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2 years ago

2Na + 2H2O → 2NaOH + H2

Chemistry

2 answers:

Zarrin [17]2 years ago

5 0

The initial quantity of sodium metal is 17.25

Alinara [238K]2 years ago

5 0

Answer: 13 grams

Explanation:

$2Na+2H_2O\rightarrow 2NaOH+H_2$

According to avogadro's law, 1 mole of every substance occupies 22.4 Liters at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

According to stoichiometry,

1 mole of $H_2$ is produced from 2 moles of sodium

or 22.4 L of $H_2$ at STP is produced from =2 moles of sodium

thus 6.30 L of $H_2$ at STP is produced from = $\frac{2}{22.4}\times 6.30=0.56moles$ of sodium

To calculate the mass of sodium metal used

$\text{mass of sodium used}=moles\times {\text {Molar mass}}=0.56\times 23=13grams$

Thus the initial quantity of sodium metal used is 13 grams.

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A 23.0g sample of a compound contains 12.0g of C, 3.0g of H, and 8.0g of O.What the empirical formula of the compound

Kryger [21]

Answer:

The empirical formula of compound is C₂H₆O.

Explanation:

Given data:

Mass of carbon = 12 g

Mass of hydrogen = 3 g

Mass of oxygen = 8 g

Empirical formula of compound = ?

Solution:

First of all we will calculate the gram atom of each elements.

no of gram atom of carbon = 12 g / 12 g/mol = 1 g atoms

no of gram atom of hydrogen = 3 g / 1 g/mol = 3 g atoms

no of gram atom of oxygen = 8 g / 16 g/mol = 0.5 g atoms

Now we will calculate the atomic ratio by dividing the gram atoms with the 0.5 because it is the smallest number among these three.

C:H:O = 1/0.5 : 3/0.5 : 0.5/0.5

C:H:O = 2 : 6 : 1

The empirical formula of compound will be C₂H₆O

5 0

2 years ago

0.15 gm of metallic oxide was dissolved in 100 ml of 0.1 N H2SO4 and 25.8 ml of 0.095N NaOH were used to neutralise the remainin

Mila [183]

Answer:

Explanation:

25.8 ml of .095 N NaOH is needed to neutralise the remaining acid

equivalent of NaOH used = 25.8 x .095 / 1000 = .002451 gm equivalent .

acid remaining = .002451 gm equivalent .

acid initially taken = 100 ml of .1 N / 1000 = . 01 gm equivalent

acid reacted with metal = .01 -.002451 = .007549 gm equivalent

This must have reacted with same gram equivalent of metal oxide

.007549 gm equivalent = .15 gm of metal oxide

1 gm equivalent = 19.87 gm

equivalent weight of metal = 19.87 - equivalent weight of oxygen

= 19.87 - 8 = 11.87 .

7 0

2 years ago

30. how many grams of boric acid, b(oh)3 (fm 61.83), should be used to make 2.00 l of 0.050 0 m solution? what kind of fl ask is

grigory [225]

Answer:

Mass = 6.183 g

Solution:

Step 1: Calculate number of moles of Boric acid using following formula,

Molarity = Moles ÷ Volume

Solving for Moles,

Moles = Molarity × Volume

Putting Values,

Moles = 0.05 mol.L⁻¹ × 2.0 L

Moles = 0.1 mol

Step 2: Calculate Mass of Boric Acid using following formula,

Moles = Mass ÷ M.mass

Solving for Mass,

Mass = Moles × M.mass

Putting values,

Mass = 0.1 mol × 61.83 g.mol⁻¹

Mass = 6.183 g

Flask used to prepare this solution is called as Volumetric flask. Take 2 L volumetric flask, add 6.183 g of Boric acid and fill it to the mark with distilled water.

7 0

2 years ago

50cm3 of sodium hydroxide solution was titrated against a solution of sulfuric acid. The concentration of the sodium hydroxide s

miskamm [114]

Answer:

49 g/L is the concentration of the acid

Explanation:

Firstly, we proceed to write the equation of reaction.

2NaOH + H2SO4 ——-> Na2SO4 + 2H2O

We can see that 1 mole of the base reacted with two moles of the acid.

kindly note that dm^3 is same as liter

Firstly, we need to get the concentration of the reacted sulphuric acid in g/L

we use the simple titration equation below;

CaVa/CbVb = Na/Nb

From the question;

Ca = ?

Va = 25 cm^3

Cb = 20 g/L

we convert this to concentration in mol/L

Mathematically, that is concentration in g/L divided by molar mass in g/mole

molar mass of NaOH = 40 g/mol

so we have; 20g/L / 40 = 0.5 mol/L

Vb = 50 cm^3

Na = 1

Nb = 2

Where C represents concentrations, V volumes and N , number of moles

Now, substitute the values;

Ca * 25/0.5 * 50 = 1/2

25Ca/25 = 0.5

So Ca = 0.5 mol/L

Now to get the concentration of H2SO4 in g/L

What we do is to multiply the concentration in mol/L by molar mass in g/mol

That would be 0.5 * 98 = 49 g/L

4 0

2 years ago

A group of students measured the average monthly temperature of 1000 cities around the world and plotted the cumulative frequenc

mr Goodwill [35]

Answer:

The MAD of city 2 is less than the MAD for city 1, which means the average monthly temperature of city 2 vary less than the average monthly temperatures for City 1.

Explanation:

For comparing the mean absolute deviations of both data sets we have to calculate the mean absolute deviation for both data sets first,

So for city 1:

Now to calculate the mean deviations mean will be subtracted from each data value. (Note: The minus sign is ignored as the deviation is the distance of value from the mean and it cannot be negative. For this purpose absolute is used)

The deviations will be added then.

So the mean absolute deviation for city 1 is 24 ..

For city 2:

Now to calculate the mean deviations mean will be subtracted from each data value. (Note: The minus sign is ignored)

The deviations will be added then.

So the MAD for city 2 is 11.33 ..

So,

The MAD of city 2 is less than the MAD for city 1, which means the average monthly temperature of city 2 vary less than the average monthly temperatures for City 1.

8 0

1 year ago