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2 years ago

You are given two fatlike solid substances and determine that sample a has a higher melting point than sample

1 answer:

5 0

Given that sample a has a higher melting point than sample
b. Therefore, sample a is a longer chain of a fatlike solid substance. It could also be that the bonds present in sample a is much stronger which will require more energy to break. Hope this answers the question.

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HElp me plzzzzz??????

iris [78.8K]

Answer:

The answer to your question is given below

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

Zn + 2HCl —> ZnCl2 + H2

Thus, we can write out the atoms present in both the reactant and the product by doing a simple head count. The atoms present are listed below:

Element >>> Reactant >>> Product

Zn >>>>>>>> 1 >>>>>>>>>> 1

H >>>>>>>>> 2 >>>>>>>>> 2

Cl >>>>>>>>> 2 >>>>>>>>> 2

4 0

2 years ago

3. After 7.9 grams of sodium are dropped into a bathtub full of water, how many grams of hydrogen gas are released?

Pavel [41]

Answer:

3) About 0.35 grams of hydrogen gas.

4) About 65.2 grams of aluminum oxide.

Explanation:

Question 3)

We are given that 7.9 grams of sodium is dropped into a bathtub of water, and we want to determine how many grams of hydrogen gas is released.

Since sodium is higher than hydrogen on the activity series, sodium will replace hydrogen in a single-replacement reaction for sodium oxide. Hence, our equation is:

$\displaystyle \text{Na} + \text{H$_2$O}\rightarrow \text{Na$_2$O}+\text{H$_2$}$

To balance it, we can simply add another sodium atom on the left. Hence:

$\displaystyle 2\text{Na} + \text{H$_2$O}\rightarrow \text{Na$_2$O}+\text{H$_2$}$

To convert from grams of sodium to grams of hydrogen gas, we can convert from sodium to moles of sodium, use the mole ratios to find moles in hydrogen gas, and then use hydrogen's molar mass to find its amount in grams.

The molar mass of sodium is 22.990 g/mol. Hence:

$\displaystyle \frac{1\text{ mol Na}}{22.990 \text{ g Na}}$

From the chemical equation, we can see that two moles of sodium produce one mole of hydrogen gas. Hence:

$\displaystyle \frac{1\text{ mol H$_2$}}{2\text{ mol Na}}$

And the molar mass of hydrogen gas is 2.016 g/mol. Hence:

$\displaystyle \frac{2.016\text{ g H$_2$}}{1\text{ mol H$_2$}}$

Given the initial value and the above ratios, this yields:

$\displaystyle 7.9\text{ g Na}\cdot \displaystyle \frac{1\text{ mol Na}}{22.990 \text{ g Na}}\cdot \displaystyle \frac{1\text{ mol H$_2$}}{2\text{ mol Na}}\cdot \displaystyle \frac{2.016\text{ g H$_2$}}{1\text{ mol H$_2$}}$

Cancel like units:

$=\displaystyle 7.9\cdot \displaystyle \frac{1}{22.990}\cdot \displaystyle \frac{1}{2}\cdot \displaystyle \frac{2.016\text{ g H$_2$}}{1}$

Multiply. Hence:

$=0.3463...\text{ g H$_2$}$

Since we should have two significant values:

$=0.35\text{ g H$_2$}$

So, about 0.35 grams of hydrogen gas will be released.

Question 4)

Excess oxygen gas is added to 34.5 grams of aluminum and produces aluminum oxide. Hence, our chemical equation is:

$\displaystyle \text{O$_2$} + \text{Al} \rightarrow \text{Al$_2$O$_3$}$

To balance this, we can place a three in front of the oxygen, four in front of aluminum, and two in front of aluminum oxide. Hence:

$\displaystyle3\text{O$_2$} + 4\text{Al} \rightarrow 2\text{Al$_2$O$_3$}$

To convert from grams of aluminum to grams of aluminum oxide, we can convert aluminum to moles, use the mole ratios to find the moles of aluminum oxide, and then use its molar mass to determine the amount of grams.

The molar mass of aluminum is 26.982 g/mol. Thus:

$\displaystyle \frac{1\text{ mol Al}}{26.982 \text{ g Al}}$

According to the equation, four moles of aluminum produces two moles of aluminum oxide. Hence:

$\displaystyle \frac{2\text{ mol Al$_2$O$_3$}}{4\text{ mol Al}}$

And the molar mass of aluminum oxide is 101.961 g/mol. Hence: $\displaystyle \frac{101.961\text{ g Al$_2$O$_3$}}{1\text{ mol Al$_2$O$_3$}}$

Using the given value and the above ratios, we acquire:

$\displaystyle 34.5\text{ g Al}\cdot \displaystyle \frac{1\text{ mol Al}}{26.982 \text{ g Al}}\cdot \displaystyle \frac{2\text{ mol Al$_2$O$_3$}}{4\text{ mol Al}}\cdot \displaystyle \frac{101.961\text{ g Al$_2$O$_3$}}{1\text{ mol Al$_2$O$_3$}}$

Cancel like units:

$\displaystyle= \displaystyle 34.5\cdot \displaystyle \frac{1}{26.982}\cdot \displaystyle \frac{2}{4}\cdot \displaystyle \frac{101.961\text{ g Al$_2$O$_3$}}{1}$

Multiply:

$\displaystyle = 65.1852... \text{ g Al$_2$O$_3$}$

Since the resulting value should have three significant figures:

$\displaystyle = 65.2 \text{ g Al$_2$O$_3$}$

So, approximately 65.2 grams of aluminum oxide is produced.

5 0

2 years ago

Read 2 more answers

What is the empirical formula for ribose?

BartSMP [9]

Answer:

The empirical formula of ribose (a sugar) is CH2O. In a separate experiment, using a mass spectrometer, the molar mass of ribose was determined to be 150 g/mol.Explanation:

5 0

2 years ago

Read 2 more answers

Consider the following reaction:

Katyanochek1 [597]

Answer:

The three statements are true

Explanation:

For the reaction:

I₂O₅(s) + 5CO(g) → I₂(s) + 5CO₂(g)

State oxidation of iodine in I₂O₅ is:

5 O²⁻ = 10⁻

As you have 2 I and the molecule has no charge, oxidation state of I is +5.

The carbon in CO has an oxidation state of +2 and in CO₂ is +4. That means the carbon is oxidized

An oxidizing agent is a substance that produce the oxidation of the agent that reacts with this one. CO is oxidized because of I₂O₅ is producing its oxidation being the oxidizing agent

Thus, the three statements are true.

3 0

3 years ago

Write the symbol of another isotope of lead with 132 neutrons.

Ivahew [28]

Answer:

Pb^+7

Explanation:

I placed the carrot there to signify the +7 is supposed to be small. It is plus seven because seven nuetrons are added

5 0

3 years ago