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Sergeu [11.5K]
3 years ago
5

A geochemist in the field takes a 46.0 mL sample of water from a rock pool lined with crystals of a certain mineral compound X.

He notes the temperature of the pool, 21·C, and caps the sample carefully. Back in the lab, the geochemist filters the sample and then evaporates all the water under vacuum. Crystals of X are left behind. The researcher washes, dries and weighs the crystals. They weigh 0.87 g. yes x10 1
Using only the information above, can you calculate the solubility of X in water at 21.° C?
a. yes
b. no
If you said yes, calculate it.
Be sure your answer has a unit symbol and the right number of significant digits.
Chemistry
1 answer:
Kobotan [32]3 years ago
6 0

Answer:

The correct answer is 1.89130 × 10⁻² g per ml.

Explanation:

Based on the given information, the volume of the water sample is 46 ml, the temperature given is 21 degree C. Weight of the compound mineral X is 0.87 grams obtained post evaporating, washing, and drying of the sample. Yes, on the basis of the given information, one can find the solubility of compound X in water at 21 degree C.  

As 46 ml of water comprise 0.87 grams of the mineral compound X. Therefore, 1 ml of the water sample will comprise,  

= 0.87/46 g of X

= 1.89130 × 10⁻² grams

Hence, the solubility of the compound X in the sample of water is 1.89130 × 10⁻² gram per ml.  

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One liter of oxygen gas at standard temperature and pressure has a mass of 1.43 g. The same volume of hydrogen gas under these c
Alchen [17]

Answer:

Indeed, the two samples should contain about the same number of gas particles. However, the molar mass of \rm O_2\; (g) is larger than that of \rm H_2\; (g) (by a factor of about 16.) Therefore, the mass of the \rm O_2\; (g) sample is significantly larger than that of the \rm H_2\; (g) sample.

Explanation:

The \rm O_2\; (g) and the \rm H_2\; (g) sample here are under the same pressure and temperature, and have the same volume. Indeed, if both gases are ideal, then by Avogadro's Law, the two samples would contain the same number of gas particles (\rm O_2\; (g) and \rm H_2\; (g) molecules, respectively.) That is:

n(\mathrm{O_2}) = n(\mathrm{H}_2).

Note that the mass of a gas m is different from the number of gas particles n in it. In particular, if all particles in this gas have a molar mass of M, then:

m = n \cdot M.

In other words,

  • m(\mathrm{O_2}) = n(\mathrm{O_2}) \cdot M(\mathrm{O_2}).
  • m(\mathrm{H_2}) = n(\mathrm{H_2}) \cdot M(\mathrm{H_2}).

The ratio between the mass of the \rm O_2\; (g) and that of the \rm H_2\; (g) sample would be:

\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})} = \frac{n(\mathrm{O_2})\cdot M(\mathrm{O_2})}{n(\mathrm{H_2})\cdot M(\mathrm{H_2})}\end{aligned}.

Since n(\mathrm{O_2}) = n(\mathrm{H}_2) by Avogadro's Law:

\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})} = \frac{n(\mathrm{O_2})\cdot M(\mathrm{O_2})}{n(\mathrm{H_2})\cdot M(\mathrm{H_2})} = \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}\end{aligned}.

Look up relative atomic mass data on a modern periodic table:

  • \rm O: 15.999.
  • \rm H: 1.008.

Therefore:

  • M(\mathrm{O_2}) = 2 \times 15.999 \approx 31.998\; \rm g \cdot mol^{-1}.
  • M(\mathrm{H_2}) = 2 \times 1.008 \approx 2.016\; \rm g \cdot mol^{-1}.

Verify whether \begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})}= \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}\end{aligned}:

  • Left-hand side: \displaystyle \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})}= \frac{1.43\; \rm g}{0.089\; \rm g} \approx 16.1.
  • Right-hand side: \displaystyle \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}= \frac{31.998\; \rm g \cdot mol^{-1}}{2.016\; \rm g \cdot mol^{-1}} \approx 15.9.

Note that the mass of the \rm H_2\; (g) sample comes with only two significant figures. The two sides of this equations would indeed be equal if both values are rounded to two significant figures.

7 0
4 years ago
Use the periodic table to determine the electron configuration for Ca and Pm in noble-gas notation Ca: [Ar]4s2 [Ar]4s1 [Ar]3s2 [
sveta [45]
Answer:

1) Ca: [Ar]4s²
2) Pm: [Xe]6s²4f⁵

Explanation:

1) Ca:

Its atomic number is 20. So it has 20 protons and 20 electrons.

Since it is in the row (period) 4 the noble gas before it is Ar, and the electron configuration is that of Argon whose atomic number is 18.

So, you have two more electrons (20 - 18 = 2) to distribute.

Those two electrons go the the orbital 4s.

Finally, the electron configuration is [Ar] 4s².

2) Pm

The atomic number of Pm is 61, so it has 61 protons and 61 electrons.

Pm is in the row (period) 6. So, the noble gas before Pm is Xe.

The atomic number of Xe is 54.

Therefore, you have to distribute 61 - 54 = 7 electrons on the orbitals 6s and 4f.

The resultant distribution for Pm is: [Xe]6s² 4f⁵.
5 0
3 years ago
Read 2 more answers
If a molecule has an empirical formula of C2H2O and a molecular mass of 84.0 g/mol, what is the molecular formula?
lord [1]

Answer:

=C₄H₄O₂

Explanation:

Given the empirical formula of a molecule, the he the quotient of the molecular mas and and the empirical mass=constant.

84.0 g/mol/mass of(C₂H₂O)=constant

=84/(12×2+1×2×16)

=84/42

=2

Therefore, the molecular formula is (C₂H₂O)₂=C₄H₄O₂

6 0
3 years ago
Read 2 more answers
What are the different phases of matter and their characteristics?
Novay_Z [31]

Answer:

The five phases of matter. There are four natural states of matter: Solids, liquids, gases and plasma. The fifth state is the man-made Bose-Einstein condensates. In a solid, particles are packed tightly together so they don't move much.

3 0
3 years ago
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What object has a mass of 10 grams and a volume of 2 mL
Elena-2011 [213]

Answer:

Destiny = 5 g/ml

Explanation:

The formula for density is d=

Knowing this, divide your mass by volume:

10 g / 2 ml = 5 g/ml

Final answer:

Destiny = 5 g/ml

3 0
3 years ago
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