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11Alexandr11 [23.1K]
3 years ago
5

The question is below

Chemistry
1 answer:
nevsk [136]3 years ago
7 0
The reducing agent is donating electrons and therefore becoming oxidised itself. In this scenario elemental zinc (Zn) is in a reduced state and is oxidised to become Zn2+, in doing so it donates electrons to Fe2+, thereby reducing it to elemental iron (Fe).

The elemental Zinc in solid state is therefore the reducing agent as it reduces Fe2+ to Fe(s).
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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
3 years ago
Read 2 more answers
In Section 5.6, we learned that triple bonds are stronger and shorter than single bonds. For example, a C−CC−C single bond has a
olya-2409 [2.1K]

Answer:the strength of a bond depends on the amount of overlap between the atoms.

Explanation:the greater the overlap the greater the strength of the bond.

6 0
3 years ago
How many elements does the periodic table have quick my wifi not working help please
dlinn [17]
There is 118 elements<span> in the periodic table</span>
3 0
3 years ago
cyclohexanol is burned as a fuel, according to the following equation, how many moles of oxygen are needed to produce 13.7 mol o
lidiya [134]

 The moles  of oxygen  that are needed  to produce 13.7 moles  of  carbon  dioxide  is  21.17  moles  of Oxygen


<u><em>calculation</em></u>

2 C₆H₁₂O  + 17 O₂ → 12 CO₂  +12 H₂O


The moles of O₂  is determined using the mole ratio

that  is for  given equation above   O₂ : Co₂  is 17 :12

therefore the moles of O ₂= 13.7   moles x 17/12 =21.17  moles

3 0
3 years ago
A 35,000 Newton car runs up a hill that is 45.5 meters high in 1.6 seconds. what power did it exert?
Evgesh-ka [11]

Answer:

995.313KW

Explanation:

the explanation is in the picture

please like and Mark as brainliest

7 0
3 years ago
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