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2 years ago

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

olution was 20g/dm3. Work out the concentration of the acid in grams per litre if it took 25cm3 of acid to completely neutralise the alkali. The relative molecular mass of sulfuric acid is 98.

1 answer:

miskamm [114]2 years ago

4 0

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

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How many grams are in 8.2 x 1022 molecules of N216?

hoa [83]

Answer:

N2I6 = 789 g

N2I6 = 8.2x1022 molecules N2I6 x 1 mole/6.02x1023 molecules = 1.36x10-1 moles = 0.136 moles

N2I6=0.136molesx789g/mole=107g=110g

----------------------

Hope this helps, have a BLESSED AND WONDERFUL DAY!

- Cutiepatutie ☺❀❤

7 0

2 years ago

Draw all of the constitutional isomers of the molecule with formula C3H5Br. Ignore geometric and stereoisomers.

Kay [80]

Answer:

-) 3-bromoprop-1-ene

-) 2-bromoprop-1-ene

-) 1-bromoprop-1-ene

-) bromocyclopropane

Explanation:

In this question, we can start with the <u>I.D.H</u> (<em>hydrogen deficiency index</em>):

$I.D.H~=~\frac{(2C)+2+(N)-(H)-(X)}{2}$

In the formula we have 3 carbons, 5 hydrogens, and 1 Br, so:

$I.D.H~=~\frac{(2*3)+2+(0)-(5)-(1)}{2}~=~1$

We have an I.D.H value of one. This indicates that we can have a cyclic structure or a double bond.

We can start with a linear structure with 3 carbon with a double bond in the first carbon and the Br atom also in the first carbon (<u>1-bromoprop-1-ene</u>). In the second structure, we can move the Br atom to the second carbon (<u>2-bromoprop-1-ene</u>), in the third structure we can move the Br to carbon 3 (<u>3-bromoprop-1-ene</u>). Finally, we can have a cyclic structure with a Br atom (<u>bromocyclopropane</u>).

See figure 1

I hope it helps!

7 0

1 year ago

A 108 49in source emits a 633-kev gamma photon and a 606-kev internal-conversion electron from the k shell. what is the binding

____ [38]

Binding energy is the energy needed to emit the electron from the shell. Using the formula below to compute for BE. Binding Energy BE = Energy of photon - Kinetic energy electron
where
Energy proton= 633 keV
KE electron = 606 keV
Binding energy BE = 27 keVThe binding energy of the k subshell is equal to 27 keV.

6 0

2 years ago

A certain gas is present in a 12.0 L cylinder at 4.0 atm pressure. If the pressure is increased to 8.0 atm the volume of the gas

diamong [38]

There are 3 parts in this question:
1) To find the initial Boyle's constant $k_{i}$
2) To find the final Boyle's constant $k_{f}$
3) To verify whether gas is obeying Boyle's law or not

Given data:
The initial volume of the cylinder(in litres) = $V_{i}$ = 12.0 L
The initial pressure(in atmospheric pressure) = $P_{i}$ = 4.0 atm

The final pressure(in atmospheric pressure) = $P_{f}$ = 8.0 atm
The final volume of the cylinder(in litres) = $V_{f}$ = 6.0 L

First you need to know what Boyle's law is:
<span>Boyle's law states that the pressure of a given mass of an ideal gas is inversely proportional to its volume at a constant temperature.
</span>
The Mathematical form of Boyle's law is:
$P = \frac{k}{V}$

Where,
P = Pressure
V = Volume of the gas
k = Boyle's constant

Now let's solve aforementioned parts one by one:
1.
The initial volume of the cylinder(in litres) = $V_{i}$ = 12.0 L
The initial pressure(in atmospheric pressure) = $P_{i}$ = 4.0 atm
The Boyle's constant = $k_{i}$ = ?

According to the Boyle's law,

$P_{i} = \frac{k_{i}}{V_{i}}$

=> $k_{i}$ = $P_{i}V_{i}$
Plug-in the values in the above equation, you would get:
$k_{i}$ = 4.0 * 12.0 = 48

Ans-1) $k_{i}$ = 48

2.
The final pressure(in atmospheric pressure) = $P_{f}$ = 8.0 atm
The final volume of the cylinder(in litres) = $V_{f}$ = 6.0 L
The Boyle's constant = $k_{f}$ = ?

According to the Boyle's law,

$P_{f} = \frac{k_{f}}{V_{f}}$

=> $k_{f}$ = $P_{f}V_{f}$
Plug-in the values in the above equation, you would get:
$k_{f}$ = 8.0 * 6.0 = 48

Ans-2) $k_{f}$ = 48

3.
In order to verify Boyle's law, the initial Boyle's constant should be EQUAL to the final Boyle's constant, meaning:

$k_{i}$ = $k_{f}$

Since,
$k_{i}$ = 48
$k_{f}$ = 48

Therefore,
48=48.

Ans-3) Hence proved: The gas IS obeying the Boyle's law.

-i

4 0

2 years ago

Read 2 more answers

A compound is made up of 28 g N, 24 g C, 48 g O, and 8 g H .What is the empirical formula?

vovikov84 [41]

Answer:

$\rm C_2H_8N_2O_3$ .

Explanation:

<h3>Step One: calculate the coefficients. </h3>

Look up the relative atomic mass of these four elements on a modern periodic table:

$\rm C$ : approximately $12$ .
$\rm H$ : approximately $1$ .
$\rm N$ : approximately $14$ .
$\rm O$ : approximately $16$ .

The relative atomic mass of an element is numerically equal to the mass (in grams, $\rm g$ ,) of one mole of atoms of this element.

For example, the relative atomic mass of $\rm C$ is approximately $12$ . Therefore, each mole of $\rm C\!$ atoms would have a mass of $12\; \rm g$ .

This sample contains $24\; \rm g$ of carbon. That would correspond to approximately $\displaystyle \left(\frac{24}{12}\right)\; \rm mol = 2\; \rm mol$ of $\rm C$ atoms.

Similarly, for the other three elements:

$\displaystyle n(\mathrm{H}) \approx \frac{8\; \rm g}{1\; \rm g \cdot mol^{-1}} = 8\; \rm mol$ .

$\displaystyle n(\mathrm{N}) \approx \frac{28\; \rm g}{14\; \rm g \cdot mol^{-1}} = 2\; \rm mol$ .

$\displaystyle n(\mathrm{O}) \approx \frac{48\; \rm g}{16\; \rm g \cdot mol^{-1}} = 3\; \rm mol$ .

Hence, the ratio between these elements in this compound would be:

$n(\mathrm{C}): n(\mathrm{H}): n(\mathrm{N}):n(\mathrm{O}) = 2: 8 : 2 : 3$ .

In the empirical formula of a compound, the coefficients should represent the smallest possible integer ratio between the number of atoms of these elements.

$n(\mathrm{C}): n(\mathrm{H}): n(\mathrm{N}):n(\mathrm{O}) = 2: 8 : 2 : 3$ is indeed the smallest possible integer ratio between the number of atoms of these elements.

<h3>Step Two: arrange the elements in an appropriate order</h3>

Apply the Hill System to arrange these four elements in the empirical formula. In the Hill System:

If carbon, $\rm C$ , is present in this compound, then:

$\rm C$ (carbon) and then $\rm H$ (hydrogen) will be the first two elements listed in the formula (ignore the hydrogen if it is not in the compound.)
The other elements in this compound will be listed in alphabetical order.

If there is no carbon $\rm C$ in this compound, then list all the elements in this compound in alphabetical order.

Both $\rm C$ (carbon) and $\rm H$ (hydrogen) are found in this compound. Therefore, the first element in the list would be $\rm C\!$ . The second would be $\rm H\!$ , followed by $\rm N\!$ and then $\rm O\!$ .

Hence, the empirical formula of this compound would be $\rm C_2H_8N_2O_3$ .

6 0

2 years ago