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Nadusha1986 [10]
3 years ago
7

Suppose 4.0 g of hydrogen reacts completely with 32.0 g of oxygen to form one product what is the mass of the product?

Chemistry
1 answer:
creativ13 [48]3 years ago
4 0

Answer: The mass of product, H_2O is, 36.0 grams.

Explanation : Given,

Mass of H_2 = 4.0 g

Mass of O_2 = 32.0 g

Molar mass of H_2 = 2 g/mol

Molar mass of O_2 = 32 g/mol

First we have to calculate the moles of H_2 and O_2.

\text{Moles of }H_2=\frac{\text{Given mass }H_2}{\text{Molar mass }H_2}

\text{Moles of }H_2=\frac{4.0g}{2g/mol}=2.0mol

and,

\text{Moles of }O_2=\frac{\text{Given mass }O_2}{\text{Molar mass }O_2}

\text{Moles of }O_2=\frac{32.0g}{32g/mol}=1.0mol

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

2H_2+O_2\rightarrow 2H_2O

From the balanced reaction we conclude that

2 mole of H_2 react with 1 mole of O_2

From this we conclude that, there is no limiting and excess reagent.

Now we have to calculate the moles of H_2O

From the reaction, we conclude that

2 moles of H_2 react to give 2 moles of H_2O

Now we have to calculate the mass of H_2O

\text{ Mass of }H_2O=\text{ Moles of }H_2O\times \text{ Molar mass of }H_2O

Molar mass of H_2O = 18 g/mole

\text{ Mass of }H_2O=(2.0moles)\times (18g/mole)=36.0g

Therefore, the mass of product, H_2O is, 36.0 grams.

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A 25.0 ml sample of 0.150 m hydrazoic acid is titrated with a 0.150 m naoh solution. what is the ph after 15.0 ml of the sodium
babunello [35]
First, we need to calculate moles of hydrazoic acid NH3:

moles NH3 = molarity * volume 

                    = 0.15 m * 0.025 L

                   =  0.00375 moles

moles NaOH = molarity * volume 

                       = 0.15 m * 0.015 L

                       = 0.00225 moles 

after that we shoul get the total volume = 0.025L + 0.015L

                                                                   = 0.04 L

So we can get the concentration of NH3 & NaOH by:

∴[NH3] = moles NH3 / total volume 

           = 0.00375 moles / 0.04 L

           = 0.09375 M

∴[NaOH] = moles NaOH / total volume 

                = 0.00225 moles / 0.04 L

                = 0.05625 M

then, when we have the value of Ka of NH3 so we can get the Pka value from:

Pka = -㏒Ka 

       = - ㏒ 1.9 x10^-5

      = 4.7 

finally, by using H-H equation we can get PH:

PH = Pka + ㏒[salt/ basic]

PH = 4.7 +㏒[0.05625/0.09375]

∴ PH = 4.48 


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3 years ago
A mixture of helium and hydrogen gases, at a total pressure of 751 mm Hg, contains 0.447 grams of helium and 0.193 grams of hydr
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Answer:

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Explanation:

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3 years ago
Can a molecule with nonpolar bonds ever be polar? Why<br> why not?
Tanya [424]

Answer:

A molecule that has only nonpolar bonds and no polar bonds cannot be polar.

Explanation:

However, a molecule that CONTAINS nonpolar bonds is different, because it can contain polar bonds. A molecule that contains nonpolar bonds can be polar as long as it also contains polar bonds.

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5 0
2 years ago
In the chemical reaction, how many grams of nitrogen are needed to completely react with 24.3 grams of hydrogen
vlabodo [156]

Answer:

112.540 gram of N2

Explanation:

From the balanced chemical equation

N2+3H2----->2NH3

It means that 1 mole of Nitrogen needs 3 moles of hydrogen

In term of mass,we can say that

6.048 gram of H2 needs 28.01 gram of N2

or, 1 gram of H2 needs (28.01/6.048) gram of N2

and 24.3gram of H2 needs (24.3*28.01/6.048) gram of N2

=112.540 gram of N2

5 0
3 years ago
2. Nitrogen has two naturally occurring
kvv77 [185]

Answer:

Suppose that:

Mass of N-14= 14

and Mass of N-15= 15

fraction N-14 x 14 + fraction N-15 x 15 = 14.007

now.. if the fractions were =, that is if fraction N-14 = fraction N-15 = 0.5, then...

0.5 x 14 + 0.5 x 15 = 14.5

more N-15 would make that number larger...

0.3 x 14 + 0.7 x 15 = 14.7

let x = abundance of N-14, then (1-x) = abundance of N-15, and we could solve for x.

(X)x14 + (1-X)x 15 = 14.007

14X + 15 - 15X = 14.007

X = 0.993

the relative abundance of N-14 = 99.3%

the relative abundance of N-15 = 0.7%

So, according to the above explanation N-14 is abundant

Explanation:

Suppose that:

Mass of N-14= 14

and Mass of N-15= 15

fraction N-14 x 14 + fraction N-15 x 15 = 14.007

now.. if the fractions were =, that is if fraction N-14 = fraction N-15 = 0.5, then...

0.5 x 14 + 0.5 x 15 = 14.5

more N-15 would make that number larger...

0.3 x 14 + 0.7 x 15 = 14.7

let x = abundance of N-14, then (1-x) = abundance of N-15, and we could solve for x.

(X)x14 + (1-X)x 15 = 14.007

14X + 15 - 15X = 14.007

X = 0.993

the relative abundance of N-14 = 99.3%

the relative abundance of N-15 = 0.7%

So, according to the above explanation N-14 is abundant

8 0
2 years ago
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