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allochka39001 [22]

3 years ago

6

The molar mass of H2O is 18.01 g/mol. The molar mass of O2 is 32.00 g/mol. What mass of H2O, in grams, must react to produce 50.

00 g of O2?

2 answers:

Annette [7]3 years ago

8 0

This reaction is called the electrolysis of water. The balanced reaction is:
2H2O = 2H2 + O2
We are given the amount of O2 produced from the electrolysis reaction. This will be the starting point of our calculation.

50.00 grams O2 ( 1 mol O2 / 32 grams O2) ( 2 mol H2O / 1 mol O2) ( 18.01 g H2O / 1 mol H2O ) = 56.28 g H2O

Tomtit [17]3 years ago

7 0

Answer:

Mass of H2O = 56.27 g

Explanation:

<u>Given:</u>

Molar mass of H2O = 18.00 g/mol

Molar mass of O2 = 32.00 g/mol

Mass of O2 produced = 50.00 g

<u>To determine:</u>

Mass of H2O reacted

<u>Explanation:</u>

Water breaks apart during electrolysis to produce O2 and H2. This can be represented as:

$2H2O \rightarrow 2H2 + O2$

Moles of O2 produced = $\frac{Mass\ of\ O2}{Molar\ Mass} = \frac{50.00}{32.00} = 1.563$

Based on the reaction stoichiometry:

2 moles of H2O produces 1 mole of O2

Therefore, moles of H2O that would react to produce 1.563 moles of O2 is:

= $\frac{1.563\ moles\ O2*2\ moles\ H2O}{1\ mole\ O2} =3.126 moles$

$Mass of H2O = Moles of H2O * Molar mass \\= 3.126 moles * 18 g/mol = 56.27 \ g$

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A solution was prepared by mixing 0.5000 m hno2 with 0.380 m no2-. ka = 4.58 x 10-4. calculate the ph of the solution. show work

adell [148]

pH of buffer can be calculated as:

pH=pKa+log[salt]/[Acid]

As ka = 4.58 x 10-4

Concentration of [Salt] that is NO2(-1)=0.380M

Concentration of [Acid] that is HNO2=0.500M

So, pH= -log(4.58*10^-4)+log((0.380)/0.500))

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So pH of solution will be 3.21

7 0

3 years ago

Please help me for question 1 and 2

likoan [24]

Answer:-

1) 6 mol

2) Mo

Explanation: -

Mass of Ozone = 48 g

Chemical formula of ozone = O3

Molar mass of Ozone O 3 = 16 x 3 = 48 g mol-1

Number of moles of ozone = Mass / molar mass

= 48 g / 48 g mol-1

= 1 mol

According to Avogadro’s law, 1 mole of a substance has 6.02 x 10^ 22 molecules.

So 1 mol of O3 has 6.02 x 10^ 22 molecules of ozone.

Now each Ozone molecule has 3 atoms of oxygen.

So, 1 mol of ozone has 3 x 6.02 x 10^22 atoms of oxygen.

Sodium must have 2 x 3 x 6.02 x 10^22 atoms as per the question.

According to Avogadro’s law, 6.02 x 10^ 22 atoms are in 1 mol of sodium

So, for 2 x 3 x 6.02 x 10^22 atoms, there should be (1/ 6.02 x 10^ 22) x 2 x 3 x 6.02 x 10^22

= 6 mol of sodium.

b)

Let the mass of M be m g

Formula of hexafluoride = MF6.

Mass of the hexafluoride = g + 6 x 19

= m + 114

Mass of M=0.250g

Moles of M = 0.250/m

Mass of MF6= 0.547g

Moles of MF6 = 0.547/ (m + 114)

We know 1 mole of M gives 1 mole of MF6.

0.250/m moles of M gives 0.250/m moles of MF6.

But number of moles of MF6 = 0.547/ (m + 114)

Thus

0.250/m = (0.547)/ (m +114))

0.250m + 0.250 x 114 = 0.547m

m = 0.250 x 114 / (0.547 -0.250)

= 96

We see from the given data that Mo is 96.

So M is Mo.

4 0

3 years ago

If one mole of a substance has a mass of 56.0 g, what is the mass of 11 nanomoles of the substance? Express your answer in nanog

9966 [12]

Answer:

616,0 ng is the right answer.

Explanation:

You should know that 1 mole = 1 .10^9 nanomoles

Get the rule of three.

1 .10^9 nanomoles ...................... 56.0 gr

11 nanomoles .....................

(11 x 56) / 1 .10^9 nanomoles = 6.16 x 10^-7 gr

Let's convert

6.16 x 10^-7 gr x 1 .10^9 = 616 ngr

8 0

3 years ago

a 2.7 L of N2 is collected at 121kpa and 288 K . if the pressure increases to 202 kpa and the temperature rises to 303 K , what

jok3333 [9.3K]

Answer:

The gas will occupy a volume of 1.702 liters.

Explanation:

Let suppose that the gas behaves ideally. The equation of state for ideal gas is:

$P\cdot V = n\cdot R_{u}\cdot T$ (1)

Where:

$P$ - Pressure, measured in kilopascals.

$V$ - Volume, measured in liters.

$n$ - Molar quantity, measured in moles.

$T$ - Temperature, measured in Kelvin.

$R_{u}$ - Ideal gas constant, measured in kilopascal-liters per mole-Kelvin.

We can simplify the equation by constructing the following relationship:

$\frac{P_{1}\cdot V_{1}}{T_{1}} = \frac{P_{2}\cdot V_{2}}{T_{2}}$ (2)

Where:

$P_{1}$ , $P_{2}$ - Initial and final pressure, measured in kilopascals.

$V_{1}$ , $V_{2}$ - Initial and final volume, measured in liters.

$T_{1}$ , $T_{2}$ - Initial and final temperature, measured in Kelvin.

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