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slega [8]
3 years ago
7

You are given mixture made of 290 grams of water and 14.2 grams of salt. Determine the % by mass of salt in the salt solution.

Chemistry
1 answer:
marishachu [46]3 years ago
3 0

Answer:

Solution is 4.67% by mass of salt

Explanation:

% by mass is the concentration that defines the mass of solute in 100g of solution.

In this case we have to find out the mass of solution with the data given:

Mass of solution = Mass of solute + Mass of solvent

Solute:  Salt → 14.2 g

Solvent: Water → 290 g

Solution's mass = 14.2 g + 290g = 304.2 g

% by mass = (mass of solute / mass of solution) . 100

(14.2 g / 304.2g) . 100 = 4.67 %

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Consider the following reaction NHAHS(s)NH3(g) + H2S(g) If a flask maintained at 302 K contains 0.196 moles of NH4HS(s) in equil
quester [9]

Answer:

Kc = 3.72 × 10⁶

Explanation:

Let's consider the following reaction:

NH₄HS(g) ⇄ NH₃(g) + H₂S(g)

At equilibrium, we have the following concentrations:

[NH₄HS] = 0.196 M (assuming a 1 L flask)

[NH₃] = 9.56 × 10² M

[H₂S] = 7.62 × 10² M

We can replace this data in the Kc expression.

Kc=\frac{[NH_{3}] \times [H_{2}S] }{[NH_{4}HS]} =\frac{9.56 \times 10^{2}  \times 7.62  \times 10^{2}}{0.196} =3.72 \times 10^{6}

7 0
3 years ago
When the temperature in a room increases from 25°C to 33°C, changes from a solid to a liquid.
mr Goodwill [35]

Answer:

true

Explanation:

Because ice melts if the temperature increasese

8 0
3 years ago
an atmosphere is considered hazardous if it contains a hazardous gas in excess of 10 percent of the hazardous material's:
svlad2 [7]

Lower flammable limit means the lowest concentration of a material that will propagate a flame.

What is hazardous atmosphere?

It is an atmosphere that may expose employees to risk of death, incapacitation, impairment of ability to self-rescue, injury, or acute illness from one or more of following causes

  • Flammable gas, vapor, or mist in excess of 10 percent of lower flammable limit (LFL)
  • Airborne combustible dust at concentration that meets or exceeds its LFL

What is lower flammable limit?

  • It means the lowest concentration of a material that will propagate a flame.
  • The LFL is usually expressed as percent by volume of material in air (or other oxidant)
  • Atmospheres with concentration of flammable vapors at or above 10 percent of lower explosive limit (LEL) are considered hazardous when located in confined spaces.
  • However, atmospheres with flammable vapors below 10 percent of LEL are not necessarily safe. Such atmospheres are too lean to burn

Learn more about lower flammable limit at brainly.com/question/2456135

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6 0
1 year ago
Which statement is true in comparison to a weak acid, a strong acid
Allushta [10]

Answer: Strong acid vs weak acid

Strong acids and strong bases refer to species that completely dissociate to form ions in solution.

Explanation: By contrast, weak acids and bases ionize only partially, and the ionization reaction is reversible. Thus, weak acid and base solutions contain multiple charged and uncharged species in dynamic equilibrium.

7 0
3 years ago
In the absence of sodium methoxide, the same alkyl bromide gives a different product. Draw an arrowpushing mechanism to account
hoa [83]

Answer:

See explanation below

Explanation:

The question is incomplete, cause you are not providing the structure. However, I found the question and it's attached in picture 1.

Now, according to this reaction and the product given, we can see that we have sustitution reaction. In the absence of sodium methoxide, the reaction it's no longer in basic medium, so the sustitution reaction that it's promoted here it's not an Sn2 reaction as part a), but instead a Sn1 reaction, and in this we can have the presence of carbocation. What happen here then?, well, the bromine leaves the molecule leaving a secondary carbocation there, but the neighbour carbon (The one in the cycle) has a more stable carbocation, so one atom of hydrogen from that carbon migrates to the carbon with the carbocation to stabilize that carbon, and the result is a tertiary carbocation. When this happens, the methanol can easily go there and form the product.

For question 6a, as it was stated before, the mechanism in that reaction is a Sn2, however, we can have conditions for an E2 reaction and form an alkene. This can be done, cause the extoxide can substract the atoms of hydrogens from either the carbon of the cycle or the terminal methyl of the molecule and will form two different products of elimination. The product formed in greater quantities will be the one where the negative charge is more stable, in this case, in the primary carbon of the methyl it's more stable there, so product 1 will be formed more (See picture 2)

For question 6b, same principle of 6a, when the hydrogen migrates to the 2nd carbocation to form a tertiary carbocation the methanol will promove an E1 reaction with the vecinal carbons and form two eliminations products. See picture 2 for mechanism of reaction.

3 0
3 years ago
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