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yan [13]
3 years ago
9

solution of 0.160 MNaOH is used to titrate 21.0 mL of a solution of H2SO4:H2SO4(aq)+2NaOH(aq)→2H2O(l)+Na2SO4(aq)If 38.8 mL of th

e NaOH solution is required to reach the endpoint, what is the molarity of the H2SO4 solution?
Chemistry
1 answer:
stealth61 [152]3 years ago
3 0

Answer:

0.0432 M H2SO4

Explanation:

First, we want to find the moles of MNaOH used. We know that Molarity x Liters = moles. 0.160M x 0.0210L = 0.00336 moles MNaOH

to find the moles of H2SO4, we can use a mol ratio.

0.00336mol MNaOH x (1Mol H2SO4 /2mol MNaOH)

= 0. 00168 mol H2SO4

I found the mol ratio by looking at the coefficients in front of the molecules I knew(MNaOH) and the molecule I needed to find(H2SO4)

then, to find Molarity, we do mol/Liters

0.00168 mol/ 0.0388L =. 0.0432 M H2SO4

You can convert mL to L by dividing by 1000

the significant figures of this problem is 3, so my final answer will also have 3 sig figs.

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Answer:cool

Explanation:radical dude

8 0
3 years ago
The ostwald process is used commercially to produce nitric acid, which is, in turn, used in many modern chemical processes. in t
Debora [2.8K]

The complete balanced chemical equation is: 
4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (g) 

In statement form: 4mol NH3 reacts with 5 mol O2 to produce 6 mol H2O 

First let us find for the limiting reactant: 
>molar mass NH3 = 17 g/mol 
moles NH3 = 54/17 = 3.18 mol NH3 
This will react with 3.18*5/4 = 3.97 mol O2 

>molar mass O2 = 32g/mol 
moles O2 = 54/32 = 1.69 mol O2 
We have insufficient O2 therefore this is the limiting reactant 

From the balanced equation: 
For every 5.0 mol O2, we get 6.0 mol H2O, therefore

moles H2O formed =  1.69 mol O2  * 6/5 = 2.025 mol
Molar mass H2O = 18g/mol 
<span>mass H2O formed = 2.025*18 = 36.45 grams H2O produced</span>

8 0
3 years ago
Read 2 more answers
Which best describes a compound such as sodium chloride?
gayaneshka [121]

Answer:

check it below

Explanation:

NaCl; Sodium Chloride is an ionic compound formed by sodium and Chlorine.

Ionic bond is very strong, It can't be separated back to sodium and chlorine just by physical change. Chemicals which are more reactive can displace ions, thus seperate it

5 0
3 years ago
Biological oxidation-reduction reactions always involve:
kondor19780726 [428]

Answer:

b. transfer of electron(s).

Explanation:

An oxidation-reduction also called a redox reaction is a

chemical reaction in which electrons are transferred of between two species of reactants. It is a chemical reaction where the oxidation number of an atom, ion, or molecule, increases or decreases by losing or gaining electrons

3 0
3 years ago
In atmospheric chemistry, the following chemical reaction converts SO2, the predominant oxide of sulfur that comes from combusti
Misha Larkins [42]

Answer:

Explanation:

From the given information;

The chemical reaction can be well presented as follows:

\mathtt{SO_{2(g)} + \dfrac{1}{2}O_{2(g)} }  ⇄ \mathtt{3SO_{2(l)}}

Now, K is known to be the equilibrium constant and it can be represented in terms of each constituent activity:

i.e

K = \dfrac{a_{so_3}}{a_{so_2} a_{o_2}^{\frac{1}{2}}}

However, since we are dealing with liquids solutions;

K = \dfrac{1}{\dfrac{Pso_2}{P^0}\Big ( \dfrac{Po_2}{P^0} \Big)^{1/2}}   since the activity of a_{so_3} is equivalent to 1

Hence, under standard conditions(i.e at a pressure of 1 bar)

K = \dfrac{1}{Pso_2Po_2^{1/2}}

(b)

From the CRC Handbook, we are meant to determine the value of the Gibb free energy by applying the formula:

\Delta _{rxn} G^o = \sum \Delta_f \ G^o (products) - \sum \Delta_fG^o (reactants) \\ \\ = (1) (-368 \ kJ/mol) - (\dfrac{1}{2}) (0) - ((1) (-300.13 \ kJ/mol)) \\ \\ = -368 \ kJ/mol + 300.13 \ kJ/mol \\ \\  \simeq -68 \ kJ/mol

Thus, for this reaction; the Gibbs frree energy = -68 kJ/mol

(c)

Le's recall that:

At equilibrium, the instantaneous free energy is usually zero &

Q(reaction quotient) is equivalent to K(equilibrium constant)

So;

\mathtt{\Delta _{rxn} G = \Delta _{rxn} G^o + RT In Q}

\mathtt{0- \Delta _{rxn} G^o = RTIn K } \\ \\ \mathtt{ \Delta _{rxn} G^o = -RTIn K }  \\ \\  K = e^{\dfrac{\Delta_{rxn} G^o}{RT}} \\ \\  K = e^{^{\dfrac{67900 \ J/mol}{8.314 \ J/mol \times 298 \ K}} }

K =7.98390356\times 10^{11} \\ \\  \mathbf{K = 7.98 \times 10^{11}}

(d)

The direction by which the reaction will proceed can be determined if we can know the value of Q(reaction quotient).

This is because;

If  Q < K, then the reaction will proceed in the right direction towards the products.

However, if Q > K , then the reaction goes to the left direction. i.e to the reactants.

So;

Q= \dfrac{1}{Pso_2Po_2^{1/2}}

Since we are dealing with liquids;

Q= \dfrac{1}{1 \times 1^{1/2}}

Q = 1

Since Q < K; Then, the reaction proceeds in the right direction.

Hence, SO2 as well O2 will combine to yield SO3, then condensation will take place to form liquid.

8 0
3 years ago
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