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ycow [4]
3 years ago
7

Assuming all volume measurements are made at the same temperature and pressure, what volume of hydrogen gas is needed to react c

ompletely with 4.55 L of oxygen gas to produce water vapor?
Chemistry
1 answer:
Leno4ka [110]3 years ago
6 0

Answer:

9.1L

Explanation:

Firstly, to solve the problem, we need a balanced chemical equation.

2H2 + O2 ———> 2H2O

2 moles of hydrogen reacted one mole of oxygen. Now, we know that at STP, one mole of a gas occupies a volume of 22.4L, now let us get the number of moles in 4.55L of oxygen.

This means 4.55/22.4 = 0.203125 mole

Since in theory we have 2 moles of hydrogen reacting one mole of oxygen. Hence the number of moles of hydrogen we have is 0.203125 * 2 = 0.40625 mole

We now proceed to get the volume of hydrogen gas. Since 1 mole is 22.4L, then

0.40625 Is 0.40625 * 22.4 = 9.1L

You might be interested in
a compound has 15.39 g of gold for every 2.77 g of chlorine. simplified there is _____ g of gold for every 1 g of chlorine
iren [92.7K]

Answer:

There is 5.56 g of gold for every 1 g of chlorine

Explanation:

The ratio is the relationship between two numbers, defined as the ratio of one number to the other. So, the ratio between two numbers a and b is the fraction \frac{a}{b}

You know that a compound has 15.39 g of gold for every 2.77 g of chlorine. This can be expressed by the ratio:

\frac{15.39 g  of gold}{2.77 g of chlorine}

The proportion is the equal relationship that exists between two reasons and is represented by:    \frac{a}{b}=\frac{c}{d}

This reads a is a b as c is a d.

To calculate the amount of gold per 1 g of chlorine, the following proportion is expressed:

\frac{15.39 g  of gold}{2.77 g of chlorine}=\frac{mass of gold}{1 g of chlorine}

Solving for the mass of gold gives:

mass of gold=1 g of chlorine*\frac{15.39 g  of gold}{2.77 g of chlorine}

mass of gold= 5.56 grams

So, <u><em>there is 5.56 g of gold for every 1 g of chlorine</em></u>

5 0
3 years ago
Calculate the mass of 25,000 molecules of nitrogen gas. (1 mole = 6.02 x 1023 molecules)
Ainat [17]

Hey there!

Molar mass N2 = 28.01 g/mol

Therefore:

28.01 g N2 -------------- 6.02*10²² molecules N2

( mass N2 ?? ) ----------- 25,000 molecules N2

mass N2 =  ( 25,000 * 28.01 ) /  ( 6.02*10²³ )

mass N2 = 700250 / 6.02*10²³

mass N2 = 1.163*10⁻¹⁸ g


Hope that helps!

7 0
3 years ago
Read 2 more answers
Question 2 (10 points)
svet-max [94.6K]
4 infiltration percolation!! I think! Correct me if I’m wrong
6 0
4 years ago
Subject: Chemistry<br><br><br>whoever does this right will give the brainliest &lt;3)
Andru [333]

Answer:

1 A

3 main types of bond are

Ionic bond ( formed due to complete transfer of electron between atoms(

Covalent bond ( formed by mutual sharing of electron)

Metalic bond ( present in the metals due to mobile electrons)

1 B bond in CaO is ionic bond formation in attached image

1 C hydrogen bond with nitrogen is covelent NH3 ammonia is formed because a bond between two non metals is expected to be covalent

More their electronegativity difference between hydrogen and nitrogen is less than 1.7 that makes it covalent

Explanation:

4 0
3 years ago
Read 2 more answers
Naturally occurring element X exists in three isotopic forms: X-28 (27.979 amu, 92.21% abundance), X-29 (28.976 amu 4.70% abunda
bezimeni [28]

<u>Answer:</u> The average atomic mass of X is 28.09 amu

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i .....(1)

  • <u>For isotope 1:</u>

Mass of isotope 1 = 27.979 amu

Percentage abundance of isotope 1 = 92.21 %

Fractional abundance of isotope 1 = 0.9212

  • <u>For isotope 2:</u>

Mass of isotope 2 = 28.976 amu

Percentage abundance of isotope 2 = 4.70 %

Fractional abundance of isotope 2 = 0.0470

  • <u>For isotope 3:</u>

Mass of isotope 3 = 29.974 amu

Percentage abundance of isotope 3 = 3.09 %

Fractional abundance of isotope 3 = 0.0309

Putting values in equation 1, we get:

\text{Average atomic mass of X}=[(27.979\times 0.9212)+(28.976\times 0.0470)+(29.974\times 0.0309)]

\text{Average atomic mass of X}=28.09amu

Hence, the average atomic mass of X is 28.09 amu

4 0
3 years ago
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