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Molodets [167]
2 years ago
7

A 3.0 L solution contains 73.5 g of H2SO4. Calculate the molar concentration of the solution.

Chemistry
1 answer:
kozerog [31]2 years ago
6 0
C_{m}=\frac{m}{MV}=\frac{73,5g}{98\frac{g}{mol}*3L}=0,25M\\\\
C_{1}V_{1}=C_{2}V_{2}\\\\
0,25M*3L=0,18M*V_{2}\\\\
V_{2}=\frac{0,25M*3L}{0,18M}\approx 4,17L\\\\
4,17L-3L=1,17L
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Quinine, an antimalarial drug, is 8.63% nitrogen. There are two nitrogen atoms per molecule. What is the molecular weight of qui
Furkat [3]

Answer:

324.18 g/mol

Explanation:

Let the molecular mass of the antimalarial drug, Quinine is x g/mol

According to question,

Nitrogen present in the drug is 8.63% of x

So, mass of nitrogen = \frac {8.63}{100}\times x

Also, according to the question,

2 atoms are present in 1 molecule of the drug.

Mass of nitrogen = 14.01 amu = 14.01 g/mol (grams for 1 mole)

So, mass of nitrogen = 14.01×2 = 28.02

These 2 must be equal so,

\frac {8.63}{100}\times x=28.02

solving for x, we get:

<u>x = 324.18 g/mol</u>

6 0
2 years ago
In the reaction C + O2 → CO2, 18 g of carbon react with oxygen to produce 72 g of carbon dioxide. What mass of oxygen would be n
bulgar [2K]
<span>Molar mass(C)= 12.0 g/mol
Molar mass (O2)=2*16.0=32.0 g/mol
Molar mass (CO2)=44.0 g/mol

18g C*1mol C/12 g C = 1.5 mol C

                                 C +     O2 →                CO2

from reaction       1 mol    1 mol              1 mol
from problem     1.5 mol   1.5 mol         1.5 mol

1.5 mol O2*32 g O2/1 mol O2 = 48 g O2

In reality this reaction requires only 48 g O2 for 18 g carbon.
And from 18 g carbon you can get only
1.5 mol CO2*44 g CO2/1 mol CO2=66 g CO2
But these problem has 72g CO2. The best that we can think, it is a mix of CO2 and O2.
So to find all amount  of O2  that was added for the reaction (probably people who wrote this problem wanted this)
we need  (the mix of 72g - mass of carbon 18 g)= 54 g.
So the only answer that is possible is 
</span><span>2.) 54 g.</span>
3 0
2 years ago
Coal gasification is a multistep process to convert coal into cleaner-burning fuels. In one step, a coal sample reacts with supe
ddd [48]

Answer :

The enthalpy of reaction is, -187.6 kJ/mol

The total heat will be, -2251 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

(a) The formation of CH_4 will be,

2C(coal)+2H_2O(g)\rightarrow CH_4(g)+CO_2(g)    \Delta H_{rxn}=?

The intermediate balanced chemical reaction will be,

(1) C(coal)+H_2O(g)\rightarrow CO(g)+H_2(g)     \Delta H_1=29.7kJ

(2) CO(g)+H_2O(g)\rightarrow CO_2(g)+H_2(g)    \Delta H_2=-41kJ

(3) CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2O(g)    \Delta H_3=-206kJ

We are multiplying equation 1 by 2 and then adding all the equations, we get :

(b) The expression for enthalpy of reaction will be,

\Delta H_{rxn}=2\times \Delta H_1+\Delta H_2+\Delta H_3

\Delta H_{rxn}=(2\times 29.7)+(-41)+(-206)

\Delta H_{rxn}=-187.6kJ/mol

Therefore, the enthalpy of reaction is, -187.6 kJ/mol

(c) Now we have to calculate the total heat.

\Delta H=\frac{q}{n}

or,

q=\Delta H\times n

where,

\Delta H = enthalpy change = -187.6 kJ/mol

q = heat = ?

n = number of moles of coal = \frac{1.00\times 1000g}{12.00g/mol}=83.33mol

Now put all the given values in the above formula, we get:

q=(-187.6kJ/mol)\times (83.33mol)=-2.251kJ

Thus, the total heat will be, -2251 kJ

4 0
2 years ago
Using the data, which of the following is the rate constant for the rearrangement of methyl isonitrile at 320 ∘C? (HINT: the act
svp [43]

Explanation:

Below is an attachment containing the solution.

3 0
1 year ago
HELP
agasfer [191]

Answer:

3.02× 10²⁴ atoms

Explanation:

Given data:

Number of nitrogen atoms = ?

Number of moles of N₂O = 2.51 mol

Solution:

1 mole contain 2 mole of nitrogen atoms.

2.51 × 2 = 5.02  mol

According to Avogadro number,

1 mole = 6.022 × 10²³ atoms

5.02  mol ×  6.022 × 10²³ atoms / 1 mol

30.2 × 10²³ atoms

3.02× 10²⁴ atoms

5 0
1 year ago
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