How much volume in mL of 0.65 M HCI solution can be made by diluting 150 mL of 9.75 M HCI

Use bond energies to calculate δhrxn for the reaction. n2(g)+3cl2(g)→2ncl3(g)

Jobisdone [24]

ΔHrxn = ΣδΗ(bond breaking) - ΣδΗ(bond making)

Bond enthalpies,
N ≡ N ⇒ 945 kJ mol⁻¹
N - Cl ⇒ 192 kJ mol⁻¹
Cl - Cl⇒ 242 kJ mol⁻¹

According to the balanced equation,
ΣδΗ(bond breaking) = N ≡ N x 1 + Cl - Cl x 3
= 945 + 3(242)
= 1671 kJ mol⁻¹
ΣδΗ(bond making) = N - Cl x 3 x 2
= 192 x 6
= 1152 kJ mol⁻¹

δHrxn = ΣδΗ(bond breaking) - ΣδΗ(bond making)
= 1671 kJ mol⁻¹ - 1152 kJ mol⁻¹
= 519 kJ mol⁻¹

3 0

3 years ago

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24 grams of CH4 was added to the above reaction. Calculate the theoretical yield of CO2. A. 66 grams B. 33 grams c. 132 grams. D

Sonbull [250]

Answer:

Option A. 66 g

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

CH₄ + 2O₂ —> CO₂ + 2H₂O

Next, we shall determine the mass of CH₄ that reacted and the mass of CO₂ produced from the balanced equation. This is illustrated below:

Molar mass of CH₄ = 12 + (4×1)

= 12 + 4 = 16 g/mol

Mass of CH₄ from the balanced equation = 1 × 16 = 16 g

Molar mass of CO₂ = 12 + (16×2)

= 12 + 32 = 44 g/mol

Mass of CO₂ from the balanced equation = 1 × 44 = 44 g

SUMMARY:

From the balanced equation above,

16 g of CH₄ reacted to produce 44 g of CO₂.

Finally, we shall determine the theoretical yield of CO₂. this can be obtained as follow:

From the balanced equation above,

16 g of CH₄ reacted to produce 44 g of CO₂.

Therefore, 24 g of CH₄ will react to produce = (24 × 44) /16 = 66 g of CO₂.

Thus, the theoretical yield of CO₂ 66 g

8 0

2 years ago

Which of the following statements about elements is not true?

guapka [62]

Answer:

Option A is not true

Explanation:

Could you please follow me and mark me as the brainliest answer

6 0

3 years ago

Write the empirical formula of at least four binary ionic compounds that could be formed from the following ions: Zn2+, Ni4+, F-

hram777 [196]

Ionic compounds are formed between oppositely charged ions.

A binary ionic compound is composed of ions of two different elements - one of which is a positive ion(metal), and the other is negative ion (nonmetal).

To write the empirical formula of binary ionic compound we must remember that one ion should be positive and other ion should be negative, then only the correct formula should be written. To write the empirical formula the charges of opposite ions should be criss-crossed.

First empirical formula of binary ionic compound is written between $Zn^{2+} (Positive ion)and F^{-} (Negative ion)$

First Formula would be $ZnF_{2}$

Second empirical formula is between $Zn^{2+}(Positive ion) and O^{2-}(Negative ion)$

Second Formula would be $Zn_{2}O_{2}$

Note : When the subscript are same they get cancel out, so $Zn_{2}O_{2}$ would be written as $ZnO$

Third empirical formula is between $Ni^{4+}(Positive ion) and F^{-}(Negative ion)$

Third Formula would be : $NiF_{4}$

Forth empirical formula is between $Ni^{4+}(Positive ion)and O^{2-}(negative ion)$

Forth Formula would be : $Ni_{2}O_{4}$ or $NiO_{2}$

Note- The subscript will be simplified and the formula will be written as $NiO_{2}$ .

The empirical formula of four binary ionic compounds are : $ZnF_{2}, ZnO, NiF_{4},NiO_{2}$

8 0

2 years ago

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Which type of molecule forms structures in cells and stores energy? (5 points)

Brums [2.3K]

Just to make sure I’m right, is number 1 miss spelled??

4 0

2 years ago