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3 years ago

What are the balanced chemical equation for these acid and base reactions?

1 answer:

6 0

It can be easily judged that only the first reaction is an acid base reaction among the three given in the question. So, we can avoid the other two reactions given in the question. Now let us focus and write down the balanced chemical equation of P-Toluidine + HCl.

C7H9N + HCl = C7H10N (+) + Cl (-)

I hope the answer has come to your help.

You might be interested in

The name given to the substance formed when a liquid is miscible in another liquid.

m_a_m_a [10]

Answer:

Solute

Explanation:

5 0

3 years ago

A sample of gas occupies a volume of 73.7 mL. As it expands, it does 133.7 J of work on its surroundings at a constant pressure

katen-ka-za [31]

In thermodynamics, work performed by a system is the energy transferred by the system to its surroundings. It can be calculated by the expression:

W = PdV

Integrating,

We will have,

W = P(V2 - V1)
133.7 (1 litre-atm / 101.325 Joule) ( 760 Torr / atm ) = 783 (V2 - .0737 )
V2 = 1.35 L

Hope this answers the question. Have a nice day.

7 0

3 years ago

When solving a problem it is important to identify your given and needed units, but it is also important to understand the relat

Angelina_Jolie [31]

Answer:

The question has some details missing. here are the details ; Given the following ;

1. 43.2 g of tablet with 20 cm3 of space

2. 5 cm3 of tablets weighs 10.8 g

3. 5 g of balsa wood with density 0.16 g/cm3

4. 150 g of iron. With density 79g/cm 3

5. 32 cm3 sample of gold with density 19.3 g/cm3

6. 18 ml of cooking oil with density 0.92 g/ml

Explanation:

Appropriate for calculating mass

32 cm3 sample of gold with density 19.3 g/cm3

18 ml of cooking oil with density 0.92 g/ml

Appropriate for calculating volume

5 g of balsa wood with density 0.16 g/cm3

150 g of iron. With density 79g/cm 3

Appropriate for calculating density

43.2 g of tablet with 20 cm3 of space

5 cm3 of tablets weighs 10.8 g

3 0

3 years ago

What is the coefficient for H2O when the equation ? Ca(OH)2(aq) + ? H3PO4(aq) → ? Ca3(PO4)2(s) + ? H2O(ℓ) is balanced using the

kari74 [83]

Answer:

Coefficient of $H_{2}O$ in the balanced equation with smallest possible integer is 6.

Explanation:

Unbalanced equation: $Ca(OH)_{2}(aq.)+H_{3}PO_{4}(aq.)\rightarrow Ca_{3}(PO_{4})_{2}(s)+H_{2}O(l)$

Balance Ca: $3Ca(OH)_{2}(aq.)+H_{3}PO_{4}(aq.)\rightarrow Ca_{3}(PO_{4})_{2}(s)+H_{2}O(l)$

Balance $PO_{4}$ : $3Ca(OH)_{2}(aq.)+2H_{3}PO_{4}(aq.)\rightarrow Ca_{3}(PO_{4})_{2}(s)+H_{2}O(l)$

Balance H and O: $3Ca(OH)_{2}(aq.)+2H_{3}PO_{4}(aq.)\rightarrow Ca_{3}(PO_{4})_{2}(s)+6H_{2}O(l)$

Balanced equation: $3Ca(OH)_{2}(aq.)+2H_{3}PO_{4}(aq.)\rightarrow Ca_{3}(PO_{4})_{2}(s)+6H_{2}O(l)$

So coefficient of $H_{2}O$ in the balanced equation with smallest possible integer is 6.

7 0

3 years ago

If 0.213 moles of argon occupies a volume of 652 mL at a particular temp and pressure, what volume would it occupy if 0.162 mole

Andreas93 [3]

Answer:

V₂ = 495.89 mL

Explanation:

Given data:

Initial number of moles = 0.213 mol

Initial volume = 652 mL

Final number of moles = 0.162 mol

Final volume = ?

Solution:

V₁/n₁ = V₂/n₂

By putting values,

652 mL/0.213 mol = V₂ /0.162 mol

V₂ = 652 mL 0.162 mol /0.213 mol

V₂ = 105.62 mL.mol /0.213 mol

V₂ = 495.89 mL

3 0

2 years ago