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3 years ago

Table 2 Heat of Reaction of NaOH (aq) and HCl(aq) Solution

Chemistry

1 answer:

kondaur [170]3 years ago

8 0

Answer:

the molarity of NaOH is 1.10. And the molarity of HCl is 1.10. And the initial Temp=0.50(°c). and The final Temp= 1.10(°c)

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Please help Chemistry question

kari74 [83]

A mixture means to elements where mixed together but a compound means they have bonded to make a different chemical

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2 years ago

Read 2 more answers

Given the equation 4Al +3O2 --> 2Al2O3 if 325 grams of Al2O3 are to be formed, determine the mass of aluminum that must be re

den301095 [7]

Answer:

172 g Al

Step-by-step explanation:

We know we will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.

M_r: 26.98 101.96

4Al + 3O₂ ⟶ 2Al₂O₃

m/g: 325

(a) Calculate the moles of Al₂O₃

n = 325 g Al₂O₃ × 1 mol Al₂O₃ /39.10 g Al₂O₃

n = 3.188 mol Al₂O₃

(b) Calculate the moles of Al

The molar ratio is (4 mol Al/2 mol Al₂O₃)

n = 3.188 mol Al₂O₃ × (4 mol Al/2 mol Al₂O₃)

n = 6.375 mol Al

(c) Calculate the mass of Al

m = 6.375 mol Al × (26.98 g Al/1 mol Al)

m = 172 g Al

Note: The answer can have only three significant figures because that is all you gave for the mass of Al₂O₃.

8 0

3 years ago

What is the electron configuration of Iron

Novay_Z [31]

The electron configuration of Iron is:

[Ar] 3d⁶4s²

5 0

2 years ago

The pressure of a gas is 750 torr when its volume is 400. ml. Calculate the

ololo11 [35]

Answer:

a. 0.66 atm

Explanation:

Pi = 750 torr = 0.99 atm. Pf = ?

Vi = 400 ml Vf = 600ml

Using PiVi = PfVf

Then 400ml × 0.99atm = Pfatm × 600ml

Solving for Pf, we have,

Pf = (400ml × 0.99atm)/600ml

Pf = 0.66atm

7 0

2 years ago

Given the following data:N2(g) + O2(g)→ 2NO(g), ΔH=+180.7kJ2NO(g) + O2(g)→ 2NO2(g), ΔH=−113.1kJ2N2O(g) → 2N2(g) + O2(g), ΔH=−163

statuscvo [17]

Answer:

ΔH = +155.6 kJ

Explanation:

The Hess' Law states that the enthalpy of the overall reaction is the sum of the enthalpy of the step reactions. To do the addition of the reaction, we first must reorganize them, to disappear with the intermediaries (substances that are not presented in the overall reaction).

If the reaction is inverted, the signal of the enthalpy changes, and if its multiplied by a constant, the enthalpy must be multiplied by the same constant. Thus:

N₂(g) + O₂(g) → 2NO(g) ΔH = +180.7 kJ

2NO(g) + O₂(g) → 2NO₂(g) ΔH = -113.1 kJ

2N₂O(g) → 2N₂(g) + O₂(g) ΔH = -163.2 kJ

The intermediares are N₂ and O₂, thus, reorganizing the reactions:

N₂(g) + O₂(g) → 2NO(g) ΔH = +180.7 kJ

NO₂(g) → NO(g) + (1/2)O₂(g) ΔH = +56.55 kJ (inverted and multiplied by 1/2)

N₂O(g) → N₂(g) + (1/2)O₂(g) ΔH = -81.6 kJ (multiplied by 1/2)

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

N₂O(g) + NO₂(g) → 3NO(g)

ΔH = +180.7 + 56.55 - 81.6

ΔH = +155.6 kJ

5 0

3 years ago