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3 years ago

How many grams of KOH are required to prepare 500. mL of 0.450 KOH solution?

Chemistry

1 answer:

Digiron [165]3 years ago

5 0

Answer:

KOH molar mass = 39 + 16 + 1 = 56g

To make 1 L of 1M soln needs 56g KOH

To make 500mL 1M needs 56/2 = 28g

To make 500mL 0.2M needs 28 x 0.2gn:

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Question 8 of 10

Harlamova29_29 [7]

Answer:

Explanation:

Because it is answered with a letter

7 0

3 years ago

2. The empirical formula of a molecule is CH2O. In an experiment, the molar mass of the molecule was determined to be 360.3 g/mo

Vanyuwa [196]

Answer:

The answer to your question is:

2.- C₁₂ H₂₄ O₁₂

Explanation:

2.-

Data

CH2O

molar mass = 360.3 g/mol

Molar mass of CH2O = 12 + 2 + 16 = 30g

Divide molar mass given by molar mass obtain

x = 360.3/30

x = 12

Finally

C₁₂ H₂₄ O₁₂

Molar mass = (12 x 12) + (24 x 1) + (16 x 12) = 144 + 24 + 192 = 360 g

3.- First we need to write the complete equation of the reaction and balanced it.

Then, we need to convert the mass given to moles of each compound.

After that, we need used rule of three calculate the amount of products based on the moles of reactants given.

Finally, convert the moles to grams.

4.-

a.- It is a relation between the mass of product obtain in an experiment and the mass of a product obtain theoretically times 100.

b.-

35 g of Mg reacted with excess O2

percent yield = 90%

Actual yield = ?

Formula

Percent yield = (actual yield/theoretical yield) x 100

Equation

2Mg + O2 ⇒ 2MgO

48.62 g of Mg ----------------- 80.62 g of MgO

35g ------------------ x

x = 58 g of MgO (Theoretical yield)

Theoretical yield = 58 g of MgO

Actual yield = percent yield x theoretical yield / 100

= 90 x 58 / 100

= 52. 23 g

5 0

3 years ago

My professor gave me two questions to solve using the Van Der Waals Equation. She told us to solve for P and the second one we h

Fed [463]

Answer:

P=atm

$b=\frac{L}{mol}$

Explanation:

The problem give you the Van Der Waals equation:

$(P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT$

First we are going to solve for P:

$(P+\frac{n^{2}a}{V^{2}})=\frac{nRT}{(V-nb)}$

$P=\frac{nRT}{(V-nb)}-\frac{n^{2}a}{v^{2}}$

Then you should know all the units of each term of the equation, that is:

$P=atm$

$n=mol$

$R=\frac{L.atm}{mol.K}$

$a=atm\frac{L^{2}}{mol^{2}}$

$b=\frac{L}{mol}$

$T=K$

$V=L$

where atm=atmosphere, L=litters, K=kelvin

Now, you should replace the units in the equation for each value:

$P=\frac{(mol)(\frac{L.atm}{mol.K})(K)}{L-(mol)(\frac{L}{mol})}-\frac{(mol^{2})(\frac{atm.L^{2}}{mol^{2}})}{L^{2}}$

Then you should multiply and eliminate the same units which they are dividing each other (Please see the photo below), so you have:

$P=\frac{L.atm}{L-L}-atm$

Then operate the fraction subtraction:

P= $P=\frac{L.atm-L.atm}{L}$

$P=\frac{L.atm}{L}$

And finally you can find the answer:

P=atm

Now solving for b:

$(P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT$

$(V-nb)=\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}$

$nb=V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}$

$b=\frac{V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}}{n}$

Replacing units:

$b=\frac{L-\frac{(mol).(\frac{L.atm}{mol.K}).K}{(atm+\frac{mol^{2}.\frac{atm.L^{2}}{mol^{2}}}{L^{2}})}}{mol}$

Multiplying and dividing units,(please see the second photo below), we have:

$b=\frac{L-\frac{L.atm}{atm}}{mol}$

$b=\frac{L-L}{mol}$

$b=\frac{L}{mol}$

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3 years ago

Formula for Tricarbon heptabromide.

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I believe it would be Au^4Cl8

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An atom has three shells with 3 electrons in the last shell. This atom could be.....

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Since the total amount of valence electrons is 3, it is in group 13 in the periodic table..therefore, it is specified as Boron.✅

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