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iogann1982 [59]
3 years ago
8

A solution contains 225 g of glucose, c6h12o6, dissolved in enough water to make 0.825 l of solution. what is the molarity of th

e solution?
Chemistry
1 answer:
Art [367]3 years ago
4 0
Molar mass glucose : <span>180.1559 g/mol
</span>
number of moles:

n  = mass solute / molar mass

n = 225 / <span>180.1559 </span>

n = 1.248 moles

therefore:

Molarity  = moles solute / volume 

Molarity =  1.248 / 0.825 => 1.513 M
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An atom of lithium (Li) forms an ionic bond with an atom of chlorine (Cl) to form lithium chloride. How are the valence electron
miskamm [114]

Answer: Option (D) is the correct answer.

Explanation:

Atomic number of lithium is 3 and electrons in its shell are distributed as 2, 1. Atomic number of chlorine is 17 and electrons in its shell are distributed as 2, 8, 7.

Thus, we can see that lithium has 1 extra electron and chlorine has deficiency of 1 electron. Therefore, in order to gain stability lithium will transfer its 1 extra electron to chlorine atom.

Thus, we can conclude that electrons are transferred from the lithium atom to the chlorine atom.

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A.) Low frequency and Long Wavelength
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Calculate the density of nitrogen gas, in grams per liter, at stp.
lesantik [10]

Standard temperature is 273 K

Standard pressure is 1 atm

We use the ideal gas equation to find out density of nitrogen gas in g/L

Ideal gas equation:

PV = nRT\\  PV = (\frac{Mass}{Molar mass)}RT\\   P(Molar mass) = (\frac{Mass}{Volume})RT\\  \frac{Mass}{Volume}=\frac{P(molar mass)}{RT} \\  Density = \frac{P(Molar mass)}{RT}

Molar mass of N_{2} = 28 g/mol

Pressure = 1 atm

Temperature = 273 K

Density = \frac{(1atm)(28 g/mol)}{(0.08206 \frac{L.atm}{mol.K})(273 K)}

= 1.25 g/L

Therefore, density of nitrogen gas at STP is 1.25 g/L

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How can the speed of solvent molecules be slowed down?
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Answer:

D

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Balance the following: <br><br> C₂ + B → B₂C₂<br><br> Ag₄ + SO₂ → AgO + S₂
jarptica [38.1K]

Answer:

The balanced reactions are:

C₂ + 2B → B₂C₂

Ag₄ + 2SO₂ → 4AgO + S₂

Explanation:

The Law of Conservation of Matter postulates that "the mass is not created or destroyed, only transformed." This means that the reagents interact with each other and form new products with physical and chemical properties different from those of the reagents because the atoms of the substances are ordered differently. But the amount of matter or mass before and after a transformation (chemical reaction) is always the same, that is, the quantities of the masses involved in a given reaction must be constant at all times, not changing in their proportions when the reaction ends. In other words, then the mass before the chemical reaction is equal to the mass after the reaction.

In other words, the law of conservation of matter states that since no atom can be created or destroyed in a chemical reaction, the number of atoms that are present in the reagents has to be equal to the number of atoms present in the products.

Then, you must balance the chemical equation. For that, you must first look at the subscripts next to each atom to find the number of atoms in the equation. If the same atom appears in more than one molecule, you must add its amounts .

The coefficients located in front of each molecule indicate the amount of each molecule for the reaction. This coefficient can be modified to balance the equation, just as you should never alter the subscripts.

By multiplying the coefficient mentioned by the subscript, you get the amount of each element present in the reaction.  

You have:

<em>C₂ + B → B₂C₂</em>

Left side: 2 C and 1 B.

Right side: 2 C and 2 B.

Carbon C is balanced since you have the same amount of the element on each side of the reaction, but boron B is not balanced. So, balancing the reaction:

<u><em>C₂ + 2B → B₂C₂</em></u>

Left side: 2 C and 2 B.

Right side: 2 C and 2 B.

Now you have the same amount of each element on each side of the reaction, so the reaction is balanced.

Now you have:

<em>Ag₄ + SO₂ → AgO + S₂</em>

Left side: 4 Ag, 1 S and 2 O.

Right side: 1 Ag, 2 S and 1 O.

Balancing the Ag:

<em>Ag₄ + SO₂ → 4AgO + S₂</em>

Left side: 4 Ag, 1 S and 2 O.

Right side: 4 Ag, 2 S and 4 O.

Balancing the O:

<u><em>Ag₄ + 2SO₂ → 4AgO + S₂</em></u>

Left side: 4 Ag, 2 S and 4 O.

Right side: 4 Ag, 2 S and 4 O.

You have the same amount of each element on each side of the reaction, so the reaction is balanced.

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3 0
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