (0.00651 mole) (0.082 L.atm/mole.K) (308.0 K) / (3.0 L) = 0.055 atm.

So, total pressure = P of N₂ + P of H₂O + P of C₂H₅OH = 0.5 atm + 0.0553 atm + 0.055 atm = 0.6103 atm.

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3 years ago

you are given a solution of lead nitrate. use any of the solution used in this experiment to form a white precipitate of lead ch

Vesnalui [34]

We have as a reagent a salt, lead nitrate (Pb(NO3)2), and an unknown solution that gives us as a product lead chloride (PbCl2). That is, the solution must contain chlorine.

If a chlorine solution is used we will have the following reaction:

$Pb(NO_{3)2}+2Cl^-\rightarrow PbCl_2+2NO^-_3$

So, with a chlorine solution, we will have a white precipitate of lead chloride.

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1 year ago