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3 years ago

Find the mass of 1.0 x 1025 formula units of iron (II) phosphide

Chemistry

1 answer:

elena55 [62]3 years ago

8 0

3295.1 g iron (II) phosphide of Fe₃P

Explanation:

To solve this question we use the Avogadro's number which tell us that in 1 mole of a substance we have 6.022 × 10²³ particles.

In our chase for iron (II) phosphide (Fe₃P) we have:

in if 1 mole of Fe₃P we have 6.022 × 10²³ formula units

then in X moles of Fe₃P we have 1.0 × 10²⁵ formula units

X = (1 × 1.0 × 10²⁵) / (6.022 × 10²³)

X = 0.1661 × 10² = 16.6 moles of Fe₃P

number of moles = mass / molecular weight

mass = number of moles × molecular weight

mass of Fe₃P = 16.6 × 198.5 = 3295.1 g

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Avogadro's number

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How many grams of KCl is needed to make .75 L of a 1 M solution of KCl?

Zielflug [23.3K]

Answer:

The answer to your question is 25.9 g of KCl

Explanation:

Data

Grams of KCl = ?

Volume = 0.75 l

Molarity = 1 M

Formula

$Molarity = \frac{number of moles}{volume}$

Solve for number of moles

$Number of moles = Molarity x volume$

Substitution

Number of moles = 1 x 0.75

Simplification

Number of moles = 0.75 moles

Molecular mass KCl = 39 + 35.5 = 34.5

Use proportions to find the grams of KCl

34.5 g of KCl ---------------- 1 mol

x ---------------- 0.75 moles

x = (0.75 x 34.5) / 1

x = 25.9 g of KCl

3 0

3 years ago

n acid with a pKa of 8.0 is present in a solution with a pH of 6.0. What is the ratio of the protonated to the deprotonated form

Nady [450]

Answer : The ratio of the protonated to the deprotonated form of the acid is, 100

Explanation : Given,

$pK_a=8.0$

pH = 6.0

To calculate the ratio of the protonated to the deprotonated form of the acid we are using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

$pH=pK_a+\log \frac{[Deprotonated]}{[Protonated]}$

Now put all the given values in this expression, we get:

$6.0=8.0+\log \frac{[Deprotonated]}{[Protonated]}$

$\frac{[Deprotonated]}{[Protonated]}=0.01$

As per question, the ratio of the protonated to the deprotonated form of the acid will be:

$\frac{[Protonated]}{[Deprotonated]}=100$

Therefore, the ratio of the protonated to the deprotonated form of the acid is, 100

5 0

2 years ago

What two things do all matter have in common?

ELEN [110]

Answer:

mass and occupies space

Explanation:

matter is made up of mass and mass takes up space

3 0

3 years ago

Read 2 more answers

You have 47.0 mL of a 2.00 M concentrated or "stock" solution that must be diluted to 0.500 M. How much water should you add?

Inessa [10]

The required volume of water to make the dilute solution of 0.5 M is 188 mL.

<h3>How do we calculate the required volume?</h3>

Required volume of water to dilute the stock solution will be calculated by using the below equation as:

M₁V₁ = M₂V₂, where

M₁ & V₁ are the molarity and volume of stock solution.
M₂ & V₂ are the molarity and volume of dilute solution.

On putting values from the question to the above equation, we get

V₂ = (2)(47) / (0.5) = 188mL

Hence required volume of water is 188 mL.

To know more about volume & concentration, visit the below link:
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6 0

2 years ago

I know the balanced equation, I just need help with the next part, any help is greatly appreciated!

baherus [9]

Answer:

20.4081633 or 20.408

Explanation:

These ones you can do the regular method but it takes to long just divide and shoot.Hope this helps.

3 0

3 years ago