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3 years ago

What affects the rates of reactions? describe each. a. b.?

1 answer:

4 0

Rates can be calculated with Arrhenius equation k = Axe^(-Ea/RT)
a. temperature affects the rate - imagine you are making coffee, so coffee crystals are boiled faster on higher temperature. Simplified but makes sense.
b. Ea is activation energy. Imagine, while preparing coffee, some of ingredients change to a different one, so there is a A -> B reaction (simplified). Now, Ea is energy barrier that stands on the arrow of this reaction, preventing A to transform to B. If Ea is small, reaction will go easy (not fast!), if Ea is large –reaction will not happen so easy (you ll have to use catalyst for example)

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The word root for a group of microorganisms living together is a. steriliz. b. prodrom. c. coloniza. d. nosocomi.

UkoKoshka [18]

The answer is C) Coloniza.

7 0

2 years ago

For each reaction, find the value of ΔSo. Report the value with the appropriate sign. (a) 3 NO2(g) + H2O(l) → 2 HNO3(l) + NO(g)

aev [14]

Answer:

ΔS° = -268.13 J/K

Explanation:

Let's consider the following balanced equation.

3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)

We can calculate the standard entropy change of a reaction (ΔS°) using the following expression:

ΔS° = ∑np.Sp° - ∑nr.Sr°

where,

ni are the moles of reactants and products

Si are the standard molar entropies of reactants and products

ΔS° = [2 mol × S°(HNO₃(l)) + 1 mol × S°(NO(g))] - [3 mol × S°(NO₂(g)) + 1 mol × S°(H₂O(l))]

ΔS° = [2 mol × 155.6 J/K.mol + 1 mol × 210.76 J/K.mol] - [3 mol × 240.06 J/K.mol + 1 mol × 69.91 J/k.mol]

ΔS° = -268.13 J/K

7 0

3 years ago

when applied to a dish soap makes grease soluble in water which explains correctly supports the role of intermolecular forces in

Temka [501]

The nonpolar end of a soap molecule attaches itself to grease.

5 0

2 years ago

Predict the charge on the monatomic ions formed from the following atoms in binary ionic compounds:(a) |(b) Sr(c) K(d) N(e) S(f)

Pavel [41]

Answer:

(a) I⁻ (charge 1-)

(b) Sr²⁺ (charge 2+)

(d) N³⁻ (charge 3-)

(e) S²⁻ (charge 2-)

(f) In³⁺ (charge 3+)

Explanation:

To predict the charge on a monoatomic ion we need to consider the octet rule: atoms will gain, lose or share electrons to complete their valence shell with 8 electrons.

(a) |

I has 7 valence electrons so it gains 1 electron to form I⁻ (charge 1-).

(b) Sr

Sr has 2 valence electrons so it loses 2 electrons to form Sr²⁺ (charge 2+).

(c) K

K has 1 valence electron so it loses 1 electron to form K⁺ (charge 1+).

(d) N

N has 5 valence electrons so it gains 3 electrons to form N³⁻ (charge 3-).

(e) S

S has 6 valence electrons so it gains 2 electrons to form S²⁻ (charge 2-).

(f) In

In has 3 valence electrons so it loses 3 electrons to form In³⁺ (charge 3+).

3 0

3 years ago

15.00 grams of Chromium react with 15.00 grams of hydrobromic acid. Calculate the theoretical yield of the reaction. At STP what

s2008m [1.1K]

Answer:

(a) 18.03 g

(b) 2.105 L

Step-by-step explanation:

We have the masses of two reactants, so this is a limiting reactant problem.

We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.

Step 1. Gather all the information in one place with molar masses above the formulas and masses below them.

M_r: 52.00 80.91 291.71

2Cr + 6HBr ⟶ 2CrBr₃ + 3H₂

Mass/g: 15.00 15.00

Step 2. Calculate the moles of each reactant

Moles of Cr = 15.00 × 1/52.00

Moles of Cr = 0.2885 mol Cr

Moles of HBr = 15.00 × 1/80.91

Moles of HBr = 0.1854 mol HBr ×

Step 3. Identify the limiting reactant

Calculate the moles of CrCl₃ we can obtain from each reactant.

From Cr:

The molar ratio of CrBr₃:Cr is 2 mol CrBr₃:2 mol Cr

Moles of CrBr₃ = 0.2885 × 2/2

Moles of CrBr₃ = 0.2885 mol CrCl₃

From HBr:

The molar ratio of CrBr₃:HBr is 2 mol CrBr₃:6 mol HBr.

Moles of CrBr₃ = 0.1854 × 2/6

Moles of CrBr₃ = 0.061 80 mol CrBr₃

The limiting reactant is HBr because it gives the smaller amount of CrBr₃.

Step 4. Calculate the theoretical yields of CrBr₃ and H₂.

Theoretical yield of CrBr₃ = 0.061 80 × 291.71/1

Theoretical yield of CrBr₃ = 18.03 g CrCl₃

The molar ratio is 3 mol H₂:6 mol HBr

Theoretical yield of H₂ = 0.1854 × 3/6

Theoretical yield of H₂ = 0.092 70 mol H₂

Step 5. Calculate the volume of H₂ at STP

STP is 1 bar and 0 °C.

The molar volume of a gas at STP is 22.71 L.

Volume = 0.092 70 × 22.71/1

Volume = 2.105 L

Step 6. Calculate the percent yield

% Yield = actual yield/theoretical yield × 100 %

Actual yield = 15.35 g

% yield = 15.35/18.03 × 100

% yield = 85.15 %

8 0

3 years ago