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2 years ago

1. How many moles of NaCl would be contained in .750 L solution with a molarity of 0.45M?

2 answers:

5 0

Formula of Molarity =

Molarity = no. of moles /volume of solution in L

Given, 0.45M= no. of moles/0.750L
Therefore, no. of moles = 0.45M x 0.750L= 0.3375

yarga [219]2 years ago

4 0

Answer:

$\boxed {\boxed {\sf 0.3375 \ mol \ NaCl}}$

Explanation:

Molarity is found by dividing the moles of solute by liters of solution.

$M=\frac{moles \ of \ solute }{liters \ of \ solution}$

We know the molarity is 0.45 M and there are 0.750 liters of solution. The solute is NaCl (sodium chloride). We can substitute the values into the formula.

$0.45 \ M = \frac{ moles \ of \ NaCl}{ 0.750 \ L }$

The molarity (M) can also be represented by mol/L

$0.45 \ mol/L = \frac{ moles \ of \ NaCl}{ 0.750 \ L }$

We are solving for the moles of solute, so we must isolate the numerator. It is being divided by 0.750 liters. The inverse operation is multiplication, so multiply both sides of the equation by 0.750 L.

$0.750 \ L * 0.45 \ mol/L = \frac{ moles \ of \ NaCl}{ 0.750 \ L }*0.750 \ L$

The liters will cancel out.

$0.750 * 0.45 \ mol = { moles \ of \ NaCl}$

$0.3375 \ mol \ = moles \ of NaCl$

There are 0.3375 moles of NaCl in a 0.750 liter solution with a molarity of 0.45 M.

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Answer:

Solids can hold their shape because their molecules are tightly packed together. ... Atoms and molecules in liquids and gases are bouncing and floating around, free to move where they want. The molecules in a solid are stuck in a specific structure or arrangement of atoms.

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3 years ago

If you have 0.50 mol of ca, how many atoms are present?

Evgen [1.6K]

Answer is: there are 3.011·10²³ atoms of calcium.

n(Ca) = 0.50 mol; amount of substance(calcium).
Na = 6.022·10²³ 1/mol; Avogadro's constant or number.
N(Ca) = n(Ca) · Na.
N(Ca) = 0.50 mol · 6.022·10²³ 1/mol.
N(Ca) = 3.011·10²³; number of calcium atoms.
The mole is an SI unit which measures the number of particles in substance. One mole is equal to 6.022·10²³ atoms.

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3 years ago

what actually is left over of water once it self ionizes? Like what exactly are the hydrogen and hydroxide ions? Do they remain

GrogVix [38]

The formula for the self ionization of water is 2H₂O(l)⇄H₃O⁺(aq)+OH⁻(aq)

The hydronium (H₃O⁺) is usually just referred to as a hydrogen ion or a proton (H⁺) and hydroxide (OH⁻) doesn't have another name that I am aware of. These ions do stay in solution. However the concentrations are really small and the equilibrium constant (K(w)) is 1×10⁻¹⁴.

I hope this helps. Let me know if anything is unclear.

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3 years ago

Identify the solute with the highest van't Hoff factor. And how do you determine which one is highest?

german

Answer : The correct option is, (A) $AlCl_3$

Explanation :

Van't Hoff factor : It is defined as the ratio of the experimental value of the colligative property to the calculated value of the colligative property.

We can determine the van't Hoff factor by the association and dissociation of the compound.

(A) $AlCl_3$

It is an electrolyte that dissociates to give aluminum ion and chloride ion.

The dissociation of $AlCl_3$ will be,

$AlCl_3\rightarrow Al^{3+}+3Cl^{-}$

So, Van't Hoff factor = Number of solute particles = $Al^{3+}+3Cl^{-}$ = 1 + 3 = 4

(B) $KI$

It is an electrolyte that dissociates to give potassium ion and iodide ion.

The dissociation of $KI$ will be,

$KI\rightarrow K^{+}+I^{-}$

So, Van't Hoff factor = Number of solute particles = $K^{+}+I^{-}$ = 1 + 1 = 2

(C) $CaCl_2$

It is an electrolyte that dissociates to give calcium ion and chloride ion.

The dissociation of $CaCl_2$ will be,

$CaCl_2\rightarrow Ca^{2+}+2Cl^{-}$

So, Van't Hoff factor = Number of solute particles = $Ca^{2+}+2Cl^{-}$ = 1 + 2 = 3

(D) $MgSO_4$

It is an electrolyte that dissociates to give magnesium ion and sulfate ion.

The dissociation of $MgSO_4$ will be,

$MgSO_4\rightarrow Mg^{2+}+SO_4^{2-}$

So, Van't Hoff factor = Number of solute particles = $Mg^{2+}+SO_4^{2-}$ = 1 + 1 = 2

(E) Non-electrolyte

The dissociation non-electrolyte is not possible. So, the Van't Hoff factor will always be, 1.

Hence, the highest van't Hoff factor of solute is, $AlCl_3$

7 0

3 years ago

When methyloxirane is treated with HBr, the bromide ion attacks the less substituted position. However, when phenyloxirane is tr

konstantin123 [22]

Answer:

See explanation and picture below

Explanation:

First, in the case of methyloxirane (Also known as propilene oxide) the mechanism that is taking place there is something similar to a Sn2 mechanism. Although a Sn2 mechanism is a bimolecular substitution taking place in only step, the mechanism followed here is pretty similar after the first step.

In both cases, the H atom of the HBr goes to the oxygen in the molecule. You'll have a OH⁺ in both. However, in the case of methyloxirane the next step is a Sn2 mechanism step, the bromide ion will go to the less substitued carbon, because the methyl group is exerting a steric hindrance. Not a big one but it has a little effect there, that's why the bromide will rather go to the carbon with more hydrogens. and the final product is formed.

In the case of phenyloxirane, once the OH⁺ is formed, the next step is a Sn1 mechanism. In this case, the bond C - OH⁺ is opened on the side of the phenyl to stabilize the OH. This is because that carbon is more stable than the carbon with no phenyl. (A 3° carbon is more stable than a 2° carbon). Therefore, when this bond opens, the bromide will go there in the next step, and the final product is formed. See picture below for mechanism and products.

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3 years ago