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4 years ago

10

For the following reaction, the reactants are favored at equilibrium. Classify each of the reactants and products based on their

strength as Bronsted-Lowry acids or bases. (CH3)2NH2 HCOO-(CH3)2NH HCOOH

1 answer:

ycow [4]4 years ago

8 0

Answer:

The answer is attached

Explanation:

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For the reaction C2H5OH + O2 --> 3CO2 + 3H2O

Lelechka [254]

The reaction given above is a combustion reaction. All combustion reactions are exothermic, meaning they give off heat when they react,

8 0

4 years ago

Need ASAP will give brainlist

Leokris [45]

Answer:

A.

The answer is A :)

7 0

2 years ago

Plz help!! I really really need the help as soon as possible thank pls!! Your help matters to me pls help it’s due tomorrow morn

Assoli18 [71]

Answer:

6.Given,

Final Velocity =60m/s

Initial Velocity= 0

Time=10 sec

A=?

A=Final Velocity- Initial Velocity/time

=60-0/10

=60/10

=6m/s ans.

Explanation:

Acceleration = Final Velocity - Initial Velocity/Time

By using this Formula we can calculate Acceleration.

5 0

3 years ago

Average Molarity for HCl is .391

Ira Lisetskai [31]

Answer:

#1: 0.00144 mmolHCl/mg Sample

#2: 0.00155 mmolHCl/mg Sample

#3: 0.00153 mmolHCl/mg Sample

Explanation:

A antiacid (weak base) will react with the HCl thus:

Antiacid + HCl → Water + Salt.

In the titration of antiacid, the strong acid (HCl) is added in excess, and you're titrating with NaOH moles of HCl that doesn't react.

Moles that react are the difference between mmoles of HCl - mmoles NaOH added (mmoles are Molarity×mL added). Thus:

Trial 1: 0.391M×14.00mL - 0.0962M×34.26mL = 2.178 mmoles HCl

Trial 2: 0.391M×14.00mL - 0.0962M×33.48mL = 2.253 mmoles HCl

Trial 3: 0.391M×14.00mL - 0.0962M×33.84mL = 2.219 mmoles HCl

The mass of tablet in mg in the 3 experiments is 1515mg, 1452mg and 1443mg.

Thus, mmoles HCl /mg OF SAMPLE<em> </em>for each trial is:

#1: 2.178mmol / 1515mg

#2: 2.253mmol / 1452mg

#3: 2.219mmol / 1443mg

<h3>#1: 0.00144 mmolHCl/mg Sample</h3><h3>#2: 0.00155 mmolHCl/mg Sample</h3><h3>#3: 0.00153 mmolHCl/mg Sample</h3>

8 0

3 years ago

Determine the heat of reaction for the process TiO2(s) + 4HCl(g) TiCl4(l) + 2H2(g) + O2(g) using the information given below: Ti

Aleks [24]

Answer : The heat of reaction for the process is, 1374.7 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.

The main chemical reaction is,

$TiO_2(s)+4HCl(g)\rightarrow TiCl_4(l)+2H_2(g)+O_2(g)$ $\Delta H_{rxn}=?$

The intermediate balanced chemical reaction will be,

(1) $Ti(s)+O_2(g)\rightarrow TiO_2(s)$ $\Delta H_1=-939.7kJ$

(2) $2HCl(g)\rightarrow H_2(g)+Cl_2(g)$ $\Delta H_2=-184.6kJ$

(3) $Ti(s)+2Cl_2(g)\rightarrow TiCl_4(l)$ $\Delta H_3=-804.2kJ$

We reversing reaction 1, 3 and multiplying reaction 2 by 2 and then adding all the equations, we get :

(1) $TiO_2(s)\rightarrow Ti(s)+O_2(g)$ $\Delta H_1=939.7kJ$

(2) $4HCl(g)\rightarrow 2H_2(g)+2Cl_2(g)$ $\Delta H_2=2\times (-184.6kJ)=-369.2kJ$

(3) $TiCl_4(l)\rightarrow Ti(s)+2Cl_2(g)$ $\Delta H_3=804.2kJ$

The expression for heat of reaction for the process is:

$\Delta H_{rxn}=\Delta H_1+\Delta H_2+\Delta H_3$

$\Delta H_{rxn}=(939.7kJ)+(-369.2kJ)+(804.2kJ)$

$\Delta H_{rxn}=1374.7kJ$

Therefore, the heat of reaction for the process is, 1374.7 kJ

3 0

3 years ago