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4 years ago

10

Write the complete balanced equation for the reaction between iron (III) oxide (Fe2O3) and water (H2O). You do not need to make

the subscripts smaller; just write them out as regular numbers. For example: Fe2O3 (4 points)

1 answer:

Alina [70]4 years ago

6 0

Fe2O3+3H2O->2Fe(OH)3

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EleoNora [17]

Answer:

conjugate base

Explanation:

7 0

3 years ago

Which of these compounds would you expect to be most soluble in water? ch3ch2cl ch3ch2ch2ch3 ch3ch2ch2f ch3ch2ch2oh?

Montano1993 [528]

Answer : The compound that would be most soluble in water is CH3CH2CH2OH

Explanation :

Water is a polar solvent and can dissolve polar molecules. This is based on the principle "Like dissolves like".

Among the given molecules, CH3CH2CH2CH3 is a hydrocarbon known as butane. All hydrocarbons are non polar. Therefore this compound will not be soluble in water.

The remaining compounds are polar, but Ch3CH2CH2OH shows greater solubility in water owing to presence of hydrogen bonding.

Hydrogen bonding is a type of intermolecular force that gets formed when a compound has hydrogen atom directly attached to highly electro-negative N, F or O atom.

When CH3CH2CH2OH is dissolved in water, it forms hydrogen bonds with water molecules. Due to this hydrogen bonding, the molecule shows greater solubility.

Therefore CH3CH2CH2OH is the most soluble compound in water

6 0

4 years ago

Read 2 more answers

What volume of a 0.00945-M solution of potassium hydroxicce would be required to titrae 50.00mL of a sample of acid rain with a

Zepler [3.9K]

<u>Answer:</u> The volume of HBr solution required is 130.16 mL

<u>Explanation:</u>

To calculate the concentration of base, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is $KOH$

We are given:

$n_1=2\\M_1=123\times 10^{-4}M=0.0123M\\V_1=50mL\\n_2=1\\M_2=0.00945M\\V_2=?mL$

Putting values in above equation, we get:

$2\times 0.0123\times 50=1\times 0.00945\times V_2\\\\V_2=130.16mL$

Hence, the volume of KOH solution required is 130.16 mL

5 0

4 years ago

Based on table N of the reference tables what is the number of hours required for 42k (potassium -42) to undergo three half-life

skelet666 [1.2K]

The number of hours required : 37.2 hours

<h3>Further explanation</h3>

Given

⁴²K (potassium -42)

Required

The number of hours

Solution

The atomic nucleus can experience decay into 2 particles or more due to the instability of its atomic nucleus.

Usually, radioactive elements have an unstable atomic nucleus.

Based on Table N(attached), the half-life for ⁴²K is 12.4 hours, which means half of a sample of ⁴²K will decay in 12.4 hours

For three half-life periods :

$\tt 3\times 12.4=37.2~hours$

3 0

3 years ago

What is an example

Paha777 [63]

Answer:

NaF sodium flouride

Explanation:

7 0

3 years ago