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zimovet [89]
3 years ago
13

How many grams of iki would it take to obtain a 100 ml solution of 0.300 m iki? how many grams of iki would it take to create a

100 ml solution of 0.600 m iki?
Chemistry
1 answer:
GalinKa [24]3 years ago
3 0

0.300 M IKI represents the concentration which is in molarity of a potassium iodide solution. This means that for every liter of solution there are 0.300 moles of potassium iodide. Knowing that molarity is a ratio of solute to solution.

By using a conversion factor:

100 ml x (1L / 1000 mL) x (0.300 mol Kl / 1 L) x (166.0g / 1 mol Kl) = 4.98 g

Therefore, in the first conversion by simply converting the unit of volume to liter, Molarity is in L where the volume is in liters. The next step is converted in moles from volume by using molarity as a conversion factor which is similar to how density can be used to convert between volume and mass. After converting to moles it is simply used as molar mass of Kl which is obtained from periodic table to convert from mole to grams.

In order to get the grams of IKI to create a 100 mL solution of 0.600 M IKI, use the same formula as above:

100 ml x (1L / 1000 mL) x (0.600 mol Kl / 1 L) x (166.0g / 1 mol Kl) = 9.96 g

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One mole of a substance is equal to 6.022 × 10²³ units of that substance (such as atoms, molecules, or ions). The number 6.022 × 10²³ is known as Avogadro's number or Avogadro's constant. The concept of the mole can be used to convert between mass and number of particles.

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3 years ago
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Compare and contrast what happens to the particles of an ionic solid and a molecular solid when each mixes with water
Setler79 [48]
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7 0
4 years ago
QUESTION 3 Consider a solution containing 0.80 M NaF and 0.80 M HF. Calculate the moles of HF and the concentration of HF after
Lisa [10]

Answer:

0.056moles HF and 0.70M

Explanation:

When a strong acid is added to a buffer, the acid reacts with the conjugate base.

In the system, NaF and HF, weak acid is HF and conjugate base is NaF. The reaction of NaF with HCl (Strong acid) is:

NaF + HCl → HF + NaCl

Initial moles of NaF and HF in 60.0mL of solution are:

NaF:

0.0600L × (0.80mol / L)= 0.048 moles NaF

HF:

0.0600L × (0.80mol / L)= 0.048 moles HF

Then, the added moles of HCl are:

0.0200L × (0.40mol / L) = 0.008 moles HCl.

Thus, after the reaction, moles of HF produced are 0.008 moles + the initial 0.048moles of HF, moles of HF are:

<em>0.056moles HF</em>

<em></em>

In 20.0mL + 60.0mL = 80.0mL = 0.0800L, molarity of HF is:

0.056mol HF / 0.0800L = <em>0.70M</em>

6 0
3 years ago
1. Metallic strontium crystallizes in a face-centered cubic lattice, with one Sr atom per lattice point. If the edge length of t
Zarrin [17]

Answer:

r=215pm

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Explanation:

From the question we are told that:

Edge length of the unit cell l=608pm

a)

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4r=\sqrt{2a}

Therefore

4r=\sqrt{2*608}

r=\frac{\sqrt{2*608}}{4}

r=215pm

b)

From the question we are told that:

Density \rho=7.297

Edge length of l=630.0 pm=>630*10^-{10}

Therefore Volume  is given as

V=l^3

V=630*10^-{10}^3

V=2.50047*10^{−22}

Generally the equation for Mass is mathematically given by

m=Volume*density

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n=\frac{M}{Molar M}

n=\frac{1.83*10^{-21}g}{55}

n=3.32*10^{-23}

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N_{Mn}=3.32*10^{-23}*6.023*10^{23}

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7 0
3 years ago
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