Which of the following is true about electrostatic forces? Check all that are true:

How many moles of hydrogen are required to make 32 moles of water?

NNADVOKAT [17]

You multiply 32 by 2, since there are two hydrogens in every water molecule.

7 0

4 years ago

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Which solution would cause blue litmus paper to turn red?

True [87]

An acidic solution would turn a blue litmus paper red due to the mobile and free hydrogen (H+) ions. These hydrogen ions affect the colour property of the litmus papers and thus the colour change. A litmus paper dropped into a solution of hydrogen ions turns red to indicate that the solution is acidic.

4 0

4 years ago

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it takes 151 kJ/mol to break an iodine-iodine single bond. calculate the maximum wavelength of light for which an iodine-iodine

Allisa [31]

Answer:

The wavelength of light require to brake an single I-I bond is 7.92 × 10⁻⁷ m

Explanation:

Amount of energy required to break the one mole of iodine-iodine single bond = 151 KJ

amount of energy to break one iodine -iodine bond = (151 KJ/mol )/ 6.02 × 10²³/mol = 2.51 × 10⁻²² KJ

or

2.51 × 10⁻¹⁹ J

Formula:

E = hc / λ

h = planck's constant = 6.626 × 10⁻³⁴ js

c = speed of light = 3 × 10⁸ m/s

λ = wavelength

Solution:

E = hc / λ

λ = hc / E

λ = (6.626 × 10⁻³⁴ js × 3 × 10⁸ m/s ) / 2.51 × 10⁻¹⁹ J

λ = 19.878 × 10⁻²⁶ j .m / 2.51 × 10⁻¹⁹ J

λ = 7.92 × 10⁻⁷ m

6 0

3 years ago

At 298 K, the reaction 2 HF (g) ⇌ H2 (g) + F2 (g) has an equilibrium constant Kc of 8.70x10-3. If the equlibrium concentrations

klio [65]

This problem is describing the equilibrium whereby hydrofluoric acid decomposes to hydrogen and fluorine gases at 298 K whose equilibrium constant is 8.70x10⁻³, the equilibrium concentrations of all the reactants are both 1.33x10⁻³ M and asks for the initial concentration of hydrofluoric acid which turns out to be 2.86x10⁻³ M.

Then, we can write the following equilibrium expression for hydrofluoric acid once the change, $x$ , has taken place:

$[HF]=[HF]_0-2x$

Now, since both products are 1.33x10⁻³ M we infer the reaction extent is also 1.33x10⁻³ M, and thus, we can calculate the equilibrium concentration of HF via the law of mass action (equilibrium expression):

$8.70x10^{-3}=\frac{(1.33x10^{-3} M)^2}{[HF]} }$