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3 years ago

What is the rate of a reaction if the value of kis 0.1, [A] is 1 M, and [B] is 2 M?

Chemistry

1 answer:

mel-nik [20]3 years ago

8 0

Answer:

D. 0.4 (mol/L)/S

Explanation:

You simply have to plug in the given values into the rate law.

Rate = k[A][B]

Rate = (0.1)(1)²(2)²

Rate = (0.1)(1)²(4)²

Rate = 0.4

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Please explain:) I would really appreciate a step by step explanation if possible.

UNO [17]

The enthalpy change of the reaction, ΔH = -311 kJ

Enthalpy change involved in the reaction of 300 g of CO = -10972.5 kJ

<h3>What is the enthalpy change for the reduction of ethyne to form ethane?</h3>

The enthalpy change for the reaction is obtained from the summation of the enthalpies of the reactions of the intermediate steps according to Hess's law.

The equation of the reaction is given below:

C₂H₂ + 2 H₂ → C₂H₆

The enthalpy of the reaction, ΔH = ΔH₁ + 2ΔH₂ + (-ΔH₃)

ΔH = {(-1299) + (2 * -286) + (1560)}Kj

ΔH = -311 kJ

The equation for the methanation reaction is given below:

3 H₂O + CO → CH₄ + H₂O

The enthalpy for the methanation reaction is as follows:

ΔH = 1.5ΔH₁ + 0.5*(-ΔH₂) + ΔH₃ + -ΔH₄

ΔH = (-483.6 * 1.5) + (0.5 * 221.0) + (-802.7) + (393.5)

ΔH = -1024.1 kJ/mol

Molar mass of CO = 28 /mol

Enthalpy change involved in the reaction of 300 g of CO = 300/28 * -1024.1 kJ/mol

Enthalpy change involved in the reaction of 300 g of CO = -10972.5 kJ

In conclusion, the enthalpy changes are calculated from the enthalpy values of the intermediate reactions.

Learn more about enthalpy changes at: brainly.com/question/26991394

#SPJ1

7 0

2 years ago

How old were you when u got your first kiss? if you still haven't gotten it how old are you now?

goldfiish [28.3K]

Answer:

I havent kiss anyone

Explanation:

cuz most men are butts

7 0

3 years ago

Read 2 more answers

Naming and Writing Formulas Assignment.

AlexFokin [52]

Answer:

Having difficulties !!! screen shotted it :)

16. H20 - Covalent

17. Mn(NO2)2 - Ionic and called Manganese(II) nitrate

18. HgO - Ionic

19. Li3N - Covalent

7 0

1 year ago

How many grams of oxygen gas are required to produce 65.75 grams of steam?

Nana76 [90]

Answer:

The mass of oxygen gas required to produce 65.75 grams of steam is approximately 162.2 grams

Explanation:

From the question, we have the following chemical reaction equation;

2C₃H₁₈(l) + 25O₂ (g) → 16CO₂(g) + 18H₂O (g)

The molar mass of oxygen, O₂ = 32 g/mol

The molar mass of steam, H₂O = 18.01528 g/mol

25 moles of oxygen are required to produce 18 moles of steam

Therefore, according to Proust's law of definite proportions;

(32 × 25) g of oxygen are required to produce (18 × 18.01528) g of steam

65.75 g of steam will be produced by (32 × 25)/(18 × 18.01528) × 65.75 g ≈ 162.2 g of oxygen O₂.

3 0

3 years ago

List six elements with only 1 electron in an s orbital

Verdich [7]

Answer:

$Lithium=Li2s^1\\Sodium=Na3s^1\\Potassium=K4s^1\\Rubidium=Rb5s^1\\Cesium=Cs6s^1\\Francium=Fr7s^1$

Explanation:

When it comes to electron configuration and orbitals, it's important to first identify what exactly we are trying to identify. Below is a given example:

$He1s^2$