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2 years ago

What properties of water are related to Oxidation?

Chemistry

1 answer:

Cloud [144]2 years ago

5 0

Answer:

water has several important physical properties. Most of the physical properties of water are quite atypical e.g molar mass is 18.0151grams per mol and melting point is 0.00 degree

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Some growth regulators are useful to farmers and gardeners.<br> O True<br> O False

lisov135 [29]

Answer:

True

Explanation:

The first artificial use of a PGR was to stimulate the production of flowers on pineapple plants. They are now used widely in agriculture. Plant hormones are also used in turf management to reduce the need to mow, to suppress seedheads, and to suppress other types of grass.

3 0

2 years ago

Which projectiles will be visibly affected by air resistance when they fall?

Anika [276]

All of the above will be affected by air resistance, but the most obvious will be the balloon or leaf.
Hope it helps somewhat!

8 0

3 years ago

Read 2 more answers

What can you calculate with the Henderson-Hasselbalch equation

Mamont248 [21]

Answer:

Actually, The Henderson - Hasselbalch equation allows you to calculate the pH of the buffer by using the pKa of the weak acid and the ratio that exists between the concentrations of the weak cid and conjugate base. The pKa of formic acid is equal to 3.75. In this case, the pH of the solution will be equal to the acid's pKa .

5 0

3 years ago

Find the mass in grams of 3.00 x 1023 molecules of F2

LUCKY_DIMON [66]

<h3>Answer:</h3>

18.9 g F₂

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table
Using Dimensional Analysis
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<h3>Explanation:</h3>

<u>Step 1: Define</u>

3.00 × 10²³ molecules F₂

<u>Step 2: Identify Conversions</u>

Avogadro's Number

Molar Mass of F₂ (Diatomic) - 38.00 g/mol

<u>Step 3: Convert</u>

Set up: $\displaystyle 3.00 \cdot 10^{23} \ molecules \ F_2(\frac{1 \ mol \ F_2}{6.022 \cdot 10^{23} \ molecules \ F_2})(\frac{38.00 \ g \ F_2}{1 \ mol F_2})$
Multiply: $\displaystyle 18.9306 \ g \ F_2$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

18.9306 g F₂ ≈ 18.9 g F₂

8 0

2 years ago

In the reaction A + B C, doubling the concentration of A doubles the reaction rate and doubling the concentration of B does not

frutty [35]

Answer:

rate = k[A]

Explanation:

The equation that relate reaction rate with reactant concentrations is known as the rate law.

for a reaction:

A + B → C

the rate law can be expressed as:

Rate = k[A]ᵃ[B]ᵇ

The proportionality constant, k, is known as the rate constant, the powers a and b is the reaction order with respect to reactants A and B, respectively.

for this reaction doubling the concentration of A doubles the reaction rate that means

Rate₂ = 2 *Rate₁ and [A]₂ = 2 [A]₁

Rate₁ = k[A]₁ᵃ[B]ᵇ → eq. 1

Rate₂ = k[A]₂ᵃ[B]ᵇ → eq. 2

Dividing eq. 2 by eq. 1 one can get

(Rate₂ / Rate₁) = (k [A]₂ᵃ[B]ᵇ) / (k[A]₁ᵃ[B]ᵇ)

using

Rate₂ = 2 *Rate₁ and [A]₂ = 2 [A]₁

∴ (2 Rate₁ / Rate₁) = ( k [2]ᵃ[B]ᵇ) / (k[1]ᵃ[B]ᵇ)

(2) = (2)ᵃ
taking log of both sides
log (2) = a Log (2)
0.693 = a * 0.693
a =1

∴ order of reaction with respect to A is first (=1) → (1)

Doubling the concentration of B does not affect the reaction rate.

that means

Rate₂ = Rate₁ and [B]₂ = 2 [B]₁

Rate₁ = k[A]ᵃ[B]₁ᵇ → eq. 1

Rate₂ = k[A]ᵃ[B]₂ᵇ → eq. 2

Dividing eq. 2 by eq. 1 one can get

(Rate₂ / Rate₁) = (k [A]ᵃ[B]₂ᵇ) / (k[A]ᵃ[B]₁ᵇ)

using

Rate₂ = Rate₁ and [B]₂ = 2 [B]₁

∴ (Rate₁ / Rate₁) = ( k [A]ᵃ[2]ᵇ) / (k[A]ᵃ[1]ᵇ)

(1) = (2)ᵇ
taking log of both sides
log (1) = b Log (2)
0 = 0.693 * b
b = 0

∴ order of reaction with respect to B is zero → (2)

So, from 1 and 2 the right choice is rate = k[A]¹[B]⁰= k[A]

6 0

2 years ago