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3 years ago

11

To interconvert the concentration units molality (m) and mass percent, you must also know the density of the solution. A) True B

) False

1 answer:

uranmaximum [27]3 years ago

7 0

Answer:

The correct answer is option false.

Explanation:

Molality of the solution defined as moles of substance present in 1 kilogram of solvent.

Moles = $\frac{Mass of solute}}{\text{molar mass of solute}}$

$Molality=\frac{Moles}{\text{Mass of solvent(kg)}}$

Mass of percent (w/w%) of the solution is defined as amount of solute present in 100 grams of solution.

$(w/w\%)_=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100$

So, if want to inter-convert molality into mass percent we can do that without knowing density of solution.

Mass of solution = Mass of solute + Mass of solvent

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How many atoms are there in 8.88 g Si?

Mariana [72]

Answer:

$\boxed {\boxed {\sf 1.90 \times 10^{23} \ atoms \ Si}}$

Explanation:

We are asked to find how many atoms are in 8.88 grams of silicon.

<h3>1. Grams to Moles </h3>

First, we convert grams to moles. We use the molar mass or the mass of 1 mole of a substance. These values are found on the Periodic Table as they are equal to the atomic masses, but the units are grams per mole instead of atomic mass units.

Look up silicon's molar mass.

Si: 28.085 g/mol

We will convert using dimensional analysis. Set up a conversion factor with the molar mass.

$\frac { 28.085 \ g \ Si}{1 \ mol \ Si}$

We are converting 8.88 grams of silicon to moles, so we multiply by this value.

$8.88 \ g \ Si *\frac { 28.085 \ g \ Si}{1 \ mol \ Si}$

Flip the fraction so the units of grams of silicon cancel.

$8.88 \ g \ Si *\frac{1 \ mol \ Si} { 28.085 \ g \ Si}$

$8.88 *\frac{1 \ mol \ Si} { 28.085 }$

$\frac {8.88} { 28.085 } \ mol \ Si$

$0.316183015845 \ mol \ Si$

<h3>2. Moles to Atoms </h3>

Next, we convert moles to atoms. We use Avogadro's Number or 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this case, the particles are atoms of silicon.

Set up another conversion factor.

$\frac {6.022 \times 10^{23} \ atoms \ Si}{1 \ mol \ Si}$

Multiply by the number of moles we calculated.

$0.316183015845\ mol \ Si *\frac {6.022 \times 10^{23} \ atoms \ Si}{1 \ mol \ Si}$

The units of moles of silicon cancel.

$0.316183015845 * \frac {6.022 \times 10^{23} \ atoms \ Si}{1}$

$0.316183015845 * {{6.022 \times 10^{23} \ atoms \ Si}$

$1.90405412 \times 10^{23} \ atoms \ Si$

<h3>3. Significant Figures</h3>

The original measurement of 8.88 grams has 3 significant figures, so our answer must have the same.

For the number we calculated, that is the hundredth place. The 4 in the thousandth place tells us to leave the 0 in the hundredth place.

$1.90 \times 10^{23} \ atoms \ Si$

<u>8.88 grams of silicon contains 1.90 ×10²³ atoms of silicon.</u>

6 0

3 years ago

Does adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent, a lesser extent, or a greater e

igor_vitrenko [27]

Answer:

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water <em><u>to the same extent</u></em> by adding 1 mol of $C_06H_{12}O_6$ to 1 kg of water.

Explanation:

1) Moles of NaCl , $n_1=1 mol$

Mass of water = m= 1 kg = 1000 g

Moles of water = $n_2=\frac{1000 g}{18 g/mol}=55.55 mol$

Vapor pressure of the solution = $p$

Vapor pressure of the pure solvent that is water = $p_o=17.5 Torr$

Mole fraction of solute(NaCl)= $\chi_1=\frac{n_1}{n_1+n_2}$

$\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}$

$\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}$

$p=17.19 Torr$

The vapor pressure for the NaCl solution at 17.19 Torr.

2) Moles of sucrose , $n_1=1mol$

Mass of water = m = 1 kg = 1000 g

Moles of water = $n_2=\frac{1000 g}{18 g/mol}=55.55 mol$

Vapor pressure of the solution = $p'$

Vapor pressure of the pure solvent that is water = $p_o=17.5 Torr$

Mole fraction of solute ( glucose)= $\chi_1=\frac{n_1}{n_1+n_2}$

$\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}$

$\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}$

$p'=17.19 Torr$

The vapor pressure for the glucose solution at 17.19 Torr.

p = p' = 17.19 Torr

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent by adding 1 mol of $C_06H_{12}O_6$ to 1 kg of water.

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3 years ago

Octylphenol ethoxylate is a detergent that is commonly used in chemistry laboratories. It is a thick, yellowish liquid. A chemis

Trava [24]

Answer:

is solvent, solute, solute

Explanation:

3 0

3 years ago

Which element would be the most conductive and

Masja [62]

Malleability described the property of physical deformation under some compressive stress; a malleable material could, for example, be hammered into thin sheets. Malleability is generally a property of metallic elements: The atoms of elemental metals in the solid state are held together by a sea of indistinguishable, delocalized electrons. This also partially accounts for the generally high electrical and thermal conductivity of metals.

In any case, only one of the elements listed here is a metal, and that’s copper. Moreover, the other elements (hydrogen, neon, and nitrogen) are gases under standard conditions, and so their malleability wouldn’t even be a sensible consideration.

8 0

3 years ago

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B) Explain the importance of photosynthesis for plants.

Sever21 [200]

Answer:

Green plants and trees use photosynthesis to make food from sunlight, carbon dioxide and water in the atmosphere: It is their primary source of energy. The importance of photosynthesis in our life is the oxygen it produces. Without photosynthesis there would be little to no oxygen on the planet.

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3 years ago