Comment se nomme le processus qui permet à une cellule- mere de donner deux cellules-filles

The. bond dissociation enthalpies of the H-H bond and the H-Cl bond are 435 kJ mol^-1 and 431 kJ mol^-1, respectively. The ΔHfO

Novay_Z [31]

The bond dissociation energy of the Cl - Cl bond is -958 kJ mol^-1.

<h3>What is the dissociation enthalpy?</h3>

Given that;

H-H bond energy = 435 kJ mol^-1

H-Cl bond energy = 431 kJ mol^-1

ΔHfO of HCL(g) = -92kJ mol^-1

Bond dissociation enthalpy of the Cl-Cl bond = x

-92 = 435 + 431 + x

x = -92 - (435 + 431)

x = -958 kJ mol^-1

Learn More about dissociation enthalpy:brainly.com/question/9998007?

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6 0

1 year ago

A sample of calcium phosphate was found to have a mass of 125.3 g. How many molecules were contained in the sample?

Viktor [21]

The answer for the following problem is mentioned below.

Therefore number of molecules(N) present in the calcium phosphate sample are 19.3 × 10^23 molecules.

Explanation:

Given:

mass of calcium phosphate ( $Ca_{3}(PO_{4} )_{2}$ ) = 125.3 grams

We know;

molar mass of calcium phosphate ( $Ca_{3}(PO_{4} )_{2}$ ) = (40×3) + 3 (31 +(4×16))

molar mass of calcium phosphate ( $Ca_{3}(PO_{4} )_{2}$ ) = 120 + 3(95)

molar mass of calcium phosphate ( $Ca_{3}(PO_{4} )_{2}$ ) = 120 +285 = 405 grams

We also know;

No of molecules at STP conditions( $N_{A}$ ) = 6.023 × 10^23 molecules

To solve:

no of molecules present in the sample(N)

We know;

$\frac{m}{M} =\frac{N} }{}$ N÷ $N_{A}$

$\frac{405}{125.3}$ = $\frac{N}{6.023*10^23}$

N =(405×6.023 × 10^23) ÷ 125.3

N = 19.3 × 10^23 molecules

Therefore number of molecules(N) present in the calcium phosphate sample are 19.3 × 10^23 molecules

3 0

3 years ago

A certain liquid X has a normal boiling point of 108.30 °C and a boiling point elevation constant Kb=1.07 °C kg/mol. A solution

Fynjy0 [20]

Answer:

34,6g of (NH₄)₂SO₄

Explanation:

The boiling-point elevation describes the phenomenon in which the boiling point of a liquid increases with the addition of a compound. The formula is:

ΔT = kb×m

Where ΔT is Tsolution - T solvent; kb is ebullioscopic constant and m is molality of ions in solution.

For the problem:

ΔT = 109,7°C-108,3°C = 1,4°C

kb = 1.07 °C kg/mol

Solving:

m = 1,31 mol/kg

As mass of X = 600g = 0,600kg:

1,31mol/kg×0,600kg = 0,785 moles of ions. As (NH₄)₂SO₄ has three ions:

0,785 moles of ions× $\frac{1(NH_{4})_{2}SO_{4}}{3Ions}$ = 0,262 moles of (NH₄)₂SO₄

As molar mass of (NH₄)₂SO₄ is 132,14g/mol:

0,262 moles of (NH₄)₂SO₄× $\frac{132,14g}{1mol}$ = 34,6g of (NH₄)₂SO₄

I hope it helps!

7 0

3 years ago

When we react a weak acid with a strong base of equal amounts and concentration, the component of the reaction that will have th

andriy [413]

Answer:

d. the conjugate base of the weak acid

Explanation:

The strong base (BOH) is completely dissociated in water:

BOH → B⁺ + OH⁻

The resulting conjugate acid (OH⁻) is a weak acid, so it remains in solution as OH⁻ ions.

By other hand, the weak acid (HA) is only slightly dissociated in water:

HA ⇄ H⁺ + A⁻

The resulting conjugate base (A⁻) is a weak base. Thus, it reacts with H⁺ ions from water to form HA, increasing the concentration of OH⁻ ions in the solution.

Therefore, the resulting solution will have a pH > 7 (basic).

8 0

2 years ago

It takes 281.7 kJ of energy to remove 1 mole of electrons from the atoms on the surface of lithium metal. If lithium metal is ir

Alex17521 [72]

Answer:

The maximum kinetic energy of electron is = 2.93 × $10^{-19}$ Joule