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GuDViN [60]
3 years ago
10

Calculate the [H+] for a solution with a pOH of 4.75

Chemistry
1 answer:
S_A_V [24]3 years ago
7 0

Answer:

Explanation:

A. 1.3 x 10-10 M

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Which element would have the lowest electronegativity? (1 point)
irinina [24]

The element that will have the lowest electronegativity is an element with a small number of valence electrons and a large atomic radius.

Electronegativity of an element is the ability or power of that element in a molecule to attract electrons to its Valence electrons.  The following are the properties of electronegativity:

  • It increases across a period from left to right of the periodic table,
  • It decreases down the periodic table groups
  • Group 1 elements are the least (lowest) electronegative elements. These elements have the lowest valence electrons with a large atomic radius.
  • Group 7 elements are the most electronegative elements.

Atomic radius of elements increase down a group because of a progressive increase in the number of shells occupied by electrons which increases the size. But it decreases across a period because electrons are accommodated within the same shell leading to greater attraction by the protons in the nucleus.

Learn more about electronegativity of elements here:

brainly.com/question/20348681

6 0
3 years ago
Find the empirical formula of a compound containing: 19.32% Ca, 34.30% Cl, and 46.38% O
Bess [88]
40×19.32/100=7.7=8×2=16Ca
35.5×34.30/100=12.1=12×2=24Cl
16×46.38/100=7.4=7×2=14O
5 0
3 years ago
Describe the cell structure of a typical plant cell
aleksley [76]

Answer:

Wrong subject my friend.

4 0
3 years ago
The first-order rate constant for the reaction of methyl chloride (CH3Cl) with water to produce methanol (CH3OH) and hydrochlori
Dvinal [7]

Answer:

K(48.5°C) = 1.017 E-8 s-1

Explanation:

  • CH3Cl + H2O → CH3OH + HCl

at T1 = 25°C (298 K) ⇒ K1 = 3.32 E-10 s-1

at T2 = 48.5°C (321.5 K) ⇒ K2 = ?

Arrhenius eq:

  • K(T) = A e∧(-Ea/RT)
  • Ln K = Ln(A) - [(Ea/R)(1/T)]

∴ A: frecuency factor

∴ R = 8.314 E-3 KJ/K.mol

⇒ Ln K1 = Ln(A) - [Ea/R)*(1/T1)]..........(1)

⇒ Ln K2 = Ln(A) - [(Ea/R)*(1/T2)].............(2)

(1)/(2):

⇒ Ln (K1/K2) = (Ea/R)* (1/T2-1/T1)

⇒ Ln (K1/K2) = (116 KJ/mol/8.3134 E-3 KJ/K.mol)*(1/321.5 K - 1/298 K)

⇒ Ln (K1/K2) = (13952.37 K)*(- 2.453 E-4 K-1)

⇒ Ln (K1/K2) = - 3.422

⇒ K1/K2 = e∧(-3.422)

⇒ (3.32 E-10 s-1)/K2 = 0.0326

⇒ K2 = (3.32 E-10 s-1)/0.0326

⇒ K2 = 1.017 E-8 s-1

7 0
3 years ago
What is the energy of a photon of infrared radiation with a frequency of 2.53 × 1012 Hz? Planck’s constant is
Novosadov [1.4K]
It is letter b or c nit d
6 0
2 years ago
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